Class 12 Engineering Chemistry - Electrochemistry

Copy and complete the following sentences ?
With platinum electrodes hydrogen is liberated at the ........... and oxygen at the .......... during the electrolysis of acidified water.

During the electrolysis of acidified water with platinum electrodes,

$H^+$ ion is reduced at cathode and so hydrogen is liberated at the cathode whereas

$O^{-2}$ ion is oxidized at anode and thus oxygen is liberated at the anode.

During the electrolysis of acidified water, what is the ratio of the volumes of hydrogen produced to the volume of oxygen ?

Electrolysis of acidified water :

$2H_2O$

$\rightarrow$

$2H_2$

$+$

$O_2$

$\implies$ Volume of hydrogen produced is twice than that of oxygen.

Give the equation for the discharge of ion at the cathode during the electrolysis of acidified water ?

Discharge of ion at cathode :

$H^+$

$+$

$e^-$

$\rightarrow$

$H$

$2H$

$+$

$2H$

$\rightarrow$

$2H_2$

Overall at cathode :

$4H^+$

$+$

$4e^-$

$\rightarrow$

$2H_2$

To carry out the so-called electrolysis of water, sulphuric acid is added to water. How does the addition of sulphuric acid produce a conducting solution ?

When sulphuric acid is added to water, water molecules get dissociated into

$H^{+}$ and

$OH^-$ ions which conducts electricity. That's how addition of sulphuric acid to water make it a conduction solution.

Pure water consists entirely of _____________ (ions / molecules).

Pure water consists entirely of

$H_2O$ molecules.

Name a solid which undergoes electrolysis when molten.

Lead bromide

$(PbBr_2)$ is a solid but undergoes electrolysis in molten state.

The whole apparatus used for electrolysis is termed as :

The whole apparatus used for electrolysis is termed as voltmeter.

Name the substance dissolved in pure water to make it a conductor of electricity.

Either dilute Hydrochloric acid, dil. sulphuric acid or common salt is dissolved in pure water to make it a conductor of electricity.

Solutions which are decomposed due to the current passed through them are called .......... .

Electrolyte is defined as the solution which gets decomposed when an electric current passed through it.

Hydrogen gas is liberated at the .......... (negative / positive) electrode called ..........

$H^+$ ions are reduced to

$H_2$ gas at the negative electrode called as cathode.

Acidified water is electrolysed by using carbon electrodes. What is produced at negative carbon electrode?

During electrolysis of acidified water,

$H^+$ ions are reduced to

$H_2$ gas at the negative carbon electrode (cathode) and hence hydrogen gas is liberated at cathode.

Acidified water is electrolysed by using carbon electrodes. What is produced at positive carbon electrode?

During electrolysis of acidified water using carbon electrodes,

$O^{2-}$ ions are oxidized at the positive carbon electrode (anode) and thus oxygen gas is liberated at anode.

Oxygen gas is liberated at the .......... (negative / positive) electrode called the .......... .

$O^{-2}$ ions are oxidized to produce

$O_2$ gas at the positive electrode called as anode.

The chemical reaction is $2H_2O\, \rightarrow\, .......... \,+\,O_2$

The balanced chemical reaction :

$2H_2O$

$\rightarrow$

$2H_2$

$+$

$O_2$

.......... (chemical / physical) reaction takes place.

Chemical oxidation-reduction reactions take place on passing an electric current through water containing a small amount of dilute sulphuric acid.

Write the equation of the reactions which take place at the cathode and anode when acidified water is electrolysed.

Cathode reaction :

$2H^+\,+\, 2e^-\, \rightarrow\, 2H\, \rightarrow\, H_2$

Anode reaction :

$2OH^-\,-\, 2e^-\, \rightarrow\, 2OH\, \rightarrow\, H_2O\,+\, O\,+\, O\,\rightarrow\, O_2$

During electrolysis the cations move to the electrode at a ___________ potential.

lower

_________ is the chemical change that takes place by the passage of current through an electrolyte.

Electrolysis

By process of electrolysis chemical energy gets converted into electrical energy.

Electrolytic cell, any device in which electrical energy is converted to chemical energy.

Whereas, Electrolysis is used to break chemical bonds via a catalyst in a solution. This converts some electrical energy (efficiency dependent) into potential chemical energy since these bonds are typically quite strong such as hydrogen and water. Heat is also given off so, electrical energy does not directly convert to chemical energy. They interact through several processes, electrolysis being one of the most common.

Rusting of iron is an example of ________ reaction.

Rusting of iron is an example of redox reaction.

Can we make use of a.c. in electrolysis? Explain briefly.

No we cannot use ac current in electrolysis. In electrolysis the ionisation of the electrolyte takes place where the positive and negative charges move towards opposite electrodes. For this the electrodes are required to be maintained at opposite polarity as long as the process continues which is possible while supplying a dc current. But if we pass an ac current the polarity of the electrodes will be continuously changing and the ions will not be attracted toward any particular electrode resulting in no ionisation. Hence electrolysis cannot take place with ac current.

$19\ g$ of molten $SnCl_{2}$ is electrolyzed for some time using inert electrodes until $0.119\ g$ of $Sn$ is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of masses of $SnCl_{2} : SnCl_{4}$ after electrolysis.

1375743_785384_ans_603657556e884034b983c3853843d6ab.jpg

How much electricity in terms of faraday is required to produce 20 grams of calcium from molten calcium chloride?

$\because W=z.Q$

$\Longrightarrow Q=\cfrac{W}{z}$

$=\left(\cfrac{20\times 2\times F}{40}\right)$

$=1F$

$\left\{ \begin{matrix} Mo(Ca)=40 \\ { n }_{ f }=2(C{ a }^{ 2+ }) \\ z={ E }/{ F=\left( \cfrac { 40 }{ 2\times F } \right) } \end{matrix} \right\}$

Write the chemistry of recharging lead storage battery highlighting all the materials that are evolved during recharging.

In lead storage battery we have,

anode - porous lead

cathode- lead dioxide

The reactions that are taking place during the recharge of lead storage battery is

$2 PbSO_4 + 2 H_2O \rightarrow Pb(s) + PbO_2 + 4 H^+ + 2SO_4^{2-}$

Sea water promotes corrosion. Why?

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An old cycle frame was left in open for a few days. A brown layer got slowly deposited on its surface and could not be removed when rubbed with sand paper. What happened actually?

It is due to corrosion of steel or iron rod is presence of moist air.

Anode:

$Fe\rightarrow { \underbrace { { Fe }^{ 3+ } }_{ \cfrac { brown }{ \left( rust \right) } } }+3e^-$

Explain the term with example:Oxidant

The species which accepts electrons from other species are

known as oxidant

For example :-

${Z_n} + A{g^ + } \to {Z_n}^{2 + } + A{g^ - }$

Here,

$A{g^ + }$ is oxidant as it accept electron from

${Z_n}$

$A{g^ + }$ reduced himself to oxidise

${Z_n}$ , therefore

$A{g^ + }$ is oxidising agent or oxidants.

How many coulomb are required for the oxidation of one mole of $H_{2}O$ to $O_{2}$ ?

As oxidation state of oxygen changes from - 2 to 0 means 2 coulomb of charge is required.

State given reason, in what state or medium does
i. $NaCL$
II. $HCL$ gas
iii. $NH_3$ gas conduct electricity.

1372787_1153191_ans_2005d0a54163410c998fec4021a089ee.jpg

What do you mean by Corrosion ?

It can be defined as the gradual eating away of a metal by chemical reaction with its environment consisting of oxygen, water vapour, acids and other chemicals.

Column II gives name of material use for device given in column I :

Manganin wire is used in resistance box because of its virtually zero temperature coefficient of resistance, which means its resistance does not fluctuate with slight change in temperature.
Fuse wire is made up of tin-lead alloy. This alloy melts with little rise of temperature. If for some reason excess current passes through a circuit, the fuse wire melts because of heat generated in the process.
Tungsten wire is used in bulb as tungsten has very high resistance and when current passes through it, it glows because of heat generation.
Glass is an example of insulator. In normal condition electricity does not flow through glass.

Statements (A, B, C, D) in list I have to be matched with statements (1,2,3,4) list II.

1. Oxalic acid, $H_2C_2O4$ , a weak polyprotic acid. A diprotic acid ionises stepwise. The first anion is a weaker base than the second, for the first ionisation is more complete than the second.

$H_2C_2O_4 \rightleftharpoons H^+ \ + \ C_2O_4^{2-}$ ,

Thus it is a weak electrolyte.

2. The nickel-cadmium battery ( $Ni-Cd$ battery or $Ni-Cd$ battery) is a type of rechargeable battery using nickel oxide hydroxide and metallic cadmium as electrodes.

3. A hydrogen-oxygen fuel cell is a nonpolluting clean fuel since the only combustion product is water.

4. Cathode: $PbO_2 (s) + 2e^- + 4 H^+ (aq ) + SO_4^{2-} (aq ) \rightarrow PbSO_4 (s) + 2 H_2O (l )$ . When a lead storage battery discharges, lead(II) sulphate forms on both electrodes.

Thus, a correct answer is A-3, B-1, C-4, D-2.

For the circuit shown in the adjoining figure, match the entries of column I with the entries of column II.

Potential difference across battery A:
The net power in the circuit is (15 - 10) V = 5 V
Total resistance in the circuit = (3 + 1 +1) ohm
So the current in the circuit = 5 / 5 amp = 1 amp.
Now, voltage drop across A = 1 V.
So PD across A will be (15 - 1) V = 14 V
Since current is moving in anti clockwise direction, there is no contribution of battery B. So, PD across battery B will be none.
In the entire circuit power is supplied by battery A as it is in higher potential. And the in entire circuit power is consumed by battery B.

Electrolysis of water is carried out in hoffman voltameter. Litmus is added to both cathodic and anodic compartments. What changes do you observe in the above process at the end of electrolysis? Justify.

In the electrolysis of water hydrogen ions are liberated as hydrogen gas at cathode so the hydroxyl ions left are basic in nature. They turn red litmus blue while hydroxyl ions go towards anode in anodic compartment and are liberated. The solution left is acidic which turns blue litmus red.

List 1 and List 2 contains four entries each. Entries of List 1 are to be matched with some entries of ListOne or more than one entries of List 1 may have the matching with the same entries of List 2.

$Pt|Fe^{3+},\space Fe^{2+}$ represents oxidation reduction half-cell.

$Pt|H_2|H^+$ represents gas-gas ion half cell.

$Pt|Hg|Hg^{2+}_2$ represents metal-metal ion half cell.

$Pb|PbSO_4|SO_4^{2-}$ represents metal sparing soluble salt half-cell.

The charge (in F) required to deposites all Al from the electrolysis of 1 mol molten Al $_2$ O $_3$ is :

$2Al^{3+} + 6e^- \rightarrow 2Al$
Hence we notice that 6 electrons are used in the reduction.
Hence for 1 mole of

$Al_2O_3$ we need

$6F$ electrons.

The quantity of charge (in Faraday) required to reduce 96 g Mg from molten solution of MgCl $_2$ is :

The molar mass of Mg is 24 g/mol.
96 g of Mg corresponds to

$\dfrac {96}{24}=4 \space$ moles.
Reduction of 1 mole of Mg requires 2 moles of electrons or 2 faraday of charge.
Hence, the reduction of 4 moles of Mg required 8 faraday of charge.

In rusting of iron, iron is oxidised and O $_2$ is reduced. The no. of electrons used during reduction O $_2$ are :

In rusting of iron, iron is oxidised and O

$_2$ is reduced. The no. of electrons used during reduction O

$_2$ are 4.

$O_2(g) +4H^+(aq) +4e^- \rightarrow 2H_2O(l)$

In a galvanic cell electrical energy is generated at the expense of chemical energy.
If true enter 1, else enter 0.

In a galvanic cell electrical energy is generated at the expense of chemical energy. In such types of cells, a spontaneous chemical reaction generates electric current.
Hence, the galvanic cells are used as a source of electric current.
They are also called voltaic cells.

The electric charge for electro deposition of 1 g equivalent of substance is $x$ faraday. What is the value of $x$ ?

The electric charge for electro deposition of 1 g equivalent of substance is one faraday. It is equal to 96500 coulombs. It corresponds to the amount of charge on 1 mole of electrons.

Presence of CO $_2$ in natural water facilitate rusting of iron.

If true enter 1, else enter 0.

Presence of

$CO_2$ in natural water facilitate rusting of iron.

$CO_2$ from the atmosphere, reacts with water to form carbonic acid.

$CO_2 + H_2O \rightarrow H_2CO_3$
Carbonic acid provides protons which are necessary for the oxidation of Fe to

$Fe^{2+}$ and further oxidation to

$Fe^{3+}$

Explain how electrolysis is an example of redox reaction?

Redox is a reaction, where one reactant oxidises and other reactant reduces.
In electrolysis like electrolysis of water,
At anode:

$2 H_2O \rightarrow O_2 + 4H^+ + 4 e^-$ oxidation
At cathode:

$4H_2O + 4 e^- \rightarrow 2H_2 + 4 OH^-$ reduction
(i) Cathode (Reducing electrode) : At cathode, the cations gain electrons to form neutral atoms. As electrons are gained, the ion is said to be reduced.
(ii) Anode (Oxidizing electrode) : At anode, the anions lose electrons to form neutral atoms. As electrons are lost, the ion is said to be oxidized.

_________ is defined as a process where materials, usually metals, deteriorate as a result of a chemical reaction with air, moisture, chemical, etc. For example, iron, in the presence of moisture, reacts with oxygen to form hydrated iron oxide.

Corrosion is defined as a process where materials, usually metals, deteriorate as a result of a chemical reaction with air, moisture, chemical, etc. For example, iron, in the presence of moisture, reacts with oxygen to form hydrated iron oxide.

$4Fe + 3O_2 + nH_2O \rightarrow \underset{Hydrated \,iron \,oxide}{2Fe_2O_3.nH_2O}$

This hydrated iron oxide is rust.

Number of Faradays (F) required to reduce 3 mol of $MnO_{4}^{\ominus}$ to $Mn^{2+}$ is:

$8H^{\oplus} + 5e^{-} + MnO_4^{\ominus} \rightarrow Mn^{2+} + 4H_2O$
1 mole of

$MnO_4^{\ominus} \equiv 5e^{-} = 5 F$
3 moles of

$MnO_4^{\ominus} = 5 \times 3 = 15 F$

How many Faradays are required to reduce 1 mol of $BrO_3^{\ominus}$ to $Br^{\ominus}$ in basic medium ?

$6e^{-} + BrO_3^{\ominus} (aq) + 3H_2O \rightarrow Br^{\ominus} (aq) + 6 \overset{\ominus}{O} H (aq)$

1 mol of

$BrO_3^{\ominus}$ = 6 F = 6 mol of electrons.

Electric Unit and Single Unit

(A) The electric unit Volt - ampere and joule second

$^{-1}$ corresponds to the single unit Watt.
(B) The electric unit Ampere - second and joule volt

$^{-1}$ corresponds to the single unit Coulomb.
(C) The electric unit Volt ampere

${-1}$ and joule ampere

${-1}$ second

$^{-1}$ corresponds to the single unit Ohm.
(D) The electric unit Joule ampere

$^{-1}$ second

$^{-1}$ corresponds to the single unit Volt.
(E) The electric unit Watt ampere

$^{-1}$ second

$^{-1}$ corresponds to the single unit Ampere.

Classify the following substances under three headings.
Strong electrolytes ,Weak electrolytes

(i) Strong electrolytes : ammonium chloride, dil. Hydrochloric acid, dil. Sulphuric acid.

(ii) Weak electrolytes : Acetic acid, ammonium hydroxide.

(iii) Non electrolytes : Distilled water, Alcohol, sugar solution

Fill in the blanks :
Electrolysis is the passage of_ (electricity / electrons) through a liquid or a solution accompanied by a _ (physical/ chemical) change.

Electrolysis is the passage of electricity through a liquid or a solution accompanied by a chemical change.

Water is broken down into its constituent elements .......... and .......... .

Electolysis of acidified water :

$2H_2O$

$\rightarrow$

$2H_2$

$+$

$O_2$

$\implies$ Water is broken down into its constituent elements hydrogen and oxygen on passing electric current.

The bulb does not glow in the set up shown in the figure. List the possible reasons. Explain your answer.

The possible reasons are:
(i) Although the solution may be conducting electricity, yet the current produced may be too small the heat the filament of the bulb and cause it to glow.
(ii) The connections may be loose.
(iii) The bulb is fused.
(iv) the cells are used up.

Paheli had heard that rainwater is as good as distilled water. So she collected some rain water in a clean glass tumbler and tested it using a tester. To her surprise she found that the compass needle showed deflection. What could be the reasons?

As rain drops fall through the air, they may dissolve ions present in air or forming acids on reaction with acidic oxides present in air, etc. This makes it a conducting solution. There are no dissolved salts present in the distilled water. Hence, rain water can allow electricity to pass through It while distilled water cannot.

What happens when
An iron rod is placed in $CuSO_{4}$ solution?

The following reactions occurs

$Fe + CuSO_{4}\rightarrow FeSO_{4} + Cu$

Why do caesium ( $Cs$ ) and potassium ( $K$ ) find applications in photoelectric cells?

The Photoelectric cells are based on the theory of photoelectric effect, where we consider that by providing sufficient energy we can excite the surface electrons so that they get replaced. So, for the element to be used in the photoelectric cell, that must have the capacity to lose electron in the even very low amount of energy supplied. Since Cs and K is the electropositive element of all so it has the minimum ionization energy and so contains the maximum capacity to lose electrons. This is why Cesium and potassium are used in photoelectric cells.

What happens when a copper rod is placed in $FeSO_{4}$ solution?

$Fe$ is more reactive than

$Cu$ hence no reaction will take place.

Differentiate between the terms strong electrolyte and weak electrolyte.
(sating any two differences)

1) Strong electrolytes allow a large amount of electricity to pass through them and hence are the good conductor of electricity whereas weak electrolytes allow small amounts to flow and are poor conductor of electricity.

2) Strong electrolytes dissociates almost completely whereas weak electrolytes dissociates partially.

Corrosion can be prevented by using _________ .

Corrosion can be prevented by painting the surface, by galvanising, and by zinc plating.

Edible oil is not allowed to stand for a long time in an iron or tin container. Give reasons.

Edible oil is not allowed to stand for a long time in an iron or tin container due to following reason. The fatty acids will react with rust and form salts which will contaminate the oil and make it rancid giving a foul odour and unpleasant taste.

What do you mean by corrosion? How can you prevent it?

When some metals are exposed to moisture, acids etc., they tarnish due to the formation of respective metal oxide on their surface. This process is called corrosion. Corrosion can be prevented by painting the surface, oiling, greasing, galvanizing, chrome plating or making alloys.

Write the SI unit of Resistivity.

The SI unit of Resistivity is

$\displaystyle \Omega m$ . The resistivity of a conductor is the resistance of the conductor that is 1 m long and 1 m

$^2$ in cross sectional area.

Write the complete chemical reaction of rusting of iron.

At anode, Fe is oxidised to Fe(II) ions and at cathode oxygen is reduced to water. The overall reaction is

$\displaystyle 2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) +2H_2O(l)$
Fe(II) ions are further oxidised to Fe(III) ions.

$\displaystyle 4Fe^{2+} (aq) + O_2(g) +4H^+(aq) \rightarrow 4Fe^{3+} (aq) +2H_2O(l)$
Fe(III) ions form an insoluble hydrated oxide which is deposited as red brown material called rust.

$\displaystyle 2Fe^{3+}(aq) +4H_2O(l) \rightarrow Fe_2O_3H_2O(s)+6H^+(aq)$

What is metallic corrosion? Give one example

Metallic corrosion is deterioration of metals by a redox process in which metals are oxidized by oxygen in presence of moisture. Examples include the rust on iron, loss of luster of silver, a green film formed on copper and brass.
Metallic corrosion is an electrochemical process.

Write three methods of preventing rusting of iron.

The rusting of iron can be prevented by the following methods:
(1) Iron surface is coated with paint. It is a temporary solution. The iron surface is shielded from oxygen and moisture.
(2) Iron is coated with another metal such as zinc. The process is called galvanising. Zinc is stronger reducing agent than iron and hence preferentially oxidised.
(3) An alloy of

$Fe$ with

$Cr$ is formed. This alloy is stainless steel. It reduces the tendency of iron to oxidise.

State the first and second law of electrolysis.

First law of electrolysis : The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passes. Thus if

$w_{g}$ of the substance is deposited on passing

$Q$ coulombs of electricity, then

$W\propto Q$ or

$W = ZQ$ where

$Z$ is a constant of proportionally called electrochemical equivalent of the substance deposited. If a current of

$C$ amperes is passed for

$t$ seconds then

$Q = C\times t$
so at

$W = Z\times C\times t$ or,

$W = \dfrac {\text {Equivalent wt. of the substance}}{96500}\times C\times t$
Second law of electrolysis : When the same quantity of electricity is passed through solutions of different electrolytes connected in the series, the weight of the substance produced at the electrodes is directly proportional to their equivalent weights.
For example, for

$CuSO_{4}$ solution and

$AgNO_{3}$ connected in the series if the same quantity of electricity is passed then,

$\dfrac {\text {Weight of Cu deposited}}{\text {Weight of Ag deposited}} = \dfrac {Eq. wt. of\ Cu}{Eq. wt. of\ Ag}$

What is corrosion? Do gold ornaments corrode? Justify.

Corrosion is the deterioration of metals by a redox process in which metals are oxidised by oxygen gas in presence of moisture. Iron rust, silver losing lustre, copper and brass coated with green film represent corrosion. Gold ornaments do not corrode. Gold is weaker reducing agent (stronger oxidising agent) and hence cannot be easily oxidised.

What is the chemical formula of rust?

Explanation

Rust is a mixture of oxides of iron formed on the surface of iron.
It is a result of a redox reaction between $Oxygen$ and $Iron$ in the presence of water/moisture.
$Oxygen$ being an excellent oxidizing agent first reacts with $Iron$ to oxidize it to a $+2$ or $+3$ state
This oxidized iron is highly reactive and thus combines with water/moisture to form hydroxides which on dehydration give oxides, thus forming a reddish-brown deposit of rust.
The formula for rust is $Fe_{2}O_{3}.nH_{2}O$ as exact value of water present in the rust is variable.

What is electrolysis? A solution of $CuSO_{4}$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

Electrolysis is a process in which an electric current is used to bring about a chemical reaction or a process in which an electrical energy is converted into a chemical energy.

Number of moles of electrons passed

$\displaystyle = \dfrac {I(A) \times t(s)}{96500}= \dfrac {1.5 \times 10 \times 60}{96500}= 0.009326$ moles of electrons

$\displaystyle Cu^{2+}+2e^- \rightarrow Cu$
The number of moles of copper deposited

$\displaystyle = \dfrac {0.009326}{2}=0.004663$ moles.
The molar mass of Cu is 63.5 g/mol.
The mass of Cu deposited

$\displaystyle =0.004663 \: mol \times 63.5 \: g/mol = 0.296 \: g$

State Faraday's law's of electrolysis. A solution of $Cu{SO}_{4}$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

Faraday's first law of electrolysis:
The amount of the substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

Thus mass of the substance produced

$\displaystyle = \dfrac {\text {I(A) } \times \text {t(s)}}{\text {96500 C/mol e}^-} \times \text {mole ratio} \times \text {molar mass}$
Here I(A) is current in ampere and t(s) is time in second.

Faraday's second law of electrolysis:
When the same amount of electricity is passed through different cells containing different electrolytes and arranged in series, the amounts of substances oxidized or reduced at the respective electrodes are directly proportional to their chemical equivalent masses.

$\displaystyle \dfrac { \text { moles of A produced}}{ \text { moles of B produced}} = \dfrac { \text { mole ratio of A half reaction }}{ \text { mole ratio of B half reaction }}$

Number of moles of electrons passed

$\displaystyle = \dfrac {I(A) \times t(s)}{96500}= \dfrac {1.5 \times 10 \times 60}{96500}= 0.009326$ moles of electrons

$\displaystyle Cu^{2+}+2e^- \rightarrow Cu$
The number of moles of copper deposited

$\displaystyle = \dfrac {0.009326}{2}=0.004663$ moles.
The molar mass of Cu is 63.5 g/mol.
The mass of Cu deposited

$\displaystyle =0.004663 \: mol \times 63.5 \: g/mol = 0.296 \: g$

What is a secondary cell?

An electric cell that can be rechargeable and can therefore be used to store electrical energy in the form of chemical energy is called secondary cell.

Some equipments and materials are given.
$Zn{SO}_{4}$ solution, $Cu{SO}_{4}$ solution, $Zn$ rod, $Cu$ rod, Voltmeter, $KCl$ solution, filter paper.
(a) Draw the diagram of the electrochemical cell which can be constructed using these equipments and materials and label the pans.
(b) Write equations of chemical reactions taking place in the two electrodes of this cell.
(Hint: reactivity $Zn> Cu$ )

(a) Electrochemical cell
(b) Reaction at the anode:

${Zn}_{(s)}+2{e}^{-}\rightarrow {Zn}^{2+}$
Reaction at the cathode:

${Cu}^{2+}+2{e}^{-}\rightarrow {Cu}_{(s)}$
Over all reaction:

${Zn}_{(s)}+{Cu}^{2+}\rightarrow {Zn}^{2+}+{Cu}_{(s)}$

$Mg,\ Al,\ Zn,\ Fe$ and $Ag$ are metals of reactivity series:
(a) Which of this metal shows the highest reactivity?
(b) If an electrochemical cell is devised by dipping Fe in $FeSO_4$ solution and Ag in $AgNO_3$ solution, which electrode is acting as cathode? Give reason.

(a) Mg has the highest reactivity.
(b) Ag can be used cathode. This is because the electrode at which reduction takes place (one which accepts electrons) is the cathode.

What do you mean by metal corrosion? Explain three methods for prevention of corrosion.

Metal corrosion:
It is deterioration of metals by a redox process in which metals are oxidized by oxygen in presence of moisture. For example, rust on iron, loss of luster of Ag, a green film formed on copper and brass.

Methods for prevention of corrosion:
(1) Galvanisation: Steel and iron is protected from rusting by coating with a thin layer of zinc.The galvanised article is protected against rusting even if the zinc coating is broken.

(2) Cathodic protection: More easily oxidized metal (such as Mg or Zn) is attached to the metal (Fe) to be protected from corrosion. Fe acts as cathode and Mg or Zn acts as sacrificial anode.

(3) Passivation: The metal surface is made inactive. The metal is treated with strong oxidizing agent such as conc

$HNO_3$ . A thin oxide layer is formed on metal surface.

(4) Painting, greasing and oiling.

(5) Chrome plating and making alloys.

The values of $E^{\circ}$ of some of the reactions are given below:
$I_{2} + 2e^{-}\rightarrow 2I^{-}; E^{\circ} = + 0.54\ volt$
$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}; E^{\circ} = + 1.36\ volt$
$Fe^{2+} + e^{-}\rightarrow Fe^{2+}; E^{\circ} = + 0.76\ volt$
$Ce^{4+} + e^{-}\rightarrow Ce^{3+}; E^{\circ} = + 1.60\ volt$
$Sn^{4+} + 2e^{-}\rightarrow Sn^{2+}; E^{\circ} = +0.15\ volt$
On the basis of the above data, answer the following questions:
(a) Whether $Fe^{3+}$ oxidises $Ce^{3+}$ or not?
(b) Whether $I_{2}$ displaces chlorine from $KCl$ ?
(c) Whether the reaction between $FeCl_{3}$ and $SnCl_{2}$ occurs or not?

(a) Chemical reaction,

$Fe^{3+} + Ce^{3+}\rightarrow Ce^{4+} + Fe^{2+}$
Two half reactions,

$Fe^{3+} + e \rightarrow Fe^{2+}; Reduction; E^{\circ} = 0.76\ volt$

$Ce^{3+} \rightarrow Ce^{4+} + e^{-}; \underline {Oxidation; E_{OX}^{\circ} = -1.60\ volt}$

$Adding; emf = -0.84\ volt$
Since, emf is negative, the reaction does not occur, i.e.,

$Fe^{3+}$ does not oxidise

$Ce^{3+}$ .
(b) Chemical reaction,

$I_{2} + 2KCl = 2KI = Cl_{2}$
Half reactions,

$I_{2} + 2e^{-}\rightarrow 2I^{-}; Reduction; E^{\circ} = 0.53\ volt$

$2Cl^{-}\rightarrow Cl_{2} + 2e^{-}; underline {Oxidation; E_{OX}^{\circ} = -1.36\ volt}$

$Adding; emf = -0.82\ volt$
Since, emf is negative, the reaction does not occur, i.e.,

$I_{2}$ does not displace

$Cl_{2}$ from

$KCl$ .
(c) Chemical reaction,

$SnCl_{2} + 2FeCl_{3}\rightarrow SnCl_{4} + 2FeCl_{2}$
Half reactions,

$Fe^{3+} + e^{-}\rightarrow Fe^{2+}; Reduction; E^{\circ} = 0.76\ volt$

$Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}; \underline {Oxidation; E^{\circ} = -0.15\ volt}$

$Adding; emf = +0.61\ volt$
Since, emf is positive, the reaction will occur.

Resistance of a solution (A) is $50\ ohm$ and that of solution (B) is $100\ ohm$ , both solutions being taken in the same conductivity cell. If equal volumes of solutions (A) and (B) are mixed, what will be the resistance of the mixture, using the same cell? Assume that there is no increase in the degree of dissociation of (A) and (B) on mixing.

Let us suppose

$K_{1}$ and

$K_{2}$ are the specific conductances of solutions

$'A'$ and

$'B'$ respectively and cell constant is

$'y'$ . We know that,
Specific conductance

$= Conductance \times Cell\ constant$
For

$(A), K_{1} = \dfrac {1}{50}\times y$
For

$(B), K_{2} = \dfrac {1}{100}\times y$
When equal volumes of

$(A)$ and

$(B)$ are mixed, the volume becomes double. Then,
Specific conductance of mixture

$= \dfrac {K_{1} + K_{2}}{2}$

$\therefore \dfrac {K_{1} + K_{2}}{2} = \dfrac {1}{2} \times y$

$\dfrac {1}{2}\left [\dfrac {y}{50} + \dfrac {y}{100}\right ] = \dfrac {1}{R}\times y$

$\dfrac {1}{100} + \dfrac {1}{200} = \dfrac {1}{R}$

$R = 200/3 = 66.66\ ohm$ .

Write short notes on :
(i)Nernst equation
(ii)Corrosion.

(i) Nernst equation : The effect of change in temperature on EMF of cell can be studied by this equation.
(a) Total change in free energy of cell can be calculated by this equations.

$\triangle G^o = -nE^oF$
(b) Total work done by cell can be calculated
Total work

$= nE^oF$ Nernst studied the effect of concentration of solution and temperature on the electrode potential and gave a equation called Nernst equation.
Nernst equation is

$E=E^o+\dfrac{RT}{nF}ln [M^{n+}_{(aq)}]$
Where, E= Electrode potential,

$E^o$ = Standard electrode potential (

$M^{n+}$ : Conc.] 1M and temp. 298K), R = Gas Constant, T = Temperature (in K), n = Valency of metal ion F = Faraday (96,500 coulomb). Concentration of pure solid and liquid are.taken as unity.

$E=E^o+\dfrac{0.059}{n}\log 10 [M^{n+}_{(aq)}]$
(ii) Corrosion : Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.
Factors affecting corrosion
(1) Nature of metal : Reactive metal corrodes readily.
(2) Impurities in metal : Impure metal corrodes quickly to a greater extent.
(3) Environment : If environment around metal contains oxygens, carbon dioxide, m oisture, saltsand acidic gases Iike

$CO_2, SO_2, SO_3$ etc. then corrosion occurs quickly.
Method to prevent corrosion
(1) Barrier protection : In this method, the surface of the metal is coated with paint; oil or grease. Due to this the surface of the metal remain unexposed to atmospheric conditions and hence corrosion is prevented. Surface of the metal can also be protected by :
(i) Coating metal surface with non-corroding metals like nickel and chromium is called electroplating :
(ii) Dipping iron article in molten metal like zinc. This process is called galvanization.
(2) Sacrificial protection : The rusting of iron can be prevented by covering the iron with more electro positive metals like zinc. Zinc metal has more tendency to get oxidized as compared to iron. So, iron articles will not be harmed till the layer of zinc present on its surface hence zinc metal is called the sacrificial metal.
(3) Antirust solution : The alkaline antirust solution are employed to prevent rusting alkaline solution prevent the availability of

$H^+$ ions.
In this method, iron articles are dipped in alkaline sodium phosphate of chromate solution. Due insoluble sticking film of iron phosphate is formed on the surface which prevents rusting.

(a) What do you understand by corrosion?
(b) Write electro-chemical theory of corrosion (Rust).
(c) Write prevention (two) of corrosion.

(a) Corrosion: Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.
(b) Electrochemical theory of corrosion: It is widely accepted electrochemical reaction. A number of galvanic cells are formed in impure iron. In rusting of iron pure iron acts as a anode while impure part of iron acts as cathode. A thin layer formed by dissolved oxygen and carbon dioxide in moisture acts as solution of electrolyte. Oxidation and reduction occur at anode and cathode, respectively.

$Fe \rightarrow {Fe}^{2+} + 2{e}^{-}$ (Anode)

$2H + 2{e}^{-} \rightarrow {H}_{2}$ (Cathode)
Thus all metals above hydrogen in electrochemical series after corrosion evolve hydrogen.
(c) Prevention of corrosion:
(i) By coating metal surface with oil or grease: A thin layer of oil or grease is coated on the surface of the metal.
(ii) By painting the metal surface: A layer of point is coated on the surface of the metal.

Explain corrosion on the basis of following points :
(i) Definition
(ii) Factors affecting any two
(iii) Prevention of corrosion any two.

(i) Definition : Corrosion is the process in which a metal deteriorates as a result of its conversion into compounds on exposure to air and water.
(ii) Factors affecting :
(1) Reactivity of metal : A reactive metal corrodes readily.
(2) Preasence of impurities : Impurities help in setting up a corrosion cell and make corrosion to occur speedly.
(iii) Preventions of corrosion :
(1) Barrier protection :
(a) By painting the surface of metal
(b) By coating the surface with thin film of oil or grease.
(c) By electroplating iron with non corrosive metal.
(2) Sacrificial protection :
In this method iron is coated with a layer of metal more active than iron.
(3) Electrical Protection.
(4) Using anti - rust solutions.

The emf of a cell corresponding to the reaction,
$Zn + 2H^{+}(aq.) \rightarrow Zn^{2+} (0.1\ M) + H_{2}(g) 1\ atm$
is $0.28$ volt at $25^{\circ}C$ . Write the half-cell reactions and calculate the $pH$ of the solution at the hydrogen electrode.
$E^{\circ}_{Zn^{2+}/Zn} = -0.76\ volt$ and $E^{\circ}_{H^{+}/H_{2}} = 0$ .

$E^{\circ}_{cell} = 0.76\ volt$

$E_{cell} = E^{\circ}_{cell} - \dfrac {0.0591}{2}\log \dfrac {[Zn^{2+}][H_{2}]}{[H^{+}]^{2}}$

$0.28 = 0.76 - \dfrac {0.0591}{2}\log \dfrac {(0.1)\times 1}{[H^{+}]^{2}}$

$\log \dfrac {0.1}{[H^{+}]^{2}} = \dfrac {2\times 0.48}{0.0591}$

$\log 0.1 - \log [H^{+}]^{2} = 16.2436$ [Since,

$-\log [H^{+}] = pH$ ]

$2\ pH = 16.2436 - \log 0.1$

$pH = \dfrac {17.2436}{2} = 8.6218$ .

What ratio of $Pb^{2+}$ to $Sn^{2+}$ concentration is needed to reverse the following cell reaction?
$Sn(s) + Pb^{2}(aq.) \rightarrow Sn^{2+}(aq.) + Pb(s)$
$E^{\circ}_{Sn^{2+}/ Sn} = -0.136\ volt$ and $E_{Pb^{2+}/ Pb}^{\circ} = -0.126\ volt$ .

In the given cell reaction,

$E_{cell}$ =

$E^0_{cell}$

$-$

$\dfrac{0.059}{2}$

$log$

$\dfrac{Sn^{2+}}{Pb^{2+}}$

$E_{cell}$ =

$E^0_{cell}$

$+$

$\dfrac{0.059}{2}$

$log$

$\dfrac{Pb^{2+}}{Sn^{2+}}$

$E_{cell}$ = 0.010 + 0.0295

$log$

$\dfrac{Pb^{2+}}{Sn^{2+}}$

when

$E_{cell}$ is negative then cell reaction reverses,

$\dfrac{Pb^{2+}}{Sn^{2+}}$ is in between 0.1 to 0.9 then, cell reaction reverse.

Calculate the charge of $6.24\times 10^{18}$ electrons (in coulomb).

$1e^-$ has

$1.6\times 10^{-19}C$ charge,

$\therefore$ Total no. of charge in

$6.24\times 10^{18}e^-={(6.24\times 10^{18})}\times 1.6\times 10^{-19}C=0.998C$

How many moles of electrons will together contribute to the charge of 289500 coulomb?

1 mole of electrone contribute to the charge of 1 F = 96500 C

1 C charge =

$\dfrac{1}{96500}$ F

289500 C charge =

$\dfrac{1}{96500}$ 289500 F

=3F

3 mole of electrone will together contribute to the charge of 289500 C

State Faraday's first law of electrolysis.

Faraday's first law of electrolysis:
The amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell. It is equal to the product of current and time.

At what $[OH^{-}]$ does the following half-reaction have a potential of $0 V$ when other species are at $1\ M$ ?
$NO_{3}^{-} + H_{2}O + 2e^{-}\rightarrow NO_{2}^{-} + 2OH^{-}, E_{cell}^{\circ} = 0.01\ V$ .

$\begin{array}{l} { E_{ cell } }=E_{ cell }^{ ^{ \circ } }-\frac { { 0.0592 } }{ 2 } \log \frac { { { { \left[ { O{ H^{ - } } } \right] }^{ 2 } }\left[ { NO_{ 2 }^{ - } } \right] } }{ { \left[ { N{ O_{ 3 } }O } \right] } } \\ 0\Rightarrow 0.01-\frac { { 0.0592 } }{ 2 } \log \frac { { { { \left[ { O{ H^{ - } } } \right] }^{ 2 } }\times 1 } }{ 1 } \\ \frac { { 0.0592 } }{ 2 } \times 2\log \left[ { O{ H^{ - } } } \right] =0.01 \\ \log \left[ { O{ H^{ - } } } \right] =0.01/0.0592=0.169 \\ \left[ { O{ H^{ - } } } \right] ={ 10^{ 0.169 } }\Rightarrow 1.476\, \, M \end{array}$

If one Faraday was to be $48250$ coulomb instead of $96500$ coulomb, what will be charge of an electron?

1 Faraday=charge of 1 mol of electron

let charge of 1 electron =

$x$

then

$N_A x=48250\ C$

$x=\frac {48250}{6.23\times 10^{ 23 }}$

$x=0.774\times 10^{ -19 }$ C

If $0.5\ L$ of a $0.6\ M\ SnSO_{4}$ solution is electrolysed for a period of $30$ min, using a current of $4.6\ A$ . If insert electrodes areas. What is the final concentration of $Sn^{2+}$ remaining in the solution?

For the deeposition of

$Sn$ electrochemical reaction is

${ Sn }^{ 2+ }+2{ e }^{ - }\rightarrow Sn$

Using Faraday's first law of electrolysis: Amount of substance deposited (m) at an electrode is directly proportional to amount of charge passed through the solution and is written as

$m=\frac { M }{ n\times 96500 } It$

where

$M=118.7 g/mol,\ n=2,\ I=4.6 A,\ t=30/times 60 s$

Amount of

$Sn$ deposited,

$m=\frac { 118.7 }{ 2\times 96500 } 4.6\times 30\times 60$

$m=5.0924\quad g=\quad 85.8\quad mM\quad$ from 0.5 L

Amount remaining in solution:

$0.6-0.0858=0.514 M$

What is an electrochemical series? How is it useful in predicting whether a metal can liberate hydrogen from acid or not?

Electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials with respect to hydrogen electrode. Standard hydrogen electrode has reduction potential of zero volt. Therefore, those species with negative standard potential will undergo oxidation and species with positive standard potential will undergo reduction. Also, higher the reduction potential higher is the ability to get reduce or stronger is the oxidising agent.

Metals with negative reduction potential will be strong reducing agent and thus will reduce the

$H^+$ ion from acid to liberate

$H_2$ gas.

Fill in the blanks by choosing the appropriate word/words from those given in the brackets:
(square pyramidal, electrical, 74, 26, $sp^3d^2$ , $sp^3d$ , chemical, 68, 32, tetrahedral, yellow, white, iodoform, Lucas).
A Galvanic cell converts energy into _ energy.

A galvanic cell converts chemical energy into electrical energy.

The resistance of a conductivity cell containing $0.001\ M\ KCl$ solution at $298\ K$ is $1500\Omega$ . What is the cell constant if conductivity of $0.001\ M\ KCl$ solution at $298\ K$ is $0.146\times 10^{-3} S\ cm^{-1}$ .

Conductivity of the cell, k

$= 0.146\times 10^-3 Scm^{-1}$

Resistance of the cell, R

$= 1500 \Omega$

$\text{Cell constant} = k\times R$

$= 0.146\times 10^{-3} \times 1500$

$= 0.129\ cm^{-1}$

Find the distance between $(345)$ plane is a cubic batteries of length $7-7\overset{o}{A}$ .

To find the distance between adjacent plane

$(h,k,l)$

$\alpha= \cfrac {a}{\sqrt {h^2+k^2+l^2}}$

$\Rightarrow a= 7\overset {o}{A}$

$\Rightarrow h=3, k=4, l=5$

$\alpha= \cfrac {7\overset {o}{A}}{\sqrt {3^2+4^2+5^2}}=\cfrac {7\overset {o}{A}}{\sqrt {50}}=0.98 \overset {o}{A}$

Thus, the distance between plane is

$0.98 \overset {o}{A}$ .

Write the equations for the reactions taking place at anode and cathode in lead storage cell.

Oxidation at anode:

$\displaystyle Pb(s) + SO_4^{2-} (aq) \to PbSO_4(s)+2e^-$

Reduction at cathode:

$\displaystyle PbO_2(s)+4H^+(aq) +SO_4^{2-} (aq) +2e^-\to PbSO_4(s)+2H_2O(l)$

Can you store copper sulphate solution in a zinc pot?

Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution.

$Zn+CuSO_4\longrightarrow ZnSO_4+Cu$

Hence, the copper sulphate solution cannot be stored in a zinc pot.

A metal carbonate X on reacting with an acid gives a gas which when passed through a solution gives the carbonate black. On the other hand, A gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives compound Z, used for disinfecting water. Identify X, Y, G and Z.

According to the problem:

Here,

$X$ is calcium carbonate

$\left( {CaC{O_3}} \right)$ .

$Y$ is slaked lime.

$Z$ is bleaching powder

$\left( {CaOC{l_2}} \right)$

$G$ is chlorine gas.

Hence the reactions are involved,

Calculate volume of the gases liberated at STP if L of $0.2$ molar solution of $CuSO_4$ is electrolysed by $5.79 A$ Current for $10000 seconds.$
$5.79\times 10000$

At Cathode :-

${ Cu }^{ 2+ }+{ 2e }^{ - }\rightarrow Cu$

at anode :

$4{ OH }^{ \left( - \right) }+{ 4e }^{ - }\rightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }$

Mol. of Faraday

$=\dfrac { I\times t }{ F } =\dfrac { 5.79\times 10000 }{ 96500 }$

$=0.6mol$

4 mol Faraday = 1 mol of

${ O }_{ 2 }$ = 22.4 lit of

${ O }_{ 2 }$

Volume of gas liberated

$=\dfrac { 1 }{ 4 } \times 22.4\times 0.6$

$=3.36lit$

In electrolysis of water:

(a) Name the gas collected at cathode and anode.
(b) Why is volume of one gas collected at one electrode is double of anode?
(c) Why are few drops of dilute $H_2SO_4$ added to water?

a) At cathode,

$H_2$ is evolved and at anode

$O_2$ is evolved.

b) At cathode,

$H^+$ ion takes two electron to convert itself into

$H_2$ gas.

$2$ moles of

$H^+$ gives

$1$ mole of

$H_2$ .

$2H^+(aq)+ 2e^- \rightarrow H_2(g)$

At anode,

$OH^-$ ion releases two electrons to convert into water and

$O_2$ .

$2$ mole

$OH^-$ gives

$0.5$ mole

$O_2$ , i.e why volume of one gas collected at one electron is double of another.

c) Acid is added to make the water conduct electricity as the distilled water is a non-conductor of electricity.

Represent the galvanic cell in which the reaction takes place.
$Zn(s)+{Cu}^{2+}(aq)\rightarrow {Zn}^{2+}(aq)+Cu(s)$

Measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials.

This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through.

If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when $Zn^{ 2+ }$ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the $Cu^{ 2+ }$ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:

$Zn_{ (S) }+Cu_{ (aq) }^{ 2+ }\longrightarrow Zn_{ (aq) }^{ 2+ }+Cu_{ (s) }$

1038852_1071716_ans_9d476824a4b842c3b2a59d4c541dcbda.jpg

Calculate the no.of colombs of electricity, required to deposit
A) $2.7\times 10^{-4}kg$ of aluminium
B) $6.5\times 10^{-5}kg$ of Zn

(A)

${ Al }^{ 3+ }+3{ e }^{ - }\rightarrow Al$

3 mol

$F\rightarrow 27g$ of

$Al$

$2.7\times { 10 }^{ -4 }kg\quad of\quad Al\rightarrow \dfrac { 3 }{ 27\left( g \right) } \times 2.7\times { 10 }^{ -4 }\times { 10 }^{ 3 }\left( g \right)$

$=0.03$ Faraday

in coloumb,

$=0.03\times 96500=2895C$

(b)

${ Zn }^{ 2+ }+{ 2e }^{ - }\rightarrow Zn$

$65g\rightarrow 2$ mol F

$6.5\times { 10 }^{ -5 }Kg=6.5\times { 10 }^{ -2 }g\rightarrow \dfrac { 2 }{ 65 } \times 6.5\times { 10 }^{ -2 }$

$=0.002$ Faraday

in coloumb

$=0.002F\times 96500$

$=193C$

Specific charges pf two particles $A$ and $B$ are in ratio $2 : 3$ , If their mass ratio $m_{A};m_{B}$ is $2:3$ , then find ratio of their charges $\left(\dfrac{q_{A}}{q_{B}}\right)$ ?

we know

specific charge

$=\dfrac {charge}{mass}=9/m$

so,

$\left (\dfrac {9A}{mA}\right)/ \left (\dfrac {9B}{mB}\right)=\dfrac {2}{3}$

$\left (\dfrac {9A}{9B}\right)=\dfrac {2}{3} \left (\dfrac {mA}{mB}\right)$

Also,

$\dfrac {9A}{9B}=\left (\dfrac {2}{3}\right) \left (\dfrac {2}{3}\right)=4:9$

$\dfrac {mA}{mB}=\dfrac {2}{3}$ [Given]

Calculate the number of coulombs required to deposit $5.4g$ of $Al$ , when the electrode reaction is ${Al}^{3+}+3{e}^{-}\rightarrow Al$
(At.mass of $Al=27,F=96500$ coulomb per mole)

${ Al }^{ 3+ }+3{ e }^{ - }\rightarrow Al$

$3mol{ e }^{ - }\rightarrow 27g$ of

$Al$

$5.4g\rightarrow \dfrac { 3 }{ 27 } \times 5.4=0.6mol\ of\ { e }^{ - }$

no. of

$F=\dfrac { Q }{ 96500 } =0.6$

$Q=0.6\times 96500=57900C$

Calculate the number of faradays required to electrolyze $6.35 \,g$ of $Cu^{+}(aq)$ ions from an aqueous solution.

${ Cu }^{ + }+{ e }^{ - }\rightarrow Cu$

1 mol

$F\rightarrow 63.5g$ of copper

1g of copper

$\rightarrow \dfrac { 1 }{ 63.5 } mol$

6.35g of copper

$\rightarrow \dfrac { 6.35 }{ 63.5 } =0.1F$ of electricity

On passing 4 amperes of current for 10min, during electrolysis of acidulated water at STP, what quantity of oxygen in $dm^3$ will be liberated?

${ H }_{ 2 }O\rightarrow { O }_{ 2 }+2{ H }^{ + }+4{ e }^{ - }$

$\therefore$ Faraday

$=\dfrac { change }{ 96500 } =\dfrac { 4\times 10\times 60 }{ 96500 } =0.0248F$

4 mol F

$\rightarrow 22.4{ dm }^{ 3 }$ of

${ O }_{ 2 }$

$0.0298F=\frac { 1 }{ 4 } \times 22.4\times 0.0248=0.14{ dm }^{ 3 }$ of oxygen

How many faradays of electricity will be required by $3\times 10^{12}$ electrons?

Moles of electrons=

$\cfrac {3 \times 10^{12}}{6.02\times 10^{23}}=4.98 \times 10^{-12}$

$1$ mole of electron require

$1$ Faradays of electricity

$\therefore 4.98 \times 10^{-12}$ Faradays of electricity is required.

If $E^{+}_{Fe^{2+}/Fe}$ is $x_1$ , $E^{+}_{Fe^{3+} / Fe}$ is $x_2$ then what will be $E^+_{Fe^+/Fe^{2+}}$ ?

By comparing Gibb's free energy,

$\therefore$

$3\times { x }_{ 2 }=E+2{ x }_{ 1 }$

$E=2{ x }_{ 1 }+3{ x }_{ 2 }$

1033350_1093167_ans_c5156e7655f246ba9437c9dec2aedeff.jpg

How many minutes are required to deliver $3.21 \times 10^{6} coulombs$ using a current of $500$ A?. That is used in the commercial production of chlorine.

$Q=i.t\quad \Longrightarrow t=\cfrac{Q}{i}$

$={\left(\cfrac{3.21\times {10^6}}{500}\right)}$

$=6420sec=107min$

What weight of Ni is plated out in an electrolysis of aqueous ${ NiSO }_{ 4 }$ solution that it takes place to deposit 2g of Ag in a silver coulometer that is arranged in series with ${ NiSO }_{ 4 }$ electrolytic cell.[atomic weight of Ag=107.8 amu,atomic weight of Ni=58.7 amu]

$\underbrace { \cfrac { { W }_{ 1 } }{ { Z }_{ 1 } } }_{ Ni } =\underbrace { \cfrac { { W }_{ 2 } }{ { Z }_{ 2 } } }_{ Ag }$ (for series)

$\therefore\quad W_1=z_1{\left(\cfrac{2}{107.8/1}\right)}$

$={\left(\cfrac{58.7}{2}\right)}$

$\times {\left(\cfrac{2}{107.8}\right)}$

$=0.54\quad g.$

$Note:\left\{ \begin{matrix} { n }_{ f }\left( { Ni }^{ 2+ } \right) =2 \\ { n }_{ f }\left( { Ag }^{ + } \right) =1 \end{matrix} \right\}$

How many coulombs of electricity is required to oxidize one mole of $Al$ to $Al^{3+}$ ?

Since 1 mole of Al to

$Al^{3+}$ oxidation requires 3 electron transfer,

So 3F of charge is required

$Q=3F=$

$(3\times96500)$ coulomb

Answer the following.
a. What is done to prevent corrosion of metals?
b. What are the metals that make the alloys brass and bronze?

a. Preventive measures for corrosion:

By painting on the surface
By alloying the layers
Add protective layers
Galvanisation
Non-corrosive layer coatings.

$b.$ Brass

$-$ Copper and zinc

Bronze

$-$ Copper and tin

0.2864g of $Cu$ was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. What is the electrochemical equivalent of copper?

Given that,

Weight of copper $=0.2864\ g$

Current $=0.5\ ampere$

Time $=30\ \operatorname{mint}$

Now we get

Quantity of charge passed

$\Rightarrow 0.5\times 30\times 60=900\ coulomb$

So, $900\ coulomb$ Deposit copper $=0.2864g$

We know that

$1\ faraday=96500\ coulomb$

Thus,

$96500$ coulomb deposit copper $=\dfrac{0.2864}{900}\times 96500=30.78\ g$

Hence, Equivalent mass of copper $=30.78g$

What is corrosion?

Corrosion is defined as the destructive & unintentional degradation of a material caused by its environment.

A common type of corrosion is rust. It is found on iron & steel structures. Iron reacts with oxygen (from air or water) to form an iron oxide which is more stable than pure iron.

Write the reaction occurring at the cathode and anode in $H_2$ AND $O_2$ fuel cell.

Image result for anode and cathode reactions for the hydrogen-oxygen fuel cell

Answer the following :
What are the constituents of Nichrome wire?
Define Corrosion

1353231_1153095_ans_cdceeb91a051495f8bf99ec9f335f093.jpg

In the adjacent diagram the electrolytic cell contains $1L$ of an aqueous $1M$ Copper (II) sulphate solution. If $0.4$ mole of electrons are passed through the cell, the molar concentration of copper ion after passing of the charge will be :

1351556_1142375_ans_05e6579556d54977a6806ae4de2bfeac.jpg

Name the process of depositing a layer of zinc on iron. Give two examples of objects on which zinc coating is done.

Galvanisation is the process of depositing a layer of zinc on iron. Rusting of an iron object can be prevented by the process of galvanization. Ex: Iron sheets used for making buckets and sheds are coated with zinc to prevent rusting.

What special name is given to the corrosion of iron?

Rusting is the special name given to corrosion of iron.

How much charge is required for the following reductions?
(i) $1\ mol$ of $Al^{3+}$ to $Al$
(ii) $1\ mol$ or $Cu^{2+}$ to $Cu$
(iii) $1\ mol$ of $MnO^{-}_4$ to $Mn^{2+}$ .

(i)

$Al^{3+} + 3e \rightarrow Al$

Charge required

$= 3F$

(ii)

$Cu^{2+} + 2e\rightarrow Cu$

Charge required

$= 2F$

(iii)

$MnO^{-}_4 + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$

Charge required

$= 5F$ .

Why is ethanol is used as a fuel?

1133152_1306426_ans_ff8bed3b1d7040ae8253a33edc3bdea7.jpg

The $E^{e}$ $(M^{2+}/M)$ value for copperis positive (+0.34V). What is possibly the reason for this ?

Copper has a high atomization energy

$\Delta_{a}H^{\Theta }$ and Low hydration energy

$\Delta_{hyd}H^{\Theta }$ . Due to which

$E^0$ value is positive.

What is Primary Battery? Give one example.

A primary cell is a kind of electrochemical battery which cannot be recharged and the chemicals are to be replaced in it regularly.

E.g.- Dry cell

Calculate the force of attraction between an electron ans a body having two proton charge when they are $0.529\times 10^{-8}\ cm$ apart. Assume charge of one electron and one proton equal to $-1.6\times 10^{-19}C$ and $+1.6\times 10^{-19}C$ respectively.

1377112_1295593_ans_f983c18536f64c1b9573a87c2bf8372f.jpg

Write the cell reaction and calculate the emf of the cell
$Pt |H_2 (g, 1 atm) | H^+ (0.5 M) || KCl (1 M) |Hg_2Cl_2 (s) | Hg (l) | Pt$ at $25^o C, \, E_{} = 0.28 V$

The half cell reactions and overall cell reaction are as follows :

$H_2(g, \, 1 atm) \rightarrow 2H^+ (0.5 M) + 2e^-$ (oxidation at anode)

$Hg_2Cl_2 (s) + 2e^- \rightarrow 2Hg(l) + 2 Cl^- (aq)$ (reduction at cathode)
_________________________________________________

$H_2 (g, \, 1 atm) + Hg_2Cl_2(s) \rightarrow 2H^+ (0.5 M) + 2Hg(l) + 2 Cl^{-} (aq)$ (overall cell reaction)

What are Galvanic cells? Explain the working of Galvanic cells with one example.

Galvanic cell

$\rightarrow$ A galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy.

Working of galvanic cell

$\rightarrow$

It involves a chemical reaction that makes the electric energy available. During a redox reaction, a galvanic cell utilizes the energy transfer between electrons to convert chemical energy into electric energy.

In a Galvanic cell, the following reaction occurs-

${Zn}_{\left( s \right)} + {{Cu}^{2+}}_{\left( aq. \right)} \longrightarrow {{Zn}^{+2}}_{\left( aq. \right)} + {Cu}_{\left( s \right)}$

Galvanic cell utilizes the ability to separate the flow of electrons in the process of oxidization and reduction. This flow of electrons is essentially called a current. Such current can be made to flow through a wire to complete a circuit and obtain its output in any device such as a television or a watch.

A galvanic cell can be made out of any two metals. These two metals can form the anode and the cathode if left in contact with each other. This combination allows the galvanic corrosion of that metal which is more anodic. A connecting circuit shall be required to allow this corrosion to take place.

E.g.- In this cell, there is a container in which a solution of concentrated Copper Sulphate

$\left( CuS{O}_{4} \right)$ is kept inside it and a copper rod is inserted inside the solution of

$CuS{O}_{4}$ which act like cathode. Inside this, the container a porous container is kept in which concentrated Sulphuric Acid

$\left( {H}_{2}S{O}_{4} \right)$ is filled in which a zinc rod is inserted in it which acts as an anode. Thus when a wire is connected through the copper rod and zinc rod an electric current starts to flow.

1188649_1469207_ans_a987cb0d4c8a4b3eba367a09a5bc332a.png

Observe the following picture and answer the following question:
What is corrosion ?

Corrosion : Corrosion is the gradual destruction of a metal due to reactions with other chemicals in its environment.

Corrosion is a natural process that converts a refined metal into a more chemically-stable form such as oxide, hydroxide, or sulfide.

A solution of ${ CuSo }_{ 4 }$ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

1359462_1512746_ans_e823f7f225134d039e589ba5544d1e40.jpg

How much amount of substance is deposited by passing one Faraday of electricity?

The equivalent weight of a substance is the amount deposited by passing

$1$ Faraday of electricity.

Given the standard electrode potentials :

$K^+/K = -2.93 \,V$ , $Ag^+ /Ag = 0.80\,V$ , $Hg^{2+} / Hg= 0.79\,V$ ,
$Mg^{2+}/Mg = -2.37\,V$ , $Cr^{3+}/Cr = -0.74\,V$

Arrange these metals in their increasing order of reducing power.

More is

$E_{RP}^{\circ}$ , more is the tendency to get reduced or more is the oxidising power or lesser is reducing power.

$Ag < Hg < Cr < Mg < K$

How many hour are required for a current of $3.0$ amperes to decompose $18\,g$ water.

1759254_1741352_ans_edfa3ed2d7dd46baa9c3c6dbd458a43c.jpg

What would be the effect on the potential of this cell if $Na_2S$ were added to the $Cd^{2+}$ half cell and CdS were precipitated? Why?

1771132_1740163_ans_d5b50f3a6f234c3b883d46e1b0edf594.jpg

Explain the term "corrosion" with an example.Write a chemical equation to show the process of corrosion of iron.

Corrosion is the process in which metals are eaten up gradually by the action of air,moisture or a chemical (such as an acid) on their surface. Rusting of iron metal is the most common form of corrosion.

During the corrosion of iron, iron metal is oxidised by the oxygen of an air in the presence of water (moisture) to form hydrated iron oxide called rust.

$4Fe+3O_2+2xH_2O \rightarrow 2Fe_2O_3.x H_2O$

What does the negative value of $E^{\circ}_{cell}$ indicate ?

Negative

$E^{\circ}_{cell}$ value means

$\Delta_{r}G^{\circ}$ will be +ve and the cell will not work.

What is meant by corrosion ? How corrosion is caused ? How it can be prevented ? What is the effect of corrosion on : (a) copper (b) Silver ?

Corrosion refers to the destructive or deterioration of a substance due to its interactions with the surrounding environment. Corrosion is a natural process and occurs when the substance is in contact with air, water, chemicals like acids etc. Corrosion can damage metals causing it to lose properties such as strength and conductivity of heat/electricity.

Corrosion can be prevented by various techniques like painting the surface of the material, galvanizing, oiling, greasing etc.

(a) Copper when in contact with moist carbon dioxide in air gains a green coloured coat over copper. This coat is nothing but copper carbonate formed by the reaction of copper and moist carbon dioxide.

(b) Silver reacts with sulphur present in the air to form a black coating of silver sulphide over the surface of silver.

(a) What is meant by corrosion? Name any two methods used for the prevention of corrosion.

(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used extracting this metal.

(a) The unwanted reaction of any metal with its surroundings that deteriorate its qualities is called corrosion. Two methods to prevent corrosion is painting the metal or alloying it with some other metal in an appropriate ratio.

(b) The first step is to separate dust like and other such physical impurities from the ore. The process is called concentration of the ore. Then following steps are carried out.

First of all sulfide ores are subjected ti roasting which converts them into metal oxides.

$MS + O_{2} \overset{\bigtriangleup }{\rightarrow} MO + SO_{2} \to$ Roasting (heating in excess air)

The metal oxide is then heated with carbon or coke which is a good reducing agent and converts oxide into pure metal.

$MO +C \overset{\bigtriangleup }{\rightarrow} CO_{2} + M \to$ Reduction with coke.

The pure metal is then refined to get an ultra-pure metal.

Define the following terms
Secondary batteries .

Secondary batteries are those batteries which can be recharged by passing electric current through them and hence can be used over again e.g., lead storage battery.

Why is alternating current used for measuring resistance of an electrolytic solution ?

Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.

What is the effect of catalyst on:
(a) Activation energy (Ea), and
(b) Gibbs energy $( \Delta G )$

(a) On adding catalyst in a reaction, the activation energy reduces and the rate of reaction increases.

(b) A catalyst does not alter the Gibbs energy (

$\Delta{G}$ ) of a reaction.

Fill in the blanks:
Iron benches kept in lawns and gardens get _(a)_. It is a _(b)_ change because a new _(c)_ is formed.

(a) rusted

(b) chemical

A bucket made of plastic does not rust like a bucket made of iron. Why?

Plastic does not rust after being left exposed to the air and water because it does not react with them. On the other hand, iron reacts with air and water, so it rusts. The rusting of iron bucket can be prevented by coating it with paints that do not allow air and water to come in contact with the iron.

Write overall cell reaction for lead storage battery when the battery is being charged .

$2 PbSO_4 (s) + 2H_2O (l) \rightarrow \, Pb(s) + PbO_2 (s) + 4H^+ (aq) + 2SO_{4}^{2-} (aq)$

Give reactions taking place at the two electrodes if these are made up of Ag .

$Ag$ electrode

Anode :

$Ag(s)$

$\rightarrow$

$Ag^{+} (aq) + e^{-}$

Cathode :

$Ag^{+} + e^-$

$\rightarrow$

$Ag(s)$

Why do copper objects develop a green coating in air?

When a copper vessel is exposed to moist air for a long time it develops a green layer on its surface. Copper corrodes by oxidation in which it reacts with oxygen in the air to form copper oxide.

Copper oxide then combines with carbon dioxide to make copper carbonate, which gives it a green colour. This process is called corrosion of copper.

The green material is a mixture of copper hydroxide (Cu(OH)

$_2$ ) and copper carbonate (CuCO

$_3$ ). The following is the reaction:

2Cu(s)+H

$_2$ O(g)+CO

$_2$ +O

$_2$

$→$ Cu(OH)

$_2$ +CuCO

$_3$ (s)

Copper(II) carbonate is a blue-green compound.

The chemistry of corrosion of iron is essentially an electrochemical phenomenon . Explain the reactions occurring during the corrosion of iron in the atmosphere .

According to electrochemical theory of rusting the impure iron surface behaves like small electrochemical cell in presence of water containing dissolves oxygen or carbon dioxide . In this cell pure iron acts as anode and impure iron surface acts as cathode. Moisture having dissolved

$CO_2$ or

$O_2$ acts as an electrolyte . The reactions are given below.

At anode :

$Fe \rightarrow \, Fe^{2+} ; R^{\circ}_{Fe^{2+} / Fe} = -0.44\,V$

At cathode :

$2H^+ + \dfrac{1}{2} O_2 + 2e^- \,\, \rightarrow \,\, H_2O ; E^{\circ}_{H^+ / O_2 / H_2O} = 1.23 \,V$

Overall reaction:

$Fe + 2H^+ + \dfrac{1}{2} O_2 \, \rightarrow \, Fe^{2+} + H_2O ; E^{\circ}_{cell} = 1.67\,V$

The

$Fe^{2+}$ ions are further oxidized by atmospheric oxygen to

$Fe^{3+}$ ions , which comes out in the form of hydrated oxide (rust).

$2Fe^{2+} + \dfrac{1}{2} O_2 + 2H_2O \rightarrow Fe_2O_3 + 4H^+$

$Fe_2O_3 + {x}H_2O \rightarrow \, Fe_2O_3. {x}H_2O$ (rust)

An electric current is passed through a conducting solution. List any three possible observations.

Bubbles of gas may be formed on the electrodes. Deposits of metal may be seen on electrodes. Change in the colour of the solution may take place. The solution may get heated.

Name the constituents of a cell.

A cell constituents of the following:

$\cdot$ Two electrodes

$\cdot$ An electrolyte in a vessel

Corrosion can be advantageous in some cases. Explain.

Corrosion of metals is an advantage as it prevents the metal underneath from further damage.
For example: On exposure to air, the surface of metal like aluminium and zinc forms layers of their oxides which are very sticky and impervious in nature and hence act as protective layer.
This layer protects the metal from further damage.

Which metals do not corrode easily?

The noble metals such as gold and platinum do not corrode easily.

Explain how the activity series accounts for the following:
Tendency to corrode

The metals lying above the hydrogen in activity series can easily react with moisture and air and corrode easily whereas the metals such as gold and platinum do not corrode easily.

State under what conditions corrosion is faster.

Conditions for increase of rate of corrosion are:
1. Presence of oxygen and moisture.
2. Metals which are placed higher in activity series corrode more easily.
3. Dissolved salts in water act as electrolyte and enhance the rate of corrosion.
4. The presence of pollutants like

$NO_{2}$ and

$CO_{2}$ increases rusting.

What do you observe, when Lead nitrate is heated?

$\text{heating Lead nitrate}$ it decomposes to give a yellow residue of Lead oxide and brown Nitrogen dioxide with evolution of Oxygen gas.

Cell 'A' has $E_{Cell} = 2V$ and cell 'B' has $E_{Cell} = 1.1V$ . Which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell?

Cell 'B' will act as electrolytic cell as it has lower emf.

$\therefore$ The electrode reactions will be:

$Zn^{2+} + 2e^- \rightarrow Zn$ at cathode

$Cu \rightarrow Cu^{2+} + 2e^-$ at anode

Represent an electrochemical cell showing indirect redox reaction between $Zn$ and $ZnSO_{4}+Cu$ solution

$Zn+CuSO_4 \to ZnSO_4 +Cu$
The electrochemical cell of this reaction is represented as follows:
Anodic representation

$-Zn / Zn^{2}(1M)$ , which is written on the left side.
Cathodic representation

$-Cu^{2+}(1M)/Cu$ , which is written on the right sie.
Salt bridge is represented by two line between the two half reactions.

$Zn / Zn^{2+}(1M) \ || \ Cu^{2+}(1M) /Cu$

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Metals in that electrode in the reactivity series is (in the Top,Bottom)

In the top
1818594_1834852_ans_8c4ade02184c48929de81f558ea37677.png

What is the reason for this?

When the

$Zn$ rod is dipped in

$CuSO_{4}$ solution, the

$Cu^{2+}$ ions in the solution get deposited at the

$Zn$ rod as

$Cu$ atoms.

What is the reason for the change in intensity of the colour of $CuSO_{4}$ solution?

The blue colour of

$CuSO_{4}$ solution is due to the presence of

$Cu^{2+}$ ions. The change in intensity of the colour of

$CuSO_{4}$ solution because when the

$Zn$ rod is dipped in

$CuSO_{4}$ solution, the

$Cu^{2+}$ ions in the solution get, deposited at the

$Zn$ rod as

$Cu$ atoms.

Sketch the cell constructed.

1815938_1834714_ans_79c810a2816d485db68a6adedca06517.png

Is this reaction oxidation or reduction? Why?

Oxidation. Because the losing of electrons is called oxidation.

What is the name of this reaction? Why?

Reduction. The gaining of electrons is called reduction.

$Zn (s)+2AgNO_3(aq)\to Zn(NO_3)_2 (aq)+2Ag(s)$
Write the chemical equation for oxidation and reduction.

Oxidation:

$Zn^o(s) \to Zn^{2+}(aq)+2e^-$

(Oxidation means losing of electrons)

Reduction:

$Ag^{1+}(aq)+1e^{-}\to Ag(s)$

(Reduction means gaining of electrons)

Complete the equation given below.

$Cu\rightarrow Cu^{2+}+2e^{-}$

Analyse the reactions and answer the following questions:
Explain the oxidation and reductions taking place here including the chemical equation?

Chemical equation: (including oxidation state)

$Mg^o (s) +Fe^{2+}SO_4^{2-}(aq)\to Mg^{2+}SO^{2-}_4 (aq) +Fe^{o}(s)$

$Anode: Mg^{o}(s) \to Mg^{2+}(aq)+2e^-$ (oxidation)

$Cathode, Fe^{2+}(aq)+2e^{-}\to Fe^{o}(s)$ (reduction)

What happened to the $Zn$ rod?

Before the experiment the

$Zn$ rod was colourless. After the experiment

$Zn$ rod became blue due to the deposition of copper.

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Direction of the flow of electron

Zn -Ag cell is also possible.

From Anode to Cathode
1818596_1834856_ans_24ca3968f88c424fbba8468566c8d96c.png

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Write the down the balanced equation taking place in both electrodes.
Galvanic Cell Electrode which Gives Electron Electrode which accept Electron

Zn-Ag cell is also possible.

Galvanic cell	Electrode which gives Electrons	Electrode which accepts Electrons
$Zn-Cu$	$An \to Zn +2e^-$	$Cu^{2+}+2e^- \to Cu$
$Cu-Ag$	$Cu\to Cu_3 +2e^-$	$Ag^{2+}+2e^- \to Ag$
$An-Ag$	$An\to Zn_2 +2e^-$	$Ag^{2+}+2e^- \to Ag$

1818597_1834857_ans_f497b854f3fe4d1b8e8a0734c8f621cb.png

Name of some scientists and their contribution are given in the following table. Match them suitably in the chronological order.

Scientist Contribution

John Dalton ----- Atomic theory

Michael Faraday ----- Law of Electrolysis

J.J. Thomson -------- Discovered electron

Rutherford --------- Planetary Model Of Atom

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride

(a) Process of accepting electrons is called

Process of accepting electrons is called as

$Reduction$

$Ag^+ +1e^-\rightarrow Ag(s)$

1755700_1834855_ans_cc9f4096e9a54b00bfb84eb3ccf62771.png

How electrical energy is produced by batteries?

$Tn$ a battery, electrical energy is produced by the chemical reaction between zinc and sulphuric acid or by zinc and carbon. Electrical energy is produced by chemical reactions. Here chemical energy is converted to electrical energy.

Iron is a metal which corrodes fast.
a. What are the factors that favour the corrosion of iron?
b. In coastal regions, copper nails are preferred to iron nails. What could be the reason?
c. Can you suggest some measure to prevent the corrosion of iron?

a. Air and water vapour in the atmosphere
b. Iron rod corrodes quickly in salt water
c. Painting on the metal, electroplating with non corrosive metals.

Complete the figure given below:

1799959_1850060_ans_d409f5d47a5342409c587aaf67f9e76f.png

What are the components affecting rusting of iron?

Components affecting rusting of iron:

Air
moisture present in atmosphere.

The standard reduction potential data at $25^{\mathrm{o}}\mathrm{C}$ is given below:

$\mathrm{E}^{\mathrm{o}} (\mathrm{F}\mathrm{e}^{3+}, \mathrm{F}\mathrm{e}^{2+})=+0.77\mathrm{V}$ ;
$\mathrm{E}^{\mathrm{o}} (\mathrm{F}\mathrm{e}^{2+}, \mathrm{F}\mathrm{e})=-0.44\mathrm{V}$
$\mathrm{E}^{\mathrm{o}} (\mathrm{C}\mathrm{u}^{2+}, \mathrm{C}\mathrm{u})=+0.34\mathrm{V}$
$\mathrm{E}^{\mathrm{o}} (\mathrm{C}\mathrm{u}^{+}, \mathrm{C}\mathrm{u})=+0.52\mathrm{V}$
$\mathrm{E}^{\mathrm{o}}[\mathrm{O}_{2}(\mathrm{g})+4\mathrm{H}^{+}+4\mathrm{e}^{-}\rightarrow 2\mathrm{H}_{2}\mathrm{0}]=+1.23\mathrm{V}$ ;
$\mathrm{E}^{\mathrm{o}}[\mathrm{O}_{2}(\mathrm{g})+2\mathrm{H}_{2}\mathrm{O}+4\mathrm{e}^{-}\rightarrow 4\mathrm{0}\mathrm{H}^{-}]=+0.40\mathrm{V}$
$\mathrm{E}^{\mathrm{o}} (\mathrm{C}\mathrm{r}^{3+}, \mathrm{C}\mathrm{r})=-0.74\mathrm{V}$ ;
$\mathrm{E}^{\mathrm{o}} (\mathrm{C}\mathrm{r}^{2+}, \mathrm{C}\mathrm{r})=-0.91\mathrm{V}$
Match $\mathrm{E}^{0}$ of the redox pair in List 1 with the values given in List 2 and select the correct answer using the code given below the lists.

$(A)$

$\Delta \mathrm{G}_{\mathrm{F}\mathrm{e}^{3+/Fe}}^{\mathrm{o}}=\Delta \mathrm{G}_{\mathrm{F}\mathrm{e}^{+3/\mathrm{F}\mathrm{e}^{2+}}}^{\mathrm{o}}+\Delta \mathrm{G}_{\mathrm{F}\mathrm{e}^{2+/\mathrm{F}\mathrm{e}}}^{\mathrm{o}}$

$\Rightarrow -3\times \mathrm{F}\mathrm{E}_{(\mathrm{F}\mathrm{e}^{+3}/\mathrm{F}\mathrm{e})}^{\mathrm{o}}=-1\times \mathrm{F}\mathrm{E}_{(\mathrm{F}\mathrm{e}^{+3}/\mathrm{F}\mathrm{e}^{+2})}^{\mathrm{o}}+(-2\times \mathrm{F}\mathrm{E}_{\mathrm{F}\mathrm{e}^{+2}/\mathrm{F}\mathrm{e}}^{\mathrm{o}})$

$\Rightarrow \mathrm{E}_{\mathrm{F}\mathrm{e}^{+3}/\mathrm{F}\mathrm{e}}^{\mathrm{o}}=-0.04\mathrm{V}$

$(B)$

$\mathrm{O}_{2}(\mathrm{g})+2\mathrm{H}_{2}\mathrm{O}+4\mathrm{e}^{-}\rightarrow4OH^- \Rightarrow \mathrm{E}^{\mathrm{o}}=0.40\mathrm{V}$ .........

$\mathrm{i}$ )

$2\mathrm{H}_{2}\mathrm{O}\rightarrow \mathrm{O}_{2}(\mathrm{g})+4\mathrm{H}^{+}+4\mathrm{e}^{-} \Rightarrow \mathrm{E}^{\mathrm{o}}=-1.23\mathrm{V}$ .....(

$\mathrm{i}\mathrm{i}$ )

So,

$4\mathrm{H}_{2}\mathrm{O}\rightleftharpoons 4\mathrm{H}^{+}+4\mathrm{O}\mathrm{H}^-$ ........ (

$iii$ )

$\mathrm{E}^{\mathrm{o}}$ for

$III^{rd}$ reduction

$=0.40-1.23=-0.83\mathrm{V}$ .

$(C)$

$\Delta \mathrm{G}^{\mathrm{o}}_{(Cu^{+2}/Cu)}=\Delta \mathrm{G}^{\mathrm{o}}_{(Cu^{+2}/Cu^{+})}+\Delta \mathrm{G}^{\mathrm{o}}_{(Cu^{+}/Cu)}$

$-2\times \mathrm{F}\mathrm{E}_{\mathrm{C}\mathrm{u}^{+2}/}^{\mathrm{o}} \mathrm{C}\mathrm{u} =-1\times \mathrm{F}\mathrm{E}_{\mathrm{C}\mathrm{u}^{+2}/}^{\mathrm{o}} \mathrm{C}\mathrm{u} ++(-1\times \mathrm{F}\times \mathrm{E}_{\mathrm{C}\mathrm{u}^{+}/Cu} ^{o} )$

$\Rightarrow \mathrm{E}_{\mathrm{C}\mathrm{u}^{+2}/\mathrm{C}\mathrm{u}}^{\mathrm{o}}=-0.18\mathrm{V}$

$(D)$

$\Delta \mathrm{G}_{\mathrm{C}\mathrm{r}^{+3}/\mathrm{C}\mathrm{r}^{+2}}^{\mathrm{o}}=\Delta \mathrm{G}_{\mathrm{C}\mathrm{r}^{+3}/\mathrm{C}\mathrm{r}}^{\mathrm{o}}+\Delta \mathrm{G}_{\mathrm{C}\mathrm{r}/\mathrm{C}\mathrm{r}^{+2}}^{\mathrm{o}}$

$-1\times \mathrm{F}\times \mathrm{E}_{\mathrm{C}\mathrm{r}^{+3}/Cr^{+2}}^{o} =-3\times \mathrm{F}\times \mathrm{E}_{\mathrm{C}\mathrm{r}^{+3}/}^{\mathrm{o}} \mathrm{C}\mathrm{r} +(-2\times \mathrm{F}\times \mathrm{E}_{\mathrm{C}\mathrm{r}/Cr^{+2}}^{\mathrm{o}} )$

$\Rightarrow \mathrm{E}_{\mathrm{C}\mathrm{r}^{+3}/\mathrm{C}\mathrm{r}^{+2}}^{\mathrm{o}}=-0.4\mathrm{V}$ .

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List The variation in conductivity of these reactions is given in List Match List 1 with List 2 and select the correct answer using the code given below the lists:

$(A)$

$(C_{2}\underset{X}{H_{5}})_{3}N+CH_{3} \underset{Y}{CO}OH \rightarrow (\mathrm{C}_{2}\mathrm{H}_{5})_{3}\mathrm{N}\mathrm{H}^{+}\mathrm{C}\mathrm{H}_{3}\mathrm{C}\mathrm{O}\mathrm{O}^{-}$
Initially, conductivity increases due to ion formation after that it becomes practically constant because X alone can not form ions. Hence, (3) is the correct match.

$(B)$

$\mathrm{K}\mathrm{I}(\underset{X}{0.1}\mathrm{M})+\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}_{3}\underset{Y}{(}0.01\mathrm{M})\rightarrow \mathrm{A}\mathrm{g}\mathrm{I}\downarrow+ KNO_{3}$
Number of ions in the solution remains constant until all the

$AgNO_{3}$ precipitated as

$AgI$ . Thereafter, conductance increases due to increases in number of ions. Hence, (4) is the correct match.

$(C)$ Initially, conductance decreases due to the decrease in the number of

$OH^{-}$ ions thereafter, it slowly increases due to the increases in number of

$H^{+}$ ions. Hence, (2) is the correct match.

$(D)$ Initially it decreases due to decrease in

$H^{+}$ ions and then increases due to the increases in

$OH^{-}$ ions. Hence, (1) is the correct match.

Match the columns.

A. The dry cell consists of a paste of manganese dioxide, ammonium chloride and carbon.
B. The lead acid storage battery is formed by dipping lead peroxide plate by dipping in sulfuric acid.
C. Zinc chloride is a strong electrolyte.
D. The secondary battery is a rechargeable battery which can be charged and recharged many times.

Calculate the value of $E^o$ (in V, upto two decimal points) from the given figure and report your answer as $100\times E^o$ .

$ClO_{3}^{-}+2H_{2}O+4e^-\rightarrow ClO^{-}+4OH^{-};\Delta G_{1}^{0}$

$ClO^{-}+H_{2}O+e^-\rightarrow \frac{1}{2}Cl_{2}+2OH^{-};\Delta G_{2}^{0}$

$\frac{1}{2}Cl_{2}+e^-\rightarrow Cl^{-};\Delta G_{3}^{0}$

From 1, 2 and 3 reactions we get

$ClO_{3}^{-}+3H_{2}O+6e^-\rightarrow Cl^{-}+6OH^{-};\Delta G^{0}$

$\therefore \Delta G^{0}=\Delta G_{1}^{0}+\Delta G_{2}^{0}+\Delta G_{3}^{0}$

$-6FE^{0}=-4F\times0.54-F\times0.45-F\times1.07$

$\therefore E^{0}=+\frac{3.68}{6}=+0.61 V$

$100\times E^o=61V$ .

The redox reaction involving the reducing power of hydrogen sulphide is:
$\quad S + 2H^+ + 2e^- \rightarrow H_2S, \quad E^{\small\circ}_{S/H_2S} = + 0.14V$
Two other half equations are:
$\quad Fe^{3+} + e^- \rightarrow Fe^{2+} \quad E^{\small\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 V$
$\quad Br_2 + 2e^- \rightarrow 2Br^- \quad E^{\small\circ}_{Br_2/Br^-} = 1.07 V$
Under standard conditions, hydrogen sulphide reacts with iron $(III)$ ions.
If this statement is true enter 1, else enter 0.

Yes,

$Fe^{3+}$ ions will oxidise

$H_2S$ to

$S$

$\quad 2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S$
Since ,

$\quad E^{\small\circ} = 0.77 - 0.14 = 0.63 V$
and thus

$\quad \triangle G^{\small\circ} = -nFE^{\small\circ}_{cell} = -ve$

Note: A cell reaction is spontaneous
if

$E^{\small\circ}_{cell} = +ve$ and

$\triangle G^{\small\circ} = -ve$

$\quad E^{\small\circ}_{cell} = E^{\small\circ}_{ox} + E^{\small\circ}_{red}$
(take values as per half-reactions)

How many faraday's are required for the reduction of $1\space mol\space C_6H_5NO_2$ into $C_6H_5NH_2$ ?

The oxidation number of

$N$ in nitrobenzene is

$+3$ .The oxidation number of

$N$ in aniline is

$-3$ . Thus, the change in the oxidation number is

$6$ .

Thus, the reduction of

$1$ mole of nitrobenzene to

$1$ mole of aniline requires

$6$ moles of electrons.

Hence, six faradays are required for the reduction of

$1\space mol\space C_6H_5NO_2$ into

$C_6H_5NH_2$ .

The correct answer is

$6$ .

E.M.F. diagram for some ions is given as:

$\displaystyle FeO_4^{2-} \xrightarrow{E^o = +2.20 V} Fe^{3+} \xrightarrow{E^o= 0.77 V} Fe^{2+} Fe^{2+} \xrightarrow{E^o= 0.445 V} Fe^0$

The value of $\displaystyle E^o_{FeO_4^{2-}/ Fe^{2+}}$ is _________(as nearest integer).

Given :

$Fe^{6+} + 3e \rightarrow Fe^{3+} ;$

$-\Delta G_1^o = 3 \times 2.20 \times F$ ...... (i)

$Fe^{3+} + e \rightarrow Fe^{2+} ;$

$-\Delta G_2^o = 1 \times 0.77 \times F$ ...... (ii)

$Fe^{2+} + 2e \rightarrow Fe^{0} ;$

$-\Delta G_3^o = 2 \times (-0.445) \times F$ ...... (iii)

By equation (i) and (ii)

$Fe^{6+} + 4e \rightarrow Fe^{2+} ;$

$-\Delta G_4^o = - \Delta G_1^o - \Delta G_2^o$

$\because 4 \times E_4^o \times F = 3 \times 2.20 \times F + (1 \times 0.77 \times F)$

$\therefore E_4^o = 1.84 V$

$Fe^{6+} + 4e \rightarrow Fe^{2+};$

$E^o = 1.84\approx 2 V$

In the manufacture of $Al,\ Al$ $_2$ O $_3$ is dissolved in $Na$ $_3$ $AlF$ $_6$ at $300 K$ and electrolysed between $Al$ and carbon electrodes following the net reaction,
$2 Al_2O_3 (solution) + 3C \rightarrow 4 Al(l) + 3 CO_2 (g)$
The minimum voltage required between the electrodes if the Gibbs energy change for the above reaction is $-1370$ $kJ mol$ $^{-1}$ is________. (write nearest integer)

The reduction half reaction is

$2Al_2O_3(aq) + 12e^- \rightarrow 4Al(l) + 6O^{2-}(aq)$
The oxidation half reaction is

$3C(aq)+6O^{2-} (aq)\rightarrow 3CO_2(g)+12e^-$

$\Delta G^o =-nFE^o$

$-1370000=12 \times 96500 \times E^o$

$E^o=1.18 \: V$

The standard electrode potential in V for the electrode $MnO^-_4 / MnO_2$ in solution is: (round off the answer to the nearest integer)
Given: $E^o_{MnO^-_4 / Mn^{2+}} = 1.51 V$ and $E^o_{MnO_2 / Mn^{2+}} = 1.23 V$

$Mn{O}^-_4 +8H^- + 5e \rightarrow Mn^{2+} + 4 H_2O; E^o_1 = 1.51 V$ ........... (i)

$\therefore - \Delta G^o_1 = 5\times 1.51 \times F$

$= 7.55 F$

$MnO_2 + 4H^+ + 2e \rightarrow Mn^{2+} + 2H_2O; E^o_2 = 1.23 V$ ........... (ii)

$\therefore - \Delta G^o_2 = 2 \times 1.23 \times F$

$= 2.46 F$

Subtracting equation (ii) from (i)

$Mn{O}^-_4 + 4H^+ + 3e \rightarrow 2H_2O + MnO_2; E^o_3 = ?$
or

$- \Delta G^o_3 = n_3 E^o_3 F$

$\therefore - \Delta G^o_3 = - \Delta G^o _1 - (- \Delta G^o_2)$

$3E^o_3 F = 7.55 F - 2.46 F$

$\therefore E^o_3 = \displaystyle \frac{5.09}{3} = 1.70 \space volt$

How much current is required to produce $Cl_{2}(g)$ at the rate of $10^{-3}\, L\, s^{-1}$ electrolysis of molten KCl? (in A)

Rate of deposition =

$10^{-3}\, L\, s^{-1}\, =\, \displaystyle \dfrac{10^{-3}}{22.4}{mols^{-1}}$

$2Cl^{\ominus}\, \rightarrow\, Cl_{2}\, +\, 2e^{-}$

$2\, F\, =\, 1\, mol\, Cl_{2}$

$\Rightarrow\, \displaystyle \dfrac{10^{-13}}{22.4}{mols^{-1}\, Cl_{2}}\, =\, \dfrac{2\, \times\, 10^{-3}}{22.4}\, F\,s^{-1}$

$\Rightarrow\, \displaystyle \dfrac{I\, \times 1s}{96500}\, =\, \dfrac{2\, \times\, 10^{-3}}{22.4}\, \Rightarrow\, I\, =\, 8.616\, A$

$m Eq\, of\, K\, deposited\, =\, Number\, of\, Faradays\, \times\, 10^{3}$

$=\, \displaystyle\, \dfrac{2\, \times\, 10^{-3}}{22.4}\, \times\, 10^{3}\, =\, 0.0892$

All the energy released from the reaction $\displaystyle X\rightarrow Y,\ \Delta_{r} G^{\circ}=-193\ kJ . mol^{-1}$ is used for oxidizing $\displaystyle M^{+}$ as $\displaystyle M^{+}\rightarrow M^{3+}+2e^{-}, E^{\circ}=-0.25V.$
Under standard conditions, the number of moles of $\displaystyle M^{+}$ oxidized when one mole of X is converted to Y is:
$\displaystyle \left [ F=96500\ C mol^{-1} \right ]$

For the reaction

$\displaystyle M^{+}\rightarrow M^{3+}+2e^{-},$ , the standard Gibbs free energy change is

$\Delta G^0 = -nFE^0 = -2 \times 96500 \times (-0.25) = 48250 \: J/mol = 48.25 \: kJ/mol$

The number of moles of

$M^+$ oxidized (when 1 mol of X is converted to Y) is

$\dfrac {193 \: kJ/mol}{48.25 \: kJ/mol} = 4$

The resistance of a conductivity cell containing $0.001M\ KCl$ solution at $298\ K$ is $1500 \Omega$ . What is the cell constant if conductivity of $0.001M\ KCl$ solution at $298\ K$ is $0.146 \times 10^{-3} S\ cm^{-1}$ ?

Resistance

$\displaystyle = 1500 \: ohm$

Conductance is the reciprocal of resistance.
Conductance

$\displaystyle = \dfrac {1}{1500} \: S$

Cell constant is the ratio of conductivity to conductance.
Cell constant

$\displaystyle = \dfrac {0.146 \times 10^{-3} \: S/cm}{\dfrac {1}{1500} S} = 0.219 \: /cm$

Depict the galvanic cell in which the reaction $Zn(s) + 2Ag^{+} (aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)$ takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

The galvanic cell in which the given reaction takes place is depicted as:

$\displaystyle Zn(s)| Zn^{2+} (aq) || Ag^+ (aq) |Ag(s)$

(i) The negatively charged electrode is the zinc electrode. It acts as an anode.

(ii) In the external circuit, the current will flow from silver to zinc.

(iii) Oxidation at anode:

$\displaystyle Zn(s) \rightarrow Zn^{2+}(aq) +2e^-$

Reduction at cathode:

$\displaystyle Ag^+(aq) + e^- \to Ag(s)$

Using the standard electrode potentials given in Table $3.1$ , predict if the reaction between the following is feasible:
(i) $Fe^{3+}(aq)$ and $I^{-}(aq)$
(ii) $Ag^{+} (aq)$ and $Cu(s)$
(iii) $Fe^{3+} (aq)$ and $Br^{-} (aq)$
(iv) $Ag(s)$ and $Fe^{3+} (aq)$
(v) $Br_{2} (aq)$ and $Fe^{2+} (aq)$

(i) When emf is positive, the reaction is feasible.

$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{I_2} \\ = 0.77-0.54 = 0.23 \: V$
Reaction is feasible.

(ii)

$\displaystyle E^o_{cell} = E^o_{Ag} -E^o_{Cu} \\ = 0.80 - 0.34 = 0.46 \: V$
Reaction is feasible.

(iii)

$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Br_2} \\ = 0.77-1.09=-0.32 \ V$
Reaction is not feasible.

(iv)

$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Ag} \\ = 0.77-0.80 = -0.03 \ V$
Reaction is not feasible.

(v)

$\displaystyle E^o_{cell} = E^o_{Br_2} - E^0_{Fe} \\ = 1.09-0.77=0.32\ V$
Reaction is feasible.

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.
(ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes.
(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.
(iv) An aqueous solution of $CuCl_{2}$ with platinum electrodes

(i) An aqueous solution of

$AgNO_{3}$ with silver electrodes.

At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Silver anode will dissolve to form silver ions in the solution.

$\displaystyle Ag \rightarrow Ag^+ + e^-$

(ii) An aqueous solution of

$AgNO_{3}$ with platinum electrodes.

At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.

$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$

(iii) A dilute solution of

$H_{2}SO_{4}$ with platinum electrodes.

At cathode:

$\displaystyle 2H^+ + 2e^- \rightarrow H_2(g)$

At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.

$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$

(iv) An aqueous solution of

$CuCl_{2}$ with platinum electrodes

At cathode: Cupric ions will be reduced in preference to protons

$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu$

At anode: Chloride ions will be oxidized in preference to hydroxide ions

$\displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-$

Arrange the following metals in the order in which they displace each other from the solution of their salts.
$Al, Cu, Fe, Mg$ and $Zn$

The order in which the metals displace each other from the solution of their salts is

$Mg, Al, Zn, Fe, Cu$ . Thus, more electropositive

$Mg$ can displace less electropositive

$Cu$ .

Draw the diagram of an electrolytic cell of copper chloride solution.

Iron rod is immersed in $KClKCl$ solution such that half its length is exposed to air and the other half immersed in $KClKCl$ solution. The part corroded faster is

The part of the rod that is dipped in KCLKCl solution is corroded faster as KCL is corrosive in nature

Give reasons:
On the basis of $E^o$ values, $O_2$ gas should be liberated at anode but it is $Cl_2$ gas which is liberated in the electrolysis of aqueous NaCl.

It is due to the phenomenon of over-voltage. Experimentally, it is observed that the actual voltage required for electrolysis is greater than that calculated from standard potentials. This addition voltage required is called over-voltage. It is required because the rate of transfer of electrons at the interface of electrode and solutions for both half reactions is slow. The over-voltage required for formation of oxygen is much larger than that required for the formation of chlorine. Hence, chlorine gas which is liberated in the electrolysis of aqueous NaCl.

$V_1$ L of solution A (resistance 500ohm) is mixed with $V_2$ L of solution B (Resistance 100ohm) resistance of final solution is 80 then $V_2$ / $V_1$ :

Since both solution has different density so they separated parallely.

$\dfrac{V_{1}}{R_{1}}+\dfrac{V_{2}}{R_{2}}=\dfrac{V_{1}+V_{2}}{Reg.}$

$\dfrac{V_{1}}{50}+\dfrac{V_{2}}{100}=\dfrac{V_{1}+V_{2}}{80}$

$80(100V_{1}+50 V_{2})=5000 V_{1}+5000V_{2}$

$8000V_{1}+4000V_{2}=5000V_{1}+5000V_{2}$

$(8000-5000)V_{1}=(5000-1000)V_{2}$

$3V_{1}=V_{2}$

$\dfrac{V_{2}}{V_{1}}=3$

How much $O_2$ gas will be collected at the anode at $300$ K temperature and $1$ bas pressure if $2.5$ ampere electric current is passed for one hour in electrolysis of aqueous solution of $Na_2SO_4$ . (F $=96500$ Coulomb) [ $1$ mole gas volume is $22.4$ litre at STP].

No of coulombs = current in amperes

$\times$ time in second

no of coulombs =

$0.50\times30\times900$

the first release of oxygen from water molecule

$2H_2O (l)\rightarrow O_2 + 4H^+(aq)+ 4e^-$

the alternative way of release of oxygen from water molecule

$4OH^- (aq)\rightarrow2H_2O(l)+ O_2(g)+ 4e^-$

release of one mole of

$O_2$ involve 4 mole of electron

$4\times96500$ coulomb gives

$22.4dm^3$ O_2 $$at STP

volume of oxygen

$= 900\times22.4/4\times96500$

$= 0.052$

How will you show that air and moisture both are required for rusting of iron?

oxygen is a very good oxidising agent,

so,

$Fe \rightarrow Fe^{2+}$ +

$2e^-$

$O_2 + H_2O + 4H^+ \rightarrow 4(OH)^-$

$Fe^{2+} + 2(OH)^- \rightarrow Fe(OH)_2$

$Fe^{2+} + 2H_2O \rightarrow Fe(OH)_2 + 2H^+$

hence, we can say that air and moisture both are required for rusting of iron.

${ Cu }^{ 2+ }+{ e }^{ - }\longrightarrow Cu^{ + }\quad { E }^{ o }=0.15\ V$
Calculate ${ E }^{ o }$ of the half cell reaction Also predict whether ${ Cu }^{ + }$ undergoes disproportionation or not.

Addiing the

$\Delta G$ of the reactions

$nF{ E }^{ 0 }=nF{ E }_{ 1 }+nF{ E }_{ 2 }$

$2\times F\times 0.34=1\times F\times 0.15+1\times F\times { E }_{ 2 }$

$0.68=0.15+{ E }_{ 2 }$

${ E }_{ 2 }=0.53V$

1030327_1072582_ans_3b7a4044004942b69c5cde4d7bf2ef35.jpg

Why do silver articles become black after sometime when exposed to air?

The silver article becomes black after sometimes when exposed to air because silver reacts with sulphur which is present in the atmosphere and then forms silver sulphide. Thus the layer of this silver sulphide is formed on the surface of the silver article so they appear as dull or black.

The voltage of a cell whose half cell reactions are given below is:
${Mg}^{2+}+2{e}^{-}\rightarrow Mg(s);E=-2.37V$
${Cu}^{2+}+2{e}^{-}\rightarrow Cu(s); E=+0.34V$

1341987_1137911_ans_38ecfcf6ce7d4806a4adc6f3a7d7d212.jpg

Can absolute electrode potential of an electrode is measured ?

No, the difference in potential between two electrodes can be measured only. As the oxidation or reduction cannot occur alone. So, standard electrode is taken to measure electrode potential of any electrode.

$e g-Z n^{2+}(a q)+2 e^{-} \rightarrow Z_{n}(s)$

$H^{+}(a q)+2 e^{-} \rightleftharpoons H_{2}(g)$

Standard electrode = Hydrogen electrode

$E^{\circ}H^{+} / H_{2}=E_{H_{2}} / H^{+}=0$

Ammonium perchlorate ${NH}_{4}{ClO}_{4}$ , used in the solid fuel in the booster rockets on the space shuttle , is prepared of sodium perchlorate, ${NaClO}_{4}$ which is produced commerically by the electrolysis of a hot, stirred solution of sodium chloride. How many Faradays are required to produce. $1.0kg$ of sodium perchlorate ?
$NaCl+4{H}_{2}O\longrightarrow {NaClO}_{4}+{4H}_{2}$

no. of eq. = no. of F

$= \dfrac{1000}{12.5} \times 8 \Rightarrow 65.3$ {Answer}

Consider a cell given below:
$Cu| Cu^{2+} || Cl^- | Cl_2, Pt$
Write the reactions that occur at anode and cathode.

1665757_1264797_ans_e7a280efe641483db9ca6d2bc287f8bc.jpg

Give reasons for the following :
In the preparation of hydrogen by electrolysis of water - the distilled water used is acidified.

During electrolysis of acidified water the ions in the water increases the conductivity of water due to which t the decomposition of water starts. The acid generates H+ ions and OH- ions with hydrogen getting formed at the cathode on reaction of H+ ions.

* Reaction in Fuel cell
(a) At cathode :
$O_{2(g)}+2H_2O(I)+4e^- \rightarrow 4OH^-_{(aq)}$
(b) Reaction at anode :
$2H_{2(g)}+4OH^-_{(aq)} \rightarrow 4H_2O_{(I)}+4e^-$

What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

1871109_1421453_ans_0cf7f14205cb4bcab052eac13bf11052.jpg

Why does the conductivity of a solution decreases with dilution?

Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Hence the conductivity decreases.

Find the charge required to flow through an electrolyte to liberate one atom of (a) a monovalent material and (b) a divalent material.

$a)Monovalent \ material$

$1$ eq. mass of the substance requires

$96500$ coulombs

Since the element is monatomic, thus equivalent mass = molecular mass

$6.023\times 10^{23}$ atoms require

$96500C$

$1$ atoms require

$=\dfrac { 96500 }{ 6.023\times 10^{23} } C=1.6\times 10^{-19}C$

$(b)Divalent \ material$

Since the element is diatomic, thus equivalent mass

$=\dfrac{molecular \ mass}{2}$

$=\dfrac{1\times 6.023\times 10^{23}}{2}$ atoms require

$96500C$

$1$ atoms require

$=\dfrac { 96500\times 2 }{ 6.023\times 10^{23} } C=3.2\times 10^{-19}C$

The electrode reaction for charging of a lead storage battery are:
$PbSO_{4}+2e\rightarrow Pb+SO_{4}^{2-}$
$PbSO_{4}+2H_{2}O\rightarrow PbO_{2}+SO_{4}^{2-}+4H^{+}+2e$
The electrolyte in the battery is an aqueous solution of sulphuric acid. Before charging, the specific gravity of the liquid was found to be 1.10 (16% $H_{2}SO_{4}$ by wt.) After charging for $\dfrac{965}{9}$ h, the specific gravity of the liquid was found to be 1.42 (40% $H_{2}SO_{4}$ by weight). If the battery contained 2 L of the liquid and the volume remains constant during charge, the average current (in A) used charging the battery is

1968908_1681633_ans_45ec90049bf8442ba0d1b1391b9244fe.jpg

Calculate the reduction potentials for the following half cells:
(i) $Ag|{ Ag }^{ + }\left( { 10 }^{ -5 }M \right) ;{ E }_{ { Ag }^{ + },Ag }^{ o }=0.80V$
(ii) $Cu|{ Cu }^{ 2+ }(0.2M);{ E }_{ { Cu }^{ 2+ },Cu }^{ o }=0.34V$

(i)

$Ag/{ Ag }^{ + }\left( { 10 }^{ -5 }M \right) ,{ E }_{ { Ag }^{ - }/Ag }^{ o }=0.80V$

${ Ag }^{ + }+{ e }^{ - }\rightarrow Ag$

${ \varepsilon }_{ { Ag }^{ + }/Ag }={ \varepsilon }_{ { Ag }^{ - }/Ag }^{ o }-\cfrac { 2.303RT }{ nF } \log { Q }$

${ \varepsilon }_{ { Ag }^{ + }/Ag }=0.80V-\cfrac { 0.0592 }{ 1 } \log { \cfrac { 1 }{ \left[ { Ag }^{ + } \right] } }$

${ \varepsilon }_{ { Ag }^{ + }/Ag }=0.80V-0.0592\log { \left( { 10 }^{ -5 } \right) } =0.80V-5\times 0.0592=0.504V\quad$

(ii)

$Cu/{ Cu }^{ 2+ }(0.2M),\quad { \varepsilon }^{ o }_{ { Cu }^{ 2+ }/Cu}=0.34V$

${ \varepsilon }_{ { Cu }^{ 2+}/Cu }={ \varepsilon }_{ { Cu }^{ 2+ }/Cu }^{ o }-\cfrac { 2.303RT }{ nF } \log { Q } =0.34V-\cfrac { 0.0592 }{ 2 } \log { \cfrac { 1 }{ \left[ { Cu }^{ 2+ } \right] } } =0.34V-\cfrac { 0.0592 }{ 2 } \times \log { 5 } =0.32V$

The water is electrolysed in a cell , hydrogen is liberated at one electrode and oxygen is simultaneously liberated at the other . in a particular experiment hydrogen and oxygen so produced were collected together and the total volume measured 16.8 mL at NTP . How many coulombs were passed through the cell in the experiment ?

$W_{H_{2}} = Z_{H_{2}} Q...(1)$

$W_{O2} =Z_{O2} Q...(2)$

$(1)\Rightarrow W_{H_2} = \dfrac{gm\;eq\; wt\; of H_{2}}{96500}\times Q\dfrac{mole \;wt}{\dfrac{2}{96500}\times Q}$

$\Rightarrow \dfrac{W_{H2}}{M_{H2}}= \dfrac{Q}{2\times 96500} \Rightarrow n_{H2}=\dfrac{Q}{96500\times 2}$

$\Rightarrow VH\times V_{H2}=\dfrac{2}{96500\times 2} \times 22400 ml$

$for O_{2}$

$WO_{2} = \dfrac{gm eg wt O_{2}}{96500}\times Q= \dfrac{mol wt O_{2}}{\dfrac{4}{96500}}\times Q$

$\Rightarrow \dfrac{W_{02}}{MO_{2}} = \dfrac{Q}{96500\times 4} \therefore no_{2} = \dfrac{Q}{96500\times 4}$

$\Rightarrow Vo_{2}=no_{2}\times 22400 ml =\dfrac{Q\times 22400}{96500\times 4}$

$A/q , VH_{2}+VO_{2} = \times \dfrac{Q\times 22400}{96500\times 2}+\dfrac{Q\times 22400}{96500\times 4}$

$16.8 = \dfrac{Q}{965} (112+56) =\dfrac{Q}{965}\times 168$

$Q = \dfrac{16.8\times 965}{16.8} = 96.5 coulombs$

From the electrochemical series given in the text, determine the approximate value of ${E}^{o}$ for ${ X }^{ 2+ }(aq)+2e\longrightarrow X(s)$
(a) The metal $X$ dissolves in nitric acid but not in hydrochloric acid. It can displace ${Ag}^{+}$ but not ${Cu}^{2+}$
(b) The metal $X$ dissolves in hydrochloric acid producing ${H}_{2}$ but does not displace either ${Zn}^{2+}$ or ${Fe}^{2+}$

(a) Since

$X$ can displace

${Ag}^{+}$ from its salt, so it has more reducing power than

$Ag$ so Oxidation potential of

$X$ will be higher than that of

$Ag$

${ \varepsilon }_{ Ag/{ Ag }^{ 2+ } }^{ 0 }<{ \varepsilon }_{ X/{ X }^{ 2+ } }^{ 0 }\quad \quad \therefore { \varepsilon }_{{ X }^{ 2+ }/X }^{ 0 }<{ \varepsilon }_{ { Ag }^{ + }/Ag }^{ 0 }\quad$

${ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }<0.8V...(1)$

Also since

$X$ cannot displace

${Cu}^{2+}$ from its salt so it has les reducing power than

$Cu$ , so less oxidation potential so high reduction potential.

$\quad { \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }>{ \varepsilon }_{ { Cu }^{ 2+ }/Cu }^{ 0 }>0.34V....(2)$

(1) and (2)

$0.34V<{ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }\quad < 0.80V$

(b)Since

$X$ cannot displace

${Zn}^{2+}$ or

${Fe}^{2+}$

So

${ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }>{ \varepsilon }_{ { Zn }^{ 2+ }/Zn }^{ 0 }>{ \varepsilon }_{ { Fe }^{ 2+ }/Fe }^{ 0 }\quad$

${ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }>-0.76V\quad and\quad -0.44V$

${ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }$ should be greater than

$-0.44V$ for only of

${Zn}^{2+}$ or

${Fe}^{2+}$ not to displace
Also

$X$ can reduce

${H}^{+}$ to

${H}_{2}$ so it has high reducing power than

${H}_{2}$

Therefore, reduction potential will be less

$\quad { \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }<0 \Rightarrow \therefore -0.44V<{ \varepsilon }_{ { X }^{ 2+ }/X }^{ 0 }\quad \quad$

Find the amount of silver liberated at cathode if $0.500 A$ of current is passed through $AgNO_3$ electrolyte for $1$ hour. Atomic weight of silver is $107.9\,g\,mol^{-1}$

Atomic weight of silvenis =

$107.9g/mol$

$I=0.500A$

The ECE of silver

Equivalent mass of silver,

$E_{Ag}=107.9$ [As Ag is monoatomic]

$Z_{Ag}=\dfrac { E }{ f } =\dfrac { 107.9 }{ 96500 } =0.001118$

$M=Zit=0.0011178\times 0.5\times 3600=2.01g$

Calculate the potential of a silver electrode in a saturated solution of $AgBr\left( { K }_{ sp }=6\times { 10 }^{ -13 } \right)$ containing, in addition, $0.1$ mole per $KBr$ .
${ E }_{ { Ag }^{ + },Ag }^{ o }=0.80volt$

$AgBr\left( { K }_{ sp }=6\times { 10 }^{ -3 } \right)$

Initially

$(0.1M)$ from

$KBr$

${eq-}^{b}$

$-S$

$S$

$0.1+S$

$Ksp=\left[ { Ag }^{ + } \right] \left[ { Br }^{ - } \right] =S(0.1+S)=6\times { 10 }^{ -13 }\Rightarrow S\times 0.1=6\times { 10 }^{ -13 }\Rightarrow S=6\times { 10 }^{ -12 }\quad$

$\therefore \left[ { Ag }^{ + } \right] =6\times { 10 }^{ -12 }M\quad \quad Q=\cfrac { 1 }{ \left[ { Ag }^{ + } \right] } =\cfrac { 1 }{ 6\times { 10 }^{ -12 } }$

$\therefore { \varepsilon }_{ { Ag }^{ + }/Ag }={ \varepsilon }_{ { Ag }^{ + }/Ag }^{ o }-\cfrac { 0.0592 }{ 1 } \log { \cfrac { 1 }{ \left[ { Ag }^{ + } \right] } } =0.80+\cfrac { 0.0592 }{ 1 } \log { (6\times { 10 }^{ -12 }) } =0.80+0.0592(-12+\log { 6 } )=0.14V$

A $Cu$ rod is dipped in a $0.1M$ $Cu{SO}_{4}$ solution. Calculate the potential of this half cell if $Cu{SO}_{4}$ undergoes 90% dissociation at this dilution at ${25}^{o}C$ .
${ E }_{ Cu|{ Cu }^{ 2+ } }^{ o }=-0.34V$

$Cu{ SO }_{ 4 }\rightleftharpoons { Cu }^{ 2+ }+{ SO }_{ 4 }^{ 2- }$

Initially:

$0.1M$

Finall:y

$0.1(1- \frac { 90 }{ 100 } )$

$0.09$

$0.1-0.09$

${ Cu }^{ 2+ }+2{ e }^{ - }\rightarrow Cu$

${ \varepsilon }_{ { Cu }^{ 2+ }/Cu }^{ o }=-\cfrac { 2.303RT }{ nF } \log { Q } =0.34V-\cfrac { 0.0592 }{ 2 } \log { \cfrac { 1 }{ \left[ { Cu }^{ 2+ } \right] } }$

${ \varepsilon }_{ { Cu }^{ 2+ }/Cu }^{ o }=0.34V-\cfrac { 0.0592 }{ 2 } \times \log { \cfrac { 1 }{ 0.09 } } =0.34V-0.031V=0.31V$

What happens when?
Concentrated solution of $NH_4HSO_4$ in $H_2SO_4$ is electrolysed using current of high density.

1567935_1726470_ans_4a6e88c60063407a9f18e68305c2921e.jpg

Calculate the electrode potential for $\underset{(1 atm)}{(Pt)|H_2}|H^+(c=0.1 M)$ .

Given:

Conc. of

$\mathrm{[H^+]}$ = 0.1 M.

$\mathrm{E^o \ for \ Hydrogen \ Electrode = 0 \ V}$

$\mathrm{P_{H_2}= 1 \ atm}$

Reaction:

$\mathrm{2e^- + 2H^+ \to H_2}$ {Reduction} ; No. of electrons (n) =2

According to Nernst Equation:

$\mathrm{E= E^o - \cfrac{0.059}{n} \,log{\cfrac{P_{H_2}}{{[H^+]}^2}}}$

$\mathrm{\implies E= 0 - \cfrac{0.059}{2} \,log{\cfrac{1}{0.01}}}$

$\mathrm{\implies E= 0 - \cfrac{0.059}{2} \times {2}}$

$\mathrm{\implies E= - {0.059} \ V}$

Define and give an example of Half-Cell.

A half cell is one of the two electrodes in a galvanic cell or simple battery.

For example, in the Zn−CuZn−Cu battery, the two half cells make an oxidizing-reducing couple.

Consider the electrochemical cell represented by
$Mg|{ Mg }^{ 2+ }\parallel { Fe }^{ 3+ }|{ Fe }^{ 2+ }$
If $150mA$ is to be drawn from this cell for a period of $20$ minutes, what is the minimum mass for the magnesium electrode?

$Mg|{ Mg }^{ 2+ }\parallel { Fe }^{ 3+ }|{ Fe }^{ 2+ }$

$I=150mA;t=20min=20\times 60sec\quad$

$Q=I.t=150\times { 10 }^{ -3 }\times 1.2\times { 10 }^{ 3 }=180coulombs$

$W=ZQ=\cfrac { gm\ eq.wt }{ 96500 } Q=\cfrac { 12 }{ 96500 } \times 180=0.0224g$

During an electrochemical experiment, $0.2773$ g of Ag was transferred from one electrode to the other electrode in a coulometer. What electric charge did pass through the circuit?

2040937_1738659_ans_b8a66b2ee55e41209f524d174f112707.jpg

Consider an electrochemical cell, $A(s)|{ A }^{ n+ }(aq,2M)\parallel { B }^{ 2n+ }(aq,1M)|B(s).$ The value of $\Delta {H}^{o}$ for the cell reaction is twice that of $\Delta {G}^{o}$ at $300K$ . If the emf of the cell is zero, the $\Delta {S}^{o}$ (in $J{ K }^{ -1 }\ { mol }^{ -1 }$ ) of the cell reaction per mole of $B$ formed at $300K$ is ________.

(Given: $\ln { (2) } =0.7,R=8.3J{ K }^{ -1 }\ { mol }^{ -1 }.\ H,\ S,\ G$ are enthalpy, entropy and Gibbs energy, resepctively)

1606737_1737828_ans_42422886102b424fb07f85aa04e6dda1.jpg

Does the physical size of a galvanic cell govern the potential that it will deliver? What does the size affect?

No, not at all, size does not affect the voltage. It affect the current at which cell work only

If in a galvanic cell, say, Daniell cell, inert platinum is used instead of a salt bridge, will the cell still produce a potential.

No. because without single salt bridge ions accumulate at electrode and so reaction stops.

What is the meaning of a positive sign for a half-cell potential?

Half- cell potential refers to the potential developed at the electrode of each half- cell in an electro-chemical cell.

Positive sign for the half cell potential (reduction potential) signifies non spontaneous reduction reaction.

Write the half reactions and number of moles of electrons involved in the oveall cell reaction for the electrochemical cell designated by
$Pt|Ag(s)|AgCl(s)|Cl^-(c=1)|Cl_2(c=1)|C$ (graphite) $|Pt$ .

Half-cell reactions:

At cathode:

$Cl_2 +2e^- \rightarrow 2Cl^-$

At anode:

$Ag(s) + Cl^-(aq.) \rightarrow AgCl(s) +e^-$

No. of moles of electrons involved in the cell reaction is 2.

Overall cell reaction:

$2Ag + Cl_2 \rightarrow 2AgCl$

Write electrode reactions taking place in Lead Acid Accumulator .

$Pb \rightarrow Pb^{2+} + 2e^-$ (at anode)

$PbO_2 + SO_{4}^{2-} + 4H^+ + 2e^- \, \rightarrow \, PbSO_4 + 2H_2O$ ( at cathode)

What does the negative sign in the expression $E^{\circ}_{Zn^{2+} / Zn} = -0.76 \, V$ mean ?

It mean that

$Zn$ is more reactive than hydrogen . When zinc electrode will be connected to SHE ,

$Zn$ will get oxidized and

$H^+$ will get reduced.

Two half-reaction of an electrochemical cell are given below :
$MnO_{4}^{-} (aq) + 8H^+ + 5e^- \rightarrow Mn^{2+} (aq) + 4H_2O (l) ; E^{\circ} = +1.51\,V$
$Sn^{2+} (aq) \rightarrow Sn^{4+}(aq) + 2e^- , E^{\circ} = +0.15\,V$
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured .

At anode :

$[Sn^{2+}(aq)$

$\rightarrow$

$Sn^{4+} (aq) + 2e^- ] \times 5$

At cathode :

$[MnO^{-}_{4} (aq) + 8H^+(aq) + 5e^-$

$\rightarrow$

$Mn^{2+} (aq) + 4H_2O(l) ] \times 2$

Cell reaction :
_____________________________________________________________________

$2MnO_{4}^{-} (aq) + 5Sn^{2+} (aq) + 16H^+ (aq) \, \, \rightarrow \,\, 2Mn^{2+} (aq) + 5Sn^{4+} (aq) + 8H_2O(l)$
______________________________________________________________________

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.51 \,V - 0.15\,V$

$= 1.36 \,V$

As cell potential is positive therefore the reaction is product favoured.

What is primary cell ? Give an example .

A primary cell is one in which the redox reaction occurs only once and the cell becomes dead after some time and cannot be used again , e.g., dry cell .

Answer the following questions
Calculate the standard free energy change for the following reaction at $25^{\circ}\,C$ .
$Au(s) + Ca^{2+} (1 \, M) \rightarrow Au^{3+}(1\,M) + Ca(s)$
$E^{o}_{Au^{3+} / Au} = + 1.50 \,V , E^{o}_{Ca^{2+} / Ca} = - 2.87\,V$
Predict whether the reaction will be spontaneous or not at $25^{\circ}\,C$ . Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent ?

$E^{o}_{cell} = E^{o}_{Ca^{2+} / Ca} - E^{o}_{Au^{3+} / Au}$

$= (-2.87 \,V) - (1.50\,V) = -4.37\,V$

$\Delta_{r}G^{o}_{cell} = -6 \times 96500 \times (-4.37\,V)$

$= + 2530.230\,kJ/mol$

Since

$\Delta_{r}G^{o}$ is positive , therefore , reaction is non-spontaneous.

$Au^{3+} / Au$ half cell will be an oxidizing agent while

$Ca^{2+} / Ca$ half cell will be a reducing agent.

In the circuit given in the above figure, Boojho observed that copper is deposited on the electrode connected to the negative terminal of the battery. Paheli tried to repeat the same experiment. But she could find only one copper plate. Therefore she took a carbon rod as negative electrode. Will copper be still deposited on the carbon rod? Explain.

Yes, copper from the copper sulphate solution will be deposited on the carbon rod. Copper from the copper plate will be dissolved into the copper sulphate solution for electroplating.

What advantage do the fuel cells have over primary and secondary batteries?

Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Answer the following questions:

Define the electro chemical cell. What happens if external potential applied becomes greater than $E^{0}_{cell}$ of electrochemical cell?

A device which is used to convert chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell.

If external potential applied becomes greater than

$E^{\circ}_{cell}$ of electrochemical cell, the reaction gets reversed and the electrochemical cell function as an electrolytic cell.

Tarnished silver contains $Ag_2S$ . can this tarnish be removed by placing tarnished silver were in an aluminium pan containing an inert electrolytic solution such as $NaCl$ ? The standard electrode potential for half reaction :
$Ag_2S (s) + 2e^- \rightarrow 2Ag(s) + S^{2-}$ is $-0.71\, V$
and for $Al^{3+} + 3e^- \rightarrow aAl (s)$ is $-1.66\,V$

$E^{o}_{cell}$ for reaction of tarnished silver ware with aluminium pan is

$(-0.71 \,V) - (-1.66\,V)$ i.e.,

$+ 0.95\,V$

Tarnished silver ware , therefore , can be cleaned by placing it in an aluminium pan a s

$E^{o}_{cell}$ is positive.

Redraw the diagram to show the direction of electron flow.

Electrons move from Zn to Ag.
1720959_1796569_ans_4af50d7bbc22483aba2b428d786f34e5.png

Answer the following questions
A current of $1.50 \,A$ was passed through an electrolytic cell containing $AgNO_3$ solution with inert electrodes . The weight of $Ag$ deposited was $1.50\,g$ . How long did the current flow ?

$Ag^+ e^- \rightarrow \,Ag$

Quantity of charge required to deposit

$108\,g$ of silver

$= 96500\,C$

$\therefore \,$ Quantity of charge required to deposit

$1.50 \,g$ of silver =

$\dfrac{96500}{108} \times 1.50 = 1340.28\,C$

$t = \dfrac{Q}{I}$

$\therefore$ Time taken

$= \dfrac{1340.28}{1.50} = 893.52\,s$

Give reasons:
Silver articles when exposed to air gradually turn blackish.

Silver article when exposed to air gradually turn blackish because silver metal react with sulphur present in atmosphere and forms a layer of silver sulphide which is of black in colour.

What are the adverse effects of corrosion?

The adverse effects of corrosion are as follows:

1) Damage to commercial airplanes that could result in possible in-flight problems.

2) Damage to oil pipelines that could cause a costly and dangerous rupture that creates significant environmental damage.

3) Damage to bridge supports that could cause a bridge failure.

4) Release of harmful pollutants from iron corrosion that contaminates the air.

5) Costs of repairing or replacing household equipment that fails.

6) Lose of efficiency.

7) Contamination of product.

8) Damage of metallic equipment.

9) Inability to use metallic materials.

10) Lose of valuable materials such as blockage of pipes, mechanical damage of underground water pipes.

11) Accidents due to mechanical lose of metallic cars, aircraft, etc.

12) Causes pollution due to escaping products from corrosion.

13) Depletion of the natural resource.

Draw the diagram of the Apparatus used in electroplating and label the following parts:
The substance to be electroplated.

The substance to be electroplated is Copper.

$Anode$ - Impure Copper metal rod

$Cathode$ - Pure Copper metal rod

$Electrolyte$ - Acidified copper sulphate solution

$\text {Process of electroplating}$ -On the application of high voltage the pure copper metal ions from the impure copper metal rod i.e. anode dissolves into the electrolyte and the same amount of

$Cu^{2+}$ ion gets deposited over the cathode rod as shown in the diagram.
1691009_1803964_ans_0604b2ed49e64306a923b1332142a906.png

Draw the diagram of the apparatus used in the electrolysis of water. Label the following parts:
(i) Graphite rod
(ii) Cathode

1678760_1803943_ans_de8245cc1eef491a9f5dcb9208cc21d3.png

What is electrolysis ?

The decomposition of an electrolyte when electricity is passed through it is called electrolysis.

When will the cell stop functioning?

When equilibrium is attained i.e.,

$E_{cell} = 0$ .

The flow of electrons in certain galvanic cells are given below:
i) $Cu \to Ag$
ii) $Ag\to Zn$
iii) $Na\to Mg$
iv) $Fe\to K$
Choose the incorrect ones.

ii.

$Ag\to Zn$ ,

iv.

$Fe\to K$ are the incorrect ones.

Write the equations of the oxidation and reduction reactions.

Anode

$- Fe \rightarrow Fe^{2+} +2e^-$ (Oxidation)

Cathode

$-Cu^{2+}(aq)+2e^- \rightarrow Cu (s)$ (Reduction)

What are the changes that can be observed with the iron rod and the colour of copper sulphate solution ?

When iron is dipped in

$CuSO_4$ a displacement reaction takes place.

In this, iron being a more reactive metal displaces a lesser reactive metal Copper(

$Cu$ ) from its compound to form

$FeSO_4$ .

The blue colour of the solution eventually fades and turn into light green due to the formation of ferrous sulphate.

You are given a solution of $AgNO_3$ , a solution of $MgSO_4$ , a $Ag$ and a $Mg$ ribbon. How can you arrange a Galvanic cell using these? Write down the reactions taking place at the cathode and the anode.

At anode

$Mg(s)\to Mg^{2+}(aq) + 2e^-$

At cathode,

$Ag(aq)+1e^- \to Ag(s)$

1754902_1834799_ans_1116de33c1bc408da8253eb3dec69a95.png

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Write down the general names used for an electrode which gives electrons.

Zn-Ag cell is also possible.

At the anode, anions (negative ions) are forced by the electrical potential to react chemically and give off electrons (oxidation) which then flow up and into the driving circuit.

Anode gives electrons.

1755664_1834850_ans_76753096841d4135892744c14c7e7a23.png

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Complete the table based on the figure drawn.
Galvanic Cell Electrode which Gives Electron Electrode which Gain Electron

Glavanic cell	Electrode which gives Electrons	Electrode which accepts electrons
$Zn -Cu$	$Zn$	$Cu$
$Cu -Ag$	$Cu$	$Ag$
$Zn -Ag$	$Zn$	$Ag$

1818593_1834848_ans_980d6bc6c76f49ef8cd6048d2c3a8ae2.png

Certain metals are given below:
$Ag, Zn, Pb, Sn, Fe$
When a galvanic cell is constructed using these metals, which one acts only as anode? Give the reason.

$Zn$ . because

$Zn$ has a higher reactivity than the other

$4$ metals.

Explain how rusting of iron is considered as setting up of an electrochemical cell.

In rainy season and in highly humid atmosphere the iron surface has a layer of water. This water layer dissolves acidic oxides of the air like

$CO_2, SO_2$ etc. to form acids which dissociate to give

$H^+$ ions.

$H_2O + CO_2 \rightarrow H_2CO_3 \rightleftharpoons 2H^+ + CO_{3}$

$^{2-}$

In the presence of

$H^+$ ions, iron starts losing electrons at some spot to form ferrous ions, i.e., its oxidation takes place. Here it is anode.

$Fe_{(s)} \rightarrow Fe^{2+}_{(aq)} + 2e^{-}_{(anode)}$

Now the released electrons move through the metal and reach to another spot where

$H^+$ ions and the dissolved oxygen take up these electrons and reduction reaction takes place. Here it is cathode.

$O_{2(g)} + 4H^+_{(aq)} + 4e^- + 2H_2O_{(l)(cathode)}$

The overall reaction is

$2Fe_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$

Thus in this way an electrochemical cell is set up on the surface. This ferrous ion further oxidised by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxide,

$Fe_2O_3.$ x

$H_2O$ , which is rust.

Take Cupric Chloride $(CuCl_2)$ solution in a beaker. Dip two graphite rod in it. Pass $5V$ electricity through it.

Why does electricity pass through a Cupric Chloride solution?

Electrolytes are substance which conduct electricity in molten states or in aqueous solution.

In molten state ions of

$CuCl_2$ can more freely. These ions are responsible for the Conduction of electricity by electrolytes.

Take Cupric Chloride $(CuCl_2)$ solution in a beaker. Dip two graphite rod in it. Pass $5V$ electricity through it.

Write one word for the process of chemical change happening in a Electrolyte while passing Electricity?

Electrolysis.

Take Cupric Chloride $(CuCl_2)$ solution in a beaker. Dip two graphite rod in it. Pass $5V$ electricity through it.
Which gas evolved out through a positive electrode? How did you identify that gas?

Chloride ions are oxidised by electron loss to form chlorine gas at the positive electrode.

Bubbles on surface of electrode indicates chlorine gas.

$2Cl^- \rightarrow Cl_2 +2e^-$

Take Cupric Chloride $(CuCl_2)$ solution in a beaker. Dip two graphite rod in it. Pass $5V$ electricity through it.
At which electrode oxidation and reduction take place?

Oxidation takes place at the positive electrode and Reduction takes place at the negative electrode.

Galvanic Cell	Electrode which Gives Electron	Electrode which Gain Electron

Electrochemistry - Class 12 Engineering Chemistry - Extra Questions

Copy and complete the following sentences ?With platinum electrodes hydrogen is liberated at the ........... and oxygen at the .......... during the electrolysis of acidified water.

During the electrolysis of acidified water, what is the ratio of the volumes of hydrogen produced to the volume of oxygen ?

Give the equation for the discharge of ion at the cathode during the electrolysis of acidified water ?

To carry out the so-called electrolysis of water, sulphuric acid is added to water. How does the addition of sulphuric acid produce a conducting solution ?

Pure water consists entirely of _____________ (ions / molecules).

Name a solid which undergoes electrolysis when molten.

The whole apparatus used for electrolysis is termed as :

Name the substance dissolved in pure water to make it a conductor of electricity.

Solutions which are decomposed due to the current passed through them are called .......... .

Hydrogen gas is liberated at the .......... (negative / positive) electrode called ..........

Acidified water is electrolysed by using carbon electrodes. What is produced at negative carbon electrode?

Acidified water is electrolysed by using carbon electrodes. What is produced at positive carbon electrode?

Oxygen gas is liberated at the .......... (negative / positive) electrode called the .......... .

The chemical reaction is 2H2O→..........+O22H_2O\, \rightarrow\, .......... \,+\,O_2

.......... (chemical / physical) reaction takes place.

Write the equation of the reactions which take place at the cathode and anode when acidified water is electrolysed.

During electrolysis the cations move to the electrode at a ___________ potential.

_________ is the chemical change that takes place by the passage of current through an electrolyte.

By process of electrolysis chemical energy gets converted into electrical energy.

Rusting of iron is an example of ________ reaction.

Can we make use of a.c. in electrolysis? Explain briefly.

19 g19\ g of molten SnCl2SnCl_{2} is electrolyzed for some time using inert electrodes until 0.119 g0.119\ g of SnSn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of masses of SnCl2:SnCl4SnCl_{2} : SnCl_{4} after electrolysis.

How much electricity in terms of faraday is required to produce 20 grams of calcium from molten calcium chloride?

Write the chemistry of recharging lead storage battery highlighting all the materials that are evolved during recharging.

Sea water promotes corrosion. Why?

An old cycle frame was left in open for a few days. A brown layer got slowly deposited on its surface and could not be removed when rubbed with sand paper. What happened actually?

Explain the term with example:Oxidant

How many coulomb are required for the oxidation of one mole of H2O H_{2}O to O2 O_{2} ?

State given reason, in what state or medium does i. NaCLNaCLII. HCLHCL gasiii. NH3NH_3 gas conduct electricity.

What do you mean by Corrosion ?

Column II gives name of material use for device given in column I :

Statements (A, B, C, D) in list I have to be matched with statements (1,2,3,4) list II.

For the circuit shown in the adjoining figure, match the entries of column I with the entries of column II.

Electrolysis of water is carried out in hoffman voltameter. Litmus is added to both cathodic and anodic compartments. What changes do you observe in the above process at the end of electrolysis? Justify.

List 1 and List 2 contains four entries each. Entries of List 1 are to be matched with some entries of ListOne or more than one entries of List 1 may have the matching with the same entries of List 2.

The charge (in F) required to deposites all Al from the electrolysis of 1 mol molten Al2_2O3_3 is :

The quantity of charge (in Faraday) required to reduce 96 g Mg from molten solution of MgCl2_2 is :

In rusting of iron, iron is oxidised and O2_2 is reduced. The no. of electrons used during reduction O2_2 are :

In a galvanic cell electrical energy is generated at the expense of chemical energy.If true enter 1, else enter 0.

The electric charge for electro deposition of 1 g equivalent of substance is xx faraday. What is the value of xx?

Presence of CO2_2 in natural water facilitate rusting of iron.If true enter 1, else enter 0.

Explain how electrolysis is an example of redox reaction?

_________ is defined as a process where materials, usually metals, deteriorate as a result of a chemical reaction with air, moisture, chemical, etc. For example, iron, in the presence of moisture, reacts with oxygen to form hydrated iron oxide.

Number of Faradays (F) required to reduce 3 mol of MnO⊖4MnO_{4}^{\ominus} to Mn2+Mn^{2+} is:

How many Faradays are required to reduce 1 mol of BrO⊖3BrO_3^{\ominus} to Br⊖Br^{\ominus} in basic medium ?

Electric Unit and Single Unit

Classify the following substances under three headings.Strong electrolytes ,Weak electrolytes

Fill in the blanks :Electrolysis is the passage of_______ (electricity / electrons) through a liquid or a solution accompanied by a _______ (physical/ chemical) change.

Water is broken down into its constituent elements .......... and .......... .

The bulb does not glow in the set up shown in the figure. List the possible reasons. Explain your answer.

Paheli had heard that rainwater is as good as distilled water. So she collected some rain water in a clean glass tumbler and tested it using a tester. To her surprise she found that the compass needle showed deflection. What could be the reasons?

What happens whenAn iron rod is placed in CuSO4CuSO_{4} solution?

Why do caesium (CsCs) and potassium (KK) find applications in photoelectric cells?

What happens when a copper rod is placed in FeSO4FeSO_{4} solution?

Differentiate between the terms strong electrolyte and weak electrolyte.(sating any two differences)

Corrosion can be prevented by using _________ .

Edible oil is not allowed to stand for a long time in an iron or tin container. Give reasons.

What do you mean by corrosion? How can you prevent it?

Write the SI unit of Resistivity.

Write the complete chemical reaction of rusting of iron.

What is metallic corrosion? Give one example

Write three methods of preventing rusting of iron.

State the first and second law of electrolysis.

What is corrosion? Do gold ornaments corrode? Justify.

What is the chemical formula of rust?

What is electrolysis? A solution of CuSO4CuSO_{4} is electrolysed for 1010 minutes with a current of 1.51.5 amperes. What is the mass of copper deposited at the cathode?

State Faraday's law's of electrolysis. A solution of CuSO4Cu{SO}_{4} is electrolysed for 1010 minutes with a current of 1.51.5 amperes. What is the mass of copper deposited at the cathode?

What is a secondary cell?

Mg, Al, Zn, FeMg,\ Al,\ Zn,\ Fe and AgAg are metals of reactivity series:(a) Which of this metal shows the highest reactivity?(b) If an electrochemical cell is devised by dipping Fe in FeSO4FeSO_4 solution and Ag in AgNO3AgNO_3 solution, which electrode is acting as cathode? Give reason.

What do you mean by metal corrosion? Explain three methods for prevention of corrosion.

Write short notes on : (i)Nernst equation(ii)Corrosion.

(a) What do you understand by corrosion?(b) Write electro-chemical theory of corrosion (Rust).(c) Write prevention (two) of corrosion.

Explain corrosion on the basis of following points :(i) Definition(ii) Factors affecting any two (iii) Prevention of corrosion any two.

Calculate the charge of 6.24×10186.24\times 10^{18} electrons (in coulomb).

How many moles of electrons will together contribute to the charge of 289500 coulomb?

State Faraday's first law of electrolysis.

At what [OH−][OH^{-}] does the following half-reaction have a potential of 0V0 V when other species are at 1 M1\ M?NO−3+H2O+2e−→NO−2+2OH−,E∘cell=0.01 VNO_{3}^{-} + H_{2}O + 2e^{-}\rightarrow NO_{2}^{-} + 2OH^{-}, E_{cell}^{\circ} = 0.01\ V.

If one Faraday was to be 4825048250 coulomb instead of 9650096500 coulomb, what will be charge of an electron?

If 0.5 L0.5\ L of a 0.6 M SnSO40.6\ M\ SnSO_{4} solution is electrolysed for a period of 3030 min, using a current of 4.6 A4.6\ A. If insert electrodes areas. What is the final concentration of Sn2+Sn^{2+} remaining in the solution?

Copy and complete the following sentences ?
With platinum electrodes hydrogen is liberated at the ........... and oxygen at the .......... during the electrolysis of acidified water.

The chemical reaction is $2H_2O\, \rightarrow\, .......... \,+\,O_2$

$19\ g$ of molten $SnCl_{2}$ is electrolyzed for some time using inert electrodes until $0.119\ g$ of $Sn$ is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of masses of $SnCl_{2} : SnCl_{4}$ after electrolysis.

How many coulomb are required for the oxidation of one mole of $H_{2}O$ to $O_{2}$ ?

State given reason, in what state or medium does
i. $NaCL$
II. $HCL$ gas
iii. $NH_3$ gas conduct electricity.

The charge (in F) required to deposites all Al from the electrolysis of 1 mol molten Al $_2$ O $_3$ is :

The quantity of charge (in Faraday) required to reduce 96 g Mg from molten solution of MgCl $_2$ is :

In rusting of iron, iron is oxidised and O $_2$ is reduced. The no. of electrons used during reduction O $_2$ are :

In a galvanic cell electrical energy is generated at the expense of chemical energy.
If true enter 1, else enter 0.

The electric charge for electro deposition of 1 g equivalent of substance is $x$ faraday. What is the value of $x$ ?

Presence of CO $_2$ in natural water facilitate rusting of iron.

If true enter 1, else enter 0.

Number of Faradays (F) required to reduce 3 mol of $MnO_{4}^{\ominus}$ to $Mn^{2+}$ is:

How many Faradays are required to reduce 1 mol of $BrO_3^{\ominus}$ to $Br^{\ominus}$ in basic medium ?

Classify the following substances under three headings.
Strong electrolytes ,Weak electrolytes

Fill in the blanks :
Electrolysis is the passage of_ (electricity / electrons) through a liquid or a solution accompanied by a _ (physical/ chemical) change.

What happens when
An iron rod is placed in $CuSO_{4}$ solution?

Why do caesium ( $Cs$ ) and potassium ( $K$ ) find applications in photoelectric cells?

What happens when a copper rod is placed in $FeSO_{4}$ solution?

Differentiate between the terms strong electrolyte and weak electrolyte.
(sating any two differences)

What is electrolysis? A solution of $CuSO_{4}$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

State Faraday's law's of electrolysis. A solution of $Cu{SO}_{4}$ is electrolysed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode?

$Mg,\ Al,\ Zn,\ Fe$ and $Ag$ are metals of reactivity series:
(a) Which of this metal shows the highest reactivity?
(b) If an electrochemical cell is devised by dipping Fe in $FeSO_4$ solution and Ag in $AgNO_3$ solution, which electrode is acting as cathode? Give reason.

Write short notes on :
(i)Nernst equation
(ii)Corrosion.

(a) What do you understand by corrosion?
(b) Write electro-chemical theory of corrosion (Rust).
(c) Write prevention (two) of corrosion.

Explain corrosion on the basis of following points :
(i) Definition
(ii) Factors affecting any two
(iii) Prevention of corrosion any two.

Calculate the charge of $6.24\times 10^{18}$ electrons (in coulomb).

At what $[OH^{-}]$ does the following half-reaction have a potential of $0 V$ when other species are at $1\ M$ ?
$NO_{3}^{-} + H_{2}O + 2e^{-}\rightarrow NO_{2}^{-} + 2OH^{-}, E_{cell}^{\circ} = 0.01\ V$ .

If one Faraday was to be $48250$ coulomb instead of $96500$ coulomb, what will be charge of an electron?

If $0.5\ L$ of a $0.6\ M\ SnSO_{4}$ solution is electrolysed for a period of $30$ min, using a current of $4.6\ A$ . If insert electrodes areas. What is the final concentration of $Sn^{2+}$ remaining in the solution?

Fill in the blanks by choosing the appropriate word/words from those given in the brackets:
(square pyramidal, electrical, 74, 26, $sp^3d^2$ , $sp^3d$ , chemical, 68, 32, tetrahedral, yellow, white, iodoform, Lucas).
A Galvanic cell converts energy into _ energy.

The resistance of a conductivity cell containing $0.001\ M\ KCl$ solution at $298\ K$ is $1500\Omega$ . What is the cell constant if conductivity of $0.001\ M\ KCl$ solution at $298\ K$ is $0.146\times 10^{-3} S\ cm^{-1}$ .

Find the distance between $(345)$ plane is a cubic batteries of length $7-7\overset{o}{A}$ .

Calculate volume of the gases liberated at STP if L of $0.2$ molar solution of $CuSO_4$ is electrolysed by $5.79 A$ Current for $10000 seconds.$
$5.79\times 10000$

In electrolysis of water:

(a) Name the gas collected at cathode and anode.
(b) Why is volume of one gas collected at one electrode is double of anode?
(c) Why are few drops of dilute $H_2SO_4$ added to water?

Represent the galvanic cell in which the reaction takes place.
$Zn(s)+{Cu}^{2+}(aq)\rightarrow {Zn}^{2+}(aq)+Cu(s)$

Calculate the no.of colombs of electricity, required to deposit
A) $2.7\times 10^{-4}kg$ of aluminium
B) $6.5\times 10^{-5}kg$ of Zn

Specific charges pf two particles $A$ and $B$ are in ratio $2 : 3$ , If their mass ratio $m_{A};m_{B}$ is $2:3$ , then find ratio of their charges $\left(\dfrac{q_{A}}{q_{B}}\right)$ ?

Calculate the number of coulombs required to deposit $5.4g$ of $Al$ , when the electrode reaction is ${Al}^{3+}+3{e}^{-}\rightarrow Al$
(At.mass of $Al=27,F=96500$ coulomb per mole)

Calculate the number of faradays required to electrolyze $6.35 \,g$ of $Cu^{+}(aq)$ ions from an aqueous solution.

On passing 4 amperes of current for 10min, during electrolysis of acidulated water at STP, what quantity of oxygen in $dm^3$ will be liberated?

How many faradays of electricity will be required by $3\times 10^{12}$ electrons?

If $E^{+}_{Fe^{2+}/Fe}$ is $x_1$ , $E^{+}_{Fe^{3+} / Fe}$ is $x_2$ then what will be $E^+_{Fe^+/Fe^{2+}}$ ?

How many minutes are required to deliver $3.21 \times 10^{6} coulombs$ using a current of $500$ A?. That is used in the commercial production of chlorine.

What weight of Ni is plated out in an electrolysis of aqueous ${ NiSO }_{ 4 }$ solution that it takes place to deposit 2g of Ag in a silver coulometer that is arranged in series with ${ NiSO }_{ 4 }$ electrolytic cell.[atomic weight of Ag=107.8 amu,atomic weight of Ni=58.7 amu]

How many coulombs of electricity is required to oxidize one mole of $Al$ to $Al^{3+}$ ?

Answer the following.
a. What is done to prevent corrosion of metals?
b. What are the metals that make the alloys brass and bronze?

0.2864g of $Cu$ was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. What is the electrochemical equivalent of copper?

Write the reaction occurring at the cathode and anode in $H_2$ AND $O_2$ fuel cell.

Answer the following :
What are the constituents of Nichrome wire?
Define Corrosion

In the adjacent diagram the electrolytic cell contains $1L$ of an aqueous $1M$ Copper (II) sulphate solution. If $0.4$ mole of electrons are passed through the cell, the molar concentration of copper ion after passing of the charge will be :

How much charge is required for the following reductions?
(i) $1\ mol$ of $Al^{3+}$ to $Al$
(ii) $1\ mol$ or $Cu^{2+}$ to $Cu$
(iii) $1\ mol$ of $MnO^{-}_4$ to $Mn^{2+}$ .

The $E^{e}$ $(M^{2+}/M)$ value for copperis positive (+0.34V). What is possibly the reason for this ?

Calculate the force of attraction between an electron ans a body having two proton charge when they are $0.529\times 10^{-8}\ cm$ apart. Assume charge of one electron and one proton equal to $-1.6\times 10^{-19}C$ and $+1.6\times 10^{-19}C$ respectively.

Write the cell reaction and calculate the emf of the cell
$Pt |H_2 (g, 1 atm) | H^+ (0.5 M) || KCl (1 M) |Hg_2Cl_2 (s) | Hg (l) | Pt$ at $25^o C, \, E_{} = 0.28 V$

Observe the following picture and answer the following question:
What is corrosion ?

A solution of ${ CuSo }_{ 4 }$ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Given the standard electrode potentials :

$K^+/K = -2.93 \,V$ , $Ag^+ /Ag = 0.80\,V$ , $Hg^{2+} / Hg= 0.79\,V$ ,
$Mg^{2+}/Mg = -2.37\,V$ , $Cr^{3+}/Cr = -0.74\,V$

Arrange these metals in their increasing order of reducing power.

How many hour are required for a current of $3.0$ amperes to decompose $18\,g$ water.

What would be the effect on the potential of this cell if $Na_2S$ were added to the $Cd^{2+}$ half cell and CdS were precipitated? Why?

What does the negative value of $E^{\circ}_{cell}$ indicate ?

(a) What is meant by corrosion? Name any two methods used for the prevention of corrosion.

(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used extracting this metal.

Define the following terms
Secondary batteries .

What is the effect of catalyst on:
(a) Activation energy (Ea), and
(b) Gibbs energy $( \Delta G )$

Fill in the blanks:
Iron benches kept in lawns and gardens get _(a)_. It is a _(b)_ change because a new _(c)_ is formed.

Explain how the activity series accounts for the following:
Tendency to corrode

Cell 'A' has $E_{Cell} = 2V$ and cell 'B' has $E_{Cell} = 1.1V$ . Which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell?

Represent an electrochemical cell showing indirect redox reaction between $Zn$ and $ZnSO_{4}+Cu$ solution

Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
Metals in that electrode in the reactivity series is (in the Top,Bottom)