Match the column I with column II and mark the appropriate choice. Column I Column II (A) Electrochemical equivalent (i) Potential difference x Quantity of charge (B) Faraday (ii) Mass of substance deposited by one coulomb of charge (C) Ampere (iii) Charge carried by one mole of electrons (D) Electrical energy (iv) One coulomb of electric charge passed through one second

(A)$$\rightarrow$$ (ii), (B)$$\rightarrow$$ (iii), (C)$$\rightarrow$$ (iv), (D)$$\rightarrow$$ (i)

(A)$$\rightarrow$$(ii), (B)$$\rightarrow$$ (i), (C)$$\rightarrow$$ (iii), (D)$$\rightarrow$$ (iv)

(A)$$\rightarrow$$ (iii), (B)$$\rightarrow$$ (iv), (C)$$\rightarrow$$ (i), (D)$$\rightarrow$$ (ii)

(A)$$\rightarrow$$ (iv), (B)$$\rightarrow$$ (i), (C)$$\rightarrow$$ (ii), (D)$$\rightarrow$$ (iii)

MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 9

Consider the following equations for a cell reaction

$A + B \rightleftharpoons C+D$

$E^0=x\,volt,k_{eq}=k_1$

$2A+2B \rightleftharpoons 2C+2D$

$E^0=y\, volt,k_{eq}=K_2$ then:

0%
$x=y,K_1=k_2$
0%
$x=2y,K_1=2k_2$
0%
$x=y,K^2_1=k_2$
0%
$x^2=y,K^2_1=k_2$

$H_{2}(g)$ and

$O_{2}(g)$ , can be produced by the electrolysis of water. What total volume (in

$L$ ) of

$O$ and

$H_{2}$ are produced at

$STP$ when a current of

$30\ A$ is passed through a

$K_{2}SO_{4}(aq)$ solution for

$193\ min.$ ?

0%
20.16
0%
40.32
0%
60.48
0%
80.64

Explanation

SOLUTION:

When 30 amperes is passed for 193 minutes, the quantity of current passed =193×60×30=347400 coulombs.

convering into faraday we get

$\rightarrow \dfrac{347400}{96500}=3.6 Faraday.$

$H_2O\xrightarrow[k_2so_4 (aq.)]{electrolysis} \dfrac{1}{2}O_2+H_2$

On passing 3.6 Faradays of electricity 3.6 moles of Hydrogen and (3.6×0.5)=1.8 moles of Oxygen will be formed.

Volume of oxygen produced at STP =1.8×22.4=40.32 liters.

hence the correct opt: B

6.5 g

$Mg$ deposits on electrolysis of

$MgCl_2$ (fused) then what will be the volume of chlorine gas at STP discharged at electrode?

0%
6.067 L
0%
3.033 L
0%
9.1 L
0%
12.133 L

Explanation

$1$ mole of

$MgCl_2$ gives

$1mol$ of

$Mg(24g)$ & one mole of

$Cl_2$

$24g$ of

$Mg=22.4dm^3$ of

$Cl_2$ gas at STP

$\therefore 6.5g$ of

$Mg=\cfrac {22.4\times 6.5}{24}$

$=6.067L$ of

$Cl_2$ gas

An electrolytic cell is composed of

$Cu$ and

$Zn$ . A current of

$9.65\ A$ is drawn from a cell for

$1$ hour. Then the loss in mass at anode and gain in mass at cathode, respectively could be:

0%
$11.77\ g, 11.43\ g$
0%
$11.77\ g, 10\ g$
0%
$22.86\ g, 23.54\ g$
0%
$23.54\ g, 22.86\ g$

Explanation

$W=\cfrac { I\times t\times M }{ nF }$

$I \rightarrow$ current

$t \rightarrow$ time

$W \rightarrow$ weight deposited

$M \rightarrow$ Molar mass

$n \rightarrow$ No. of electrons involved

$F \rightarrow$ faraday's constant

$=96500$

For Copper

$(Cu)$

$M=63.5$ g/mol

$n=2$

$I=9.65$ A

$t=1$ hr

$=3600$ sec

$W=\cfrac { 9.65\times 3600\times 63.5 }{ 2\times 96500 } \\ W=11.43g$

For Zinc

$(Zn)$

$M=65.38$ g/mol

$n=2$

$W=\cfrac { 9.65\times 3600\times 65.38 }{ 2\times 96500 } =11.77g$

For the cell prepared from electrode

$A$ and

$B$ , electrode

$A:$

$\displaystyle\frac{Cr_2O^{2-}_7}{Cr^{3+}}$ ,

$E^0_{red}=+1.33$ V and electrode

$B:$

$\displaystyle\frac{Fe^{3+}}{Fe^{2+}}$ ,

$E^0_{red}=0.77$ V, which of the following statement is not correct?

0%
The electrons will flow from $B$ to $A$ (in the outer circuit) when connections are made.
0%
The standard e.m.f. of the cell will be $0.56$ V
0%
$A$ will be positive electrode
0%
Current flows from B to A in external circuit.

Explanation

(A) The electrons will flow from B to A when

connection are made.

This statement is wrong because electrons will move

from A to B as electrons move from higher potential

$+1.33V$ to the lower potential

$0.77V.$

1113888_873242_ans_0fff5c8f45e045c5b96961c41b8c3157.jpg

A gas

$X$ at

$1\ atm$ is bubbled through a solution containing a mixture of

$1\ M$

${ Y }^{ - }$ and

$1\ M$

${ Z }^{ - }$ ions at

$25^{\circ} C$ . If the reduction potential of

$Z > Y > X$ , then _____________.

0%
Y will oxidize X and not Z
0%
Y will oxidize Z and not X
0%
Y will oxidize both X and Z
0%
Y will reduce both X and Z

Explanation

The order of reduction potential is given as :
$Z>Y>X$
Higher reduction potential means self-reduction of the species, but it oxidizes other species. This means a species with higher reduction potential acts as an oxidizing agent for species with lower reduction potential.

Lower reduction potential means self-oxidation of the species, but it reduces other species. This means a species with lower reduction potential acts as a reducing agent for species with higher reduction potential.

Thus, $Y$ will oxidize $X$ $X$ but not $Z$ $Y$ .

Hence, option (A) is correct.

Standard reduction potentials of the half reactions are given below

$\displaystyle F_{2(g)} + 2e^- \rightarrow 2F^{-}_{(aq)}$ ;

$E^o = +2.85V$

$\displaystyle Cl_{2(g)} + 2e^- \rightarrow 2Cl^{-}_{(aq)}$ ;

$E^o = +1.36V$

$\displaystyle Br_{2(l)} + 2e^- \rightarrow 2Br^{-}_{(aq)}$ ;

$E^o = +1.06V$

$\displaystyle I_{2(s)} + 2e^- \rightarrow 2I^{-}_{(aq)}$ ;

$E^o = +0.53V$
The strongest oxidising and reducing agents respectively are:

0%
$F_2$ and $I^-$
0%
$Br_2$ and $Cl^-$
0%
$Cl_2$ and $Br^-$
0%
$Cl_2$ and $I_2$

Explanation

Because of high standard reduction potential of

$F_{2}$ , it reduces to

$F^{-}$ . So it oxidises other elements strongly. So, it is the strongest agent.

$I_{2}$ has the lowest standard reduction potential here. It oxidises itself strongly and reduces others. So, it is a strong reducing agent.

Four faradays of electricity were passed through

${AgNO}_3$ ,

${CdSO}_4$ ,

${AICI}_3$ and

${PbCI}_4$ kept in four vessels using insert electrodes. The ratio of moles of Ag, Cd, AI and Pb deposited will be ?

0%
12 : 4 : 6 : 3
0%
1 : 2 : 3 : 4
0%
12 : 6 : 4 : 3
0%
4 : 3 : 2 : 1

Explanation

$AgNO_3\longrightarrow Ag^++NO^-_3\\CdSO_4\longrightarrow Cd^{2+}+{(SO_4)}^{2-}\\AlCl_3\longrightarrow Al^{3+}+3Cl^-\\PbCl_4\longrightarrow Pb^{4+}+4Cl^-$

It requires

$1F$ of current to deposit

$1$ mol of

$Ag$ .

Similarly

$2F$ of current to deposit

$1$ mol of

$Cd.$

$3F$ of current to deposit

$1$ mol of

$Al.$

$4F$ of current to deposit

$1$ mol of

$Pb.$

When

$4$ faradays of electricity were passed,

moles of

$Ag$ diposited

$=\cfrac{4}{1}=4mol.$

moles of

$Cd$ diposited

$=\cfrac{4}{2}=2mol.$

moles of

$Al$ diposited

$=\cfrac{4}{3}mol.$

moles of

$lead$ diposited

$=\cfrac{4}{4}=1mol.$

$therefore$ Ratio of moles of

$Ag,Cd,Al,Pb$ will be

$4:2:\cfrac{4}{3}:1$

$=12:6:4:3.$

The time required for a current of 3 ampere to decompose electrolytically 18 g of

$H_{2}O$ is:

0%
18 hour
0%
36 hour
0%
9 hour
0%
18 second

Explanation

$\theta =nF$

$n \rightarrow$ No. of moles

$F \rightarrow$ Faraday's constant

$\theta \rightarrow$ charge

Moles of water

$=\cfrac { 18g }{ 18g/mol } =1$ mole

$\theta=F$

$\theta=I\times t$

$I=3$ amp

$t=\cfrac { 96500 }{ 3 }$

$t=8.935\sim 9$ hour

Question No.60
Indicate the wrong statement according to Valence bond theory:

0%
A sigma bond is stronger then $\pi$ - bond
0%
p-orbitals always have only sidewise overlapping
0%
s-orbitals never form $\pi$ - bonds
0%
There can be only one sigma bond between two atoms

Explanation

By Faraday second law, if the same amount of charge is passing through the solution connected in series for the same time then the amount of substances deposited is directly proportional to their equivalent weight. In dilute

${ H }_{ 2 }{ SO }_{ 4 }$ , hydrogen is present at cathode and oxygen is present at the anode in their molecular form. Weight deposited is directly proportional to their equivalent weight.

So, the mass of hydrogen at the cathode: the mass of oxygen at anode = 2: 16 = 1: 8

So, the correct answer is option A.

Read the passage given below and answer the question:
Chemical reactions involve the interactions of atoms and molecules. A large number of atoms/molecules (approximately

$6.023 \times 10^{23}$ ) are present in few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. the following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 faraday =96500 coulombs).
The total charge (coulombs) required for complete electrolysis is:

0%
24125
0%
48250
0%
96500
0%
193000

Explanation

$2Cl^{\circleddash}\longrightarrow Cl_2+2e^-$

Moles of

$Cl^{\circleddash}=$ Moles of

$NaCl=4\times 0.5=2$

$mol$

$2$

$mol$ of

$Cl^{\circleddash}$ required

$2$

$mol$

$F$ of electricity

$\therefore$ Total charge

$=2\times 96500=193000$

$C$

1000 mL 1 M

$CuSO_4(aq)$ is electrolysed by 9.65 amp current for 100 sec using

$Pt-$ electrode. Which is incorrect statement?

0%
Blue colour intensity decreases during electrolysis
0%
Blue colour intensity remains constant if $Cu-$ electrode used
0%
$pH$ of solution is 8 after electrolysis
0%
At anode $O_2$ gas liberated during electrolysis

Explanation

${ CuSO }_{ 4 }\rightleftharpoons { Cu }^{ 2+ }+{ SO }_{ 4 }^{ 2- }$ is formed from strong acid

${ H }_{ 2 }{ SO }_{ 4 }$ and weak base

${ Cu\left( OH \right) }_{ 2 }$ . Hence, the

${ CuSO }_{ 4 }$ is acidic in nature and it's

$pH$ is thus less than

$7$ always.

$\therefore$ The incorrect statement is

$pH$ of solution after electrolysis is

$8$ .

Spacecraft like the satellites that circle the earth and are used in television transmission have electric batteries that supply power for various purposes. These batteries contain ion-exchange resins instead of a liquid electrolyte solution because:

0%
of extreme cold conditions
0%
of lack of oxygen
0%
a high electric voltage is required
0%
of zero gravity

For the cell

$M/ M^{+} || X^{-}/ X, E^{\circ}_{M^{+}/ M} = 0.44$ and

$E_{X/ X^{-}}^{\circ} = 0.33\ V$ . From the data it can be deduced that

$M + X\rightarrow M^{+} + X^{-}$ is spontaneous.

0%
$M + X\rightarrow M^{+} + X^{-}$ is spontaneous
0%
$E_{Cell} = 0.77\ V$
0%
$E_{Cell} = -0.77\ V$
0%
None

Explanation

For a reaction to be spantaneous,

${ E }_{ re }$ should be positive

Hence

${ { E }^{ 0 } }_{ { M }^{ + }/M }\quad { { -E }^{ 0 } }_{ X/{ X }^{ - } }=0.44-0.33=0.11$

${ E }_{ { M }^{ + }/{ X }^{ - } }$ is positive. We can conclude that

${ M }^{ + }+{ X }^{ - }\rightarrow { M }^{ + }+{ X }^{ - }$ is Spantaneous as

${ E }_{ cell }$ is

$0.11V$ .

Which of the following metals has been used for building boats because it has resistance to corrosion by sea water?

0%
Copper
0%
Nickel
0%
Fungsten
0%
Titanium

Explanation

Titanium is used for building boats.

Inert metals which do not react with gases in air seen as oxygen are found to be non-corrosive.

It is corrosion resistance because it forms stable, protective, oxide film when

$Ti$ is exposed to oxide film.

Iron nails rust fast when used for fixing plates or strips of aluminium on a building. Similarly, a water pipe made of iron corrodes faster when connected to a pipe of copper. This happens because of the?

0%
Greater wetting of iron
0%
Easier flow of electrons to iron
0%
Higher rusting power of iron
0%
Chemical reduction of iron

The standard reduction potential for the half-cell reaction,

$Cl_2+2e^-\rightarrow 2Cl^-$ will be:

$Pt^{2+}+2Cl^-\rightarrow Pt+Cl_2, E^o_{cell}=-0.15$ V;

$Pt^{2+}+2e^-\rightarrow Pt, E^o=1.20$ V

0%
$-1.35$ V
0%
$+1.35$ V
0%
$-1.05$ V
0%
$+1.05$ V

Explanation

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

The given reaction is,

$Pt^{2+}+2Cl^{-}\rightarrow Pt+Cl_2,E^{0}_{cell}=-0.15V$

On reversing the above reaction,

$Pt+Cl_2\rightarrow Pt^{2+}+2Cl^{-},E^{0}_{cell}=0.15V$

Also given,

$Pt^{2+}+2e^{-}\rightarrow Pt,E^{0}_{cell}=1.20V$

As we know the relation,

$E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}$

So we get,

$E^{0}_{Cl_2/2Cl^{-}}=0.15-(-1.20)=1.35V$

Hence, the correct option is

$\text{B}$

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	Electrochemical equivalent	(i)	Potential difference x Quantity of charge
(B)	Faraday	(ii)	Mass of substance deposited by one coulomb of charge
(C)	Ampere	(iii)	Charge carried by one mole of electrons
(D)	Electrical energy	(iv)	One coulomb of electric charge passed through one second

0%
(A) $\rightarrow$ (ii), (B) $\rightarrow$ (i), (C) $\rightarrow$ (iii), (D) $\rightarrow$ (iv)
0%
(A) $\rightarrow$ (ii), (B) $\rightarrow$ (iii), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (i)
0%
(A) $\rightarrow$ (iii), (B) $\rightarrow$ (iv), (C) $\rightarrow$ (i), (D) $\rightarrow$ (ii)
0%
(A) $\rightarrow$ (iv), (B) $\rightarrow$ (i), (C) $\rightarrow$ (ii), (D) $\rightarrow$ (iii)

Explanation

Electrochemical equivalent: The weight of a substance deposited or evolved during electrolysis by the passage of a specified quantity of electricity and usually expressed in grams per coulomb
Faraday: A measure of the electric charge carried by one mole of electrons, used in electrolysis as the quantity of charge required to deposit or liberate one gram equivalent weight of a substance.
Ampere: By definition, if 1 Coulomb of Charge flows through a Conductor in 1 second then the magnitude of the current is said to be 1 A.
Electrical Energy: Electrical energy is the work done by electric charge. If current $i$ ampere flows through a conductor or through any other conductive element of potential difference V volts across it, for time $t$ second. We know $Q=I\times t$ and hence electrical energy becomes, Potential difference $\times$ quantity of charge

How much electricity in terms of Faraday is required to produce

$100$ g of Ca from molten

$CaCl_2$ ?

0%
$1$ F
0%
$2$ F
0%
$3$ F
0%
$5$ F

Explanation

We have to note that one mole of electron charge is equivalent to one Faraday.

Given reaction,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

Ca is undergoing oxidation i.e,

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

We can observe that 40g of Ca takes

$2e^{-}$ charge,

So,1 mole of

$Ca(40g)\equiv 2F$

Now, 100g of

$Ca\equiv 5F$

Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on:

0%
the nature and structure of the metal
0%
the number of valence electrons per atom
0%
change in temperature
0%
all of these

Which of the following is the correct cell representation for the given cell reaction?

$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$

0%
$Zn|Zn^{2+}||H^+|H_2$
0%
$Zn|Zn^{2+}||H^+, H_2|Pt$
0%
$Zn|ZnSO_4||H_2SO_4|Zn$
0%
$Zn|H_2SO_4||ZnSO_4|H_2$

Explanation

In the given reaction the oxidation state of

$Zn$ is changing from 0 to +2 so it is undergoing oxidation. Whereas the oxidation state of

$H^{+}$ is changing from +1 to 0 which implies it is undergoing reduction. And in cell representation oxidation is represented in left part and the reduction in right. It has to be remembered that

$H_2$ electrode has Pt too in it.

So, the correct representation of the cell is,

$Zn|Zn^{2+}||H^{+},H_2|Pt$

Hence option (B) is correct

A galvanic cell has electrical potential of

$1.1$ V. If an opposing potential of

$1.1$ V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

0%
The concentration of reactants becomes unity and current flows from cathode to anode
0%
The cell does not function as a galvanic cell and zinc is deposited on zinc plate
0%
The reaction stops and no current flows through the cell
0%
The reaction continuous but current flows in opposite direction

Explanation

We can study the effect of opposing potential applied to a galvanic cell as follows.

1) When

$E_{external}< 1.1V$
In a galvanic cell, if an external opposite potential is applied and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V.
2) When

$E_{external}= 1.1V$
When the external voltage is 1.1V the reaction stops altogether and no current flows through the cell.
3) When

$E_{external}> 1.1V$
Any further increase in the external potential again starts the reaction but in the opposite direction. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions.

Hence option (A) is correct.

A current of

$1.40$ ampere is passed through

$500$ mL of

$0.180$ M solution of zinc sulphate for

$200$ seconds. The molarity of

$Zn^{2+}$ ions after deposition of zinc is:

0%
$0.154$ M
0%
$0.177$ M
0%
$2$ M
0%
$0.180$ M

Explanation

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

$Faraday=\cfrac{Charge}{96500}=\cfrac{1.40\times 200}{96500}=2.90\times 10^{-3}$

$mol$

$\therefore Zn$ deposited

$=\cfrac{2.90\times 10^{-3}}{2}=1.45\times 10^{-3}$

$mol$

Now,

$mol$ of

$Zn^{2+}$ initially

$=0.180\times 0.5$

$Molarity\times V=0.09$

$mol$

$mol$ of

$Zn^{2+}$ left after deposition

$=0.09-0.00145=0.08855$

$mol$

$Molarity=\cfrac{Moles}{Volume}=\cfrac{0.08855}{0.5}=0.177$

$M$

Hence option (B) is correct.

How long would it take to deposit

$50$ g of

$Al$ from an electrolytic cell containing

$Al_2O_3$ using a current of

$105$ ampere?

0%
$1.54$ h
0%
$1.42$ h
0%
$1.32$ h
0%
$2.15$ h

Explanation

Faraday's First Law of Electrolysis states that only, According to this law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

The reaction of Al is,

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

Now as n=3, Equivalent weight of Al=

$\dfrac{27}{3}=9$

As per Faraday's first law,

$W=Z\times I\times t\implies Z=\dfrac{Eq. wt}{96500}$

$\implies t=\dfrac{W\times96500}{Eq. wt\times1}=\dfrac{50\times96500}{9\times105}=5102.2$ or

$1.42$

$hours$

If a current of

$1.5$ ampere flows through a metallic wire for

$3$ hours, then how many electrons would flow through the wire?

0%
$2.25\times 10^{22}$ electrons
0%
$1.13\times 10^{23}$ electrons
0%
$1.01\times 10^{23}$ electrons
0%
$4.5\times 10^{23}$ electrons

Explanation

According to this Faraday's first law, the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

So,

$Q=I \times t=1.5\times3\times 60 \times 60=16200 C$

Charge on one electron

$=1.601\times10^{-19}C$

Now,

16200C charge is on

$\dfrac{1\times16200}{1.601\times10^{-19}}$ electrons

$=1.01\times10^{23}$ .

When water is added to an aqueous solution of an electrolyte, what is the change in specific conductivity of the electrolyte?

0%
Conductivity decreases
0%
Conductivity increases
0%
Conductivity remains same
0%
Conductivity does not depend on number of ions

Explanation

Factors affecting electrolytic conductance are:

Concentration of ions: The sole reason for the conductivity of electrolytes is the ions present in them. The conductivity of electrolytes increases with an increase in the concentration of ions as there will be more charge carriers if the concentration of ions is more and hence the conductivity of electrolytes will be high.

Nature of electrolyte: Electrolytic conduction is significantly affected by the nature of electrolytes. The degree of dissociation of electrolytes determines the concentration of ions in the solution and hence the conductivity of electrolytes. Substances such as $CH_3COOH$ , with a small degree of separation, will have less number of ions in the solution and hence their conductivity will also below, and these are called weak electrolytes. Strong electrolytes such as $KNO_3$ have a high degree of dissociation and hence their solutions have a high concentration of ions and so they are good electrolytic conductance.

Temperature: Temperature affects the degree to which an electrolyte gets dissolved in solution. It has been seen that higher temperature enhances the solubility of electrolytes and hence the concentration of ions which results in an increased electrolytic conduction.

When water is added to an aqueous solution the number of ions per unit volume decreases i.e. the concentration of ions decreases and hence thereby conductivity gets decreased.

How many coulombs of electricity is required to reduce

$1$ mole of

$Cr_2O^{2-}_7$ in acidic medium?

0%
$4\times 96500$ C
0%
$6\times 96500$ C
0%
$2\times 96500$ C
0%
$1\times 96500$ C

Explanation

In acidic medium dichromate ions undergoes reduction and the reaction is,

$Cr_2O_7+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_2O$

So, 1 mole of

$Cr_2O^{2-}_{7}$ ion requires 6 moles of electrons i.e.

$6\times96500$ C of electricity

The given figure shows the corrosion of iron in atmosphere.

Fill in the blanks by choosing an appropriate option.
At a particular spot of an object made of iron,

$\underline{(i)}$ of iron to ferrous ion takes place and that spot behaves as

$\underline{(ii)}$ . Electrons released at

$\underline{(i)}$ spot move through the metal and go to another spot on the metal and reduce oxygen in presence of

$H^+$ . This spot behave as

$\underline{(iii)}$ . The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust,

$\underline{(iv)}$ and with further production of

$\underline{(v)}$ ions.

0%
(i)- Oxidation, (ii)- Anode, (iii)-Cathode, (iv)- $Fe_2O_3.xH_2O$ , (v)-Hydrogen
0%
(i)- Reduction, (ii)-Cathode, (iii)-Anode, (iv)- $Fe_3O_4$ , (v)-Hydrogen
0%
(i)-Oxidation, (ii)-Cathode, (iii)-Anode, (iv)- $Fe_2O_3xH_2O$ , (v)-Hydrogen
0%
(i)-Oxidation, (ii)-Anode, (iii)-Cathode, (iv)- $Fe_2O_3\cdot H_2O$ , (v)-Ferrous

Explanation

At a particular spot of an object made of iron, oxidation of iron to ferrous ion takes place and that spot behaves as an anode.

Electrons released at oxidation spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H

$^+$ . This spot behaves as a cathode. The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust,

$Fe_2O_3.xH_2O$ and with further production of hydrogen ions.

$Fe\longrightarrow \underbrace{Fe^{2+}+2e^-}_{(Ferrous)}$ (Oxidation)

Oxidation takes place at the anode whereas reduction takes place at the cathode.

$4Fe^{2+}+3O_2+2xH_2O\longrightarrow \underbrace{2Fe_2O_3\cdot xH_2O}_{Rust}$

Hydrogen ions are produced.

The electrode potential is the tendency of metal:

0%
to gain electrons
0%
to lose electrons
0%
to either lose or gain electrons
0%
none of these

Which of the following reactions does not take place during rusting?

0%
$H_2CO_3\rightleftharpoons 2H^++CO^{2-}_3$
0%
$4Fe^{2+}+O_{2(dry)}\rightarrow Fe_2O_3$
0%
$4Fe^{2+}+O_2+4H_2O\rightarrow 2Fe_2O_3+8H^+$
0%
$Fe_2O_3+xH_2O\rightarrow Fe_2O_3\cdot xH_2O$

Explanation

Rust is an iron oxide, a usually red oxide formed by the redox reaction of iron and oxygen in the presence of water or air moisture.

The following redox reaction also occurs in the presence of water and is crucial to the formation of rust:

$4Fe^{2+}+O_2\rightarrow 4Fe^{3+}+2O^{2-}$

But in the given reaction $Fe^{2+}$ is reacted with dry oxygen. And hence option B is wrong.

Various methods of prevention of rust are-

Oiling - for example bicycle chains
Greasing - for example nut and bolts
Painting - for example car body panels
Galvanising is a method of rust prevention. The iron or steel object is coated in a thin layer of zinc.

Which metal is usefull for protection of iron by corrosion?

0%
Aluminium
0%
Sodium
0%
Zinc
0%
Copper

During electrolysis of a solution of

$AgNO_3$ ,

$9650$ coulombs of charge is passed through the solution. What will be the mass of silver deposited on the cathode?

0%
$108$ g
0%
$10.8$ g
0%
$1.08$ g
0%
$216$ g

Explanation

According to this Faraday's first law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

The given reaction is,

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

Amount of Ag deposited by 1F(96500) is 108g

So, amount of Ag deposited by

$9650C=\dfrac{108}{96500}\times 9650=10.8g$

Which of the statements about solutions of electrolysis is not correct?

0%
Conductivity of solution depends upon size of ions
0%
Conductivity depends upon viscosity of solution
0%
Conductivity does not depend upon solvation of ions present in solution
0%
Conductivity of solution increases with temperature

How many electrons are there in one coulomb of electricity?

0%
$6.023 \times 10^{23}$
0%
$1.64 \times 10^{ 24}$
0%
$6.24 \times 10^{ 18}$
0%
$6.24 \times 10^{ 24}$

Explanation

Change on

$1e^-=1.602\times 10^{-19}$

$C$

If total charge

$=1$

$C$

$\therefore$ Total no. of

$e^-=\cfrac{1}{1.602\times 10^{-19}}=6.24\times 10^{18}e^-$

Hence, the correct option is

$(C)$ .

Which of the following metal do no evolve

${ H }_{ 2 }$ gas from dil. acid?

0%
$A\left( { E }_{ { A }^{ + }/A }^{ o }=0.25V \right)$
0%
$B\left( { E }_{ B/{ B }^{ + } }^{ 0 }=+0.22V \right)$
0%
$C\left( { E }_{ { C }^{ + }/C }^{ 0 }=0.30V \right)$
0%
$D\left( { E }_{ D/{ D }^{ + } }^{ 0 }=+0.44V \right)$

Explanation

If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas.
Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species.
Here the standard electrode potential is reduction potential and hence option A doesn't give $H_2$ gas when reacted with dilute acid.

For an electrochemical reaction occuring in a galvanic cell.

$2{ Fe }_{ \left( aq \right) }^{ +3 }+{ Zn }_{ \left( s \right) }\rightarrow { Zn }_{ \left( aq \right) }^{ +2 }+2{ Fe }_{ \left( aq \right) }^{ +2 }$
if concentration of

${ Fe }^{ +2 }$ is increased.

0%
${E}_{cell}$ will remain unchanged
0%
${E}_{cell}$ will decrease
0%
$pH$ of the solution will change
0%
less non P-V work will be obtained

Explanation

In given cell reaction we have,

at anode

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

at cathode

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

$E_{cell} = E^0_{cell}$ -

$\dfrac{0.0591}{2}$

$log$

$\dfrac{Zn^{2+} (Fe^{2+})^2}{(Fe^{3+})^2}$

so, on increasing the concentration of

$Fe^{2+}$ ions

$E_{cell}$ will decreases.

The approximate time duration in hours to electroplate $30\space g$ of calcium from molten calcium chloride using a current of $5 \space A$ is (At. mass of $Ca = 40g/mol$ )

0%
80
0%
10
0%
16
0%
8

Explanation

$\begin{array}{l}\text{Equivalent weight of a metal is defined as }\\ \text { the mass of the metal that react with 1} \\ \text { mole of } \mathrm{H}^{+} \text {lons. } \\ \qquad 1 \mathrm{~F}=96500 \text { coulomb } \end{array}$

$\text { Eq.wt }=\dfrac{\text { Molecular mass }}{\text { valency factor }}$

$\begin{array}{l} =\dfrac{40}{2} = 20 \\\text{20g of Ca to 1F }\\ 1 g - \dfrac{1}{20}\\ 30g-\dfrac{1}{20} \times 30=1.5 \mathrm{~F} \end{array}$

$\begin{array}{c} q=i t \\ 1.5 \times 96500=5 \times t \\ t=8.041 \\ \approx. 8sec \end{array}$

Consider the following equations for a cell reaction :

$A + B \rightarrow C + D \,; E^o = x \,volt, \Delta G = \Delta G_1$

$2A +2 B \rightarrow 2C +2D \,; E^o = x \,volt, \Delta G = \Delta G_2$

0%
$x = y, \Delta G_1 = \Delta G_2$
0%
$x > y, \Delta G_1 > \Delta G_2$
0%
$x = y, \Delta G_2 = 2\Delta G_1$
0%
$x < y, \Delta G_2 = 2\Delta G_1$

Explanation

$x=y, \Delta G_{2}=2\Delta G_{1}$ because

$\Delta G$ is an expensive property and eletrode potential is an intensive property i.e, electrote potential doesnt denpends upon the coefficient of the reactant and product in the equation of reaction

$\because \Delta G=-n FE_{cell}$ where

$n=no$ , of electrons used in the reaction which depends upon the coefficient of the reactant

In which of the following conditions is the potential for the following half-cell reaction maximum?

$2H^+ + 2e^{-} \longrightarrow H_2$

0%
$1.0 \,M \,HCl$
0%
A solution having $pH = 4$
0%
Pure water
0%
$1.0 \,M \,NaOH$

Explanation

In the given options,

(A). Concentration of

${ H }^{ + }$ in 1.0 M HCl =

$1M$

(B). Since,

$pH$ of solution is 4.

So, concentration of

${ H }^{ + }$ =

${ 10 }^{ -4 }M$

(C).

$pH$ of pure water = 7

So, concetration of

${ H }^{ + }$ =

${ 10 }^{ -7 }M$

(D). Concentration of

${ H }^{ + }$ in 1.0 M

${ H }^{ + }$ is much less than the all the above options.

Since, concentration of

${ H }^{ + }$ is highest in option A. So, potential of option A is maximum .

So, correct answer is option A.

Given that

$E^o_{Ag^+/Ag} = + 0.8V$ and

$E^o_{Ni^{+2}/Ni} = 0.25 V$ . Which of the following statement is true with regard to this?

0%
$Ag^+$ is an oxidizing agent while $Ni^{2+}$ is a reducing agent
0%
$Ag^{2+}$ can be reduced by silver metal
0%
$Ag^+$ is a better oxidizing agent than $Ni^{+2}$ and $Ni$ is a better reducing agent that $Ag$
0%
$Ni^{2+}$ is a better oxidizing agent than $Ag^+$ and $Ag$ is better reducing agent that $Ni$

Explanation

In the given problem,

${ E }_{ { Ag }^{ + }/Ag }^{ 0 }\quad =\quad +0.8\quad V\\ { E }_{ { Ni }^{ 2+ }/Ni }^{ 0 }\quad =\quad +0.25\quad V$

Here,

${ E }_{ { Ag }^{ + }/Ag }^{ 0 }>{ E }_{ { Ni }^{ 2+ }/Ni }^{ 0 }$ .
So,

${ Ag }^{ + }$ can be easily reduced than

${ Ni }^{ 2+ }$ .

So,

${ Ag }^{ + }$ acts as an oxidizing agent and

${ Ni }^{ 2+ }$ acts as a reducing agent.

So, correct answer is option A.

Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always true?

0%
$E^o_{cell} > 0, \Delta G^o < 0$ , and $Q < K$
0%
$E^o_{cell} > 0, \Delta G^o < 0$ , and $Q > K$
0%
$E^o_{cell} > 0, \Delta G^o > 0$ , and $Q > K$
0%
$E_{cell} > 0, \Delta G^o < 0$ , and $Q < K$

Explanation

For a reaction to be spontaneous we have to know that always

$\Delta G^0<0$ always.

We know for a galvanic cell,

$\Delta G=-nFE^0_{cell}$

So for

$\Delta G^0<0$ then

$E^{0}_{cell}>0$ .

We know that reaction will be spontaneous

$Q<K$

And hence option D is correct answer.

Amount of charge is required to convert 17 gm

$H_2O_2$ into

$O_2$ :-

0%
1 F
0%
2 F
0%
6 F
0%
None of these

Explanation

Solution -

$H_{2}O_{2}\rightarrow 2H_{2}O+O_{2}$

mole of

$H_{2}O_{2} = \dfrac{17}{34} = 0.5\ mol$

For 0.5 mole, amount of change required

$= 0.5\times 2\times 6.023\times 10^{23}\times 1.6\times 10^{-19} C$

$= 6.023\times 10^{4}\times 1.6 C$

$= 9.6\times 10^{4}C$

$\Rightarrow 96500C = 1F$

How many electrons flow when a current of 5 amperes passed through a solution for 193 seconds?

0%
$6.022\times { 10 }^{ 23 }$
0%
$6.022\times { 10 }^{ 22 }$
0%
$6.022\times { 10 }^{ 25 }$
0%
$6.022\times { 10 }^{ 21 }$

Explanation

Current, I = 5 A

Time, t = 193 sec

We know that,

Q = It

Plug the values,we get

= 5 A × 193 s

= 965 Coulombs

Number of electrons = total charge / charge on 1 electrons

$\dfrac{965}{1.6 \times {10} ^ {-19}}$

$6.022 \times {10} ^{21}$ electrons

If equal quantities of electricity are passed through three voltmeter containing

$FeSO_4$ ,

$Fe_2(SO_4)_3$ and

$Fe(NO_3)_2$ , then which of the following is not true?

0%
Amount of iron deposited in $FeSO_4$ and $Fe_2(SO_4)_3$ is equal
0%
Amount of iron deposited in $Fe(NO_3)_3$ is $\dfrac{2}{3}$ of the amount of iron deposited in $FeSO_4$
0%
Amount of iron deposited in $Fe_2(SO_4)_3$ and $Fe(NO_3)_3$ are equal
0%
Same gas will evolve in all three cases at anode

Explanation

(1)

$FeSO_4\rightarrow Fe^{2+}+SO_4^{2-}$

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

$\therefore 2F$ charge is needed for

$1$ mole deposition of

$Fe$ .

(2)

$Fe_2(SO_4)_3\rightarrow 2Fe^3+3SO_4^{2-}$

$Fe^{3+}+3e^- \rightarrow Fe_{(s)}$

$3F \rightarrow 1$ mole

$2F$ charge is needed for

$\cfrac {2}{3}$ mole deposition of

$Fe$ .

The gas liberated in all these cases is the same and that is

$H_2$ gas.

Therefore, option A is correct.

Consider the reaction :

$Cr_2O^{2-}_{7} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ .

What is the quantity of electricity in coulombs needed to reduce 1 mole of

$Cr_2O^{2-}_{7}$ ?

0%
$579000\ C$
0%
$679000\ C$
0%
$879000\ C$
0%
$979000\ C$

Explanation

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+14{ H }^{ + }+6{ e }^{ - }\rightarrow 2{ Cr }^{ 3+ }+7{ H }_{ 2 }O$

For complete reduction of

$1$ mole of

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- },6{ e }^{ - }$ is needed.

Hence, Quantity of electricity

$=6\times96500=579000$

$C$ .

Hence, option A is correct.

In which one of the following one faraday of electricity will liberate 1/2 gram atom of the metal ?

0%
$AlCl_3$
0%
$CuSO_4$
0%
$FeCl_3$
0%
$NaCl$

Explanation

$CuSO_{4} \rightarrow Cu^{2+}+SO_{4}^{2}$

ralency

$= 2$

$\therefore$ Ans

$= CuSO_{4}$

The passage of a constant current through a solution of dilute

$H_2SO_4$ with 'Pt' electrode liberated

$336cm^2$ of a mixture of

$H_2$ and

$O_2$ at S.T.P. The quantity of electricity that was passed is:

0%
96500 C
0%
1965 C
0%
1930 C
0%
0.01 Faraday

Explanation

$2H^+ + 2e \rightarrow H_2(8)$

$\overset{\circleddash }{40H} \rightarrow O_2(8)+H_2O(8)+4e^-$

For ;

$1\ F \rightarrow H_2$ liberated

$=\dfrac{22.4}{2}=11.2\ lit$

$O_2$ liberated

$=\dfrac{22.4}{4}=5.6\ lit$

$1\ F \rightarrow$ volt mixture

$=(11.2+5.6)\,lit$

$=16.8\ lit$

$16.8\,lit \rightarrow 1\ F$

$1\, lit \rightarrow \dfrac{1}{16.8}\ F$

$1000\ cc\rightarrow \dfrac{1}{16.8}\ F$

$16.8 \, lit \rightarrow 1\ F$

$1 \, lit \rightarrow \dfrac{1}{16.8}\ F$

$1000\ cc \rightarrow \dfrac{1}{16.8}\ F$

$1\ cc \rightarrow \dfrac{1}{16.8 \times 1000}\ F$

$336\ cc \rightarrow \dfrac{336}{16.8 \times 1000}\ F$

$=\dfrac{20}{1000}\ F$

$=\dfrac{20}{1000}\times 96500\ C$

$=1930\ C$

The time required to produce 2F of electricity with a current of 2.5 amperes is:

0%
21.44 h
0%
1200 min
0%
50000 s
0%
1.5 h

Explanation

$\left\{\begin{matrix} 1F = 96500c \\ 1A = 1c/s \end{matrix}\right.$

to produce 2F from 2.5 A,

$\dfrac{2\times 96500}{2.5\times 3600}hrs = 21.44\,hrs$

1178185_1037009_ans_18f5f7fb111b45c6b8bf3f213aaa776c.jpeg

The equilibrium constant for the following reaction at

$25^o$ C is

$2.9\times 10^9$ . Calculate standard voltage of the cell.

$Cl_{2(g)}+Br^-_{(aq)}\rightleftharpoons 2Cl^-_{(aq)}$

0%
$0.18$ V
0%
$0.22$ V
0%
$0.26$ V
0%
$0.28$ V

Explanation

$K_{eq}= 2.9 \times 10^9$

At equilibrium,

$\Delta G^o=-RT lnK_{eq}$

$\therefore -nFE^o=-RTlnK_{eq}$

$\therefore E^o=\cfrac {RTlnK_{eq}}{nF}=\cfrac {8.314 \times 298 \times ln(2.9\times 10^9)}{2 \times 96500}$

$\therefore E^o=0.28V$

Therefore, standard voltage of the cell is

$0.28V$

A solution of

$Ni(NO_3)_2$ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

0%
1.622 gm
0%
1.722 gm
0%
1.822 gm
0%
1.922 gm

Explanation

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

$t=20 \ min=20 \times 60 \ sec$

$n=2$

$i=5 \ A$

Atomic Mass of

$Ni=58.7$

$m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t$

$= \cfrac {58.7}{2 \times 96500} \times 20 \times 60 \times 5$

$= \cfrac {352200}{2 \times 96500}=1.824$

The mass of

$Ni$ deposited at the cathode is

$1.824 \ g$

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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