MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 4

For which of these redox pairs, the reduction potential vary with pH

0%
$AmO_2^{2+} / AmO_2^{+}$
0%
$AmO_2^{2+} / Am^{4+}$
0%
$MnO_4^-/ Mn^{2+}$
0%
$Zn^{2+} / Zn$

Explanation

The half reaction involves H

$^+$ only in

$4H^+ + AmO_2^{2+} + 2e \rightarrow Am^{4+} + 2H_2O$
and

$MnO_4^- + 8H^+ + 5e \rightarrow Mn^{2+} + 4H_2O$
(according to oxygen balance)

The passage of electricity in the Daniell cell when

$Zn$ and

$Cu$ electrodes are connected:

0%
from $Cu$ to $Zn$ inside the cell
0%
from $Cu$ to $Zn$ outside the cell
0%
from $Zn$ to $Cu$ outside the cell
0%
none of the above

Molten

$AlCl_3$ is electrolysed with the current of

$0.5$ empere to produced 27 g

$Al$ . Number of Farady used are :

0%
$3$
0%
$5$
0%
$6$
0%
None of these

Explanation

$Al^{3+} + 3e^- \rightarrow Al$
From Faraday's laws of electrolysis, we know that

$W= \frac{EQ}{96500}$ .

$\frac{Q}{96500} =\dfrac {W}{E}$ .

$E = 27/3 = 9$ .

$\frac {Q}{96500} = 27/9 = 3F$ .
Hence option A is the right answer.

An apparatus used for the measurement of the quantity of electricity passed is known as a:

0%
calorimeter
0%
cathetometer
0%
coulometer
0%
colorimeter

In electrolysis experiment, after some times, the battery connection is reversed. Then

0%
The electrolysis will stop
0%
the rate of liberation of material at electrodes will increase
0%
the rate of liberation will decrease
0%
rate of liberation will remain unchanged
0%
more heat will be produced at the electrodes

The charge carriers in metallic conductors and in electrolytes are respectively:

0%
both ions
0%
both electrons
0%
electrons and ions
0%
ions and electrons

Which of the following statements is/are incorrect ?

0%
Rust is $Fe_{2}O_{3}$
0%
Zn - Cu cell is called Daniell cell.
0%
Saline water slows down rusting.
0%
Pure metals undergo corrosion faster than impure metals.

Explanation

Correct statements are given below:

a. Rust is

$Fe_{2}O_{3}.xH_{2}O$ .
b. Correct.
c. Saline water fastens the rusting.
d. Pure metals undergo corrosion slower than impure metals.

Hence, options A, C and D are correct options

Iron can be prevented from rusting by:

0%
connecting iron to more electropositive metal a case of cathodic protection.
0%
connecting iron to more electropositive metal a case of anodic protection.
0%
connecting iron to less electropositive metal a case of anodic protection.
0%
connecting iron to less electropositive metal a case of cathodic protection.

Electrolysis of aqueous solutions of which of the following substances results in only the decomposition of water ?

0%
Potassium chloride
0%
Zinc sulphate
0%
Potassium hydroxide
0%
Sodium phosphate

In the atmosphere of industrial smog, copper corrodes to form:

0%
Basic copper carbonate
0%
Copper sulphide
0%
Basic copper sulphate
0%
Copper oxide

Explanation

Reactions taking place are:

$8Cu\, +\, 6H_{2}O\, +\, 2SO_{2}\, +\, 5O_{2}\, \rightarrow\, 2[CuSO_{4}.3Cu(OH)_{2}]$

$2Cu\, +\, H_{2}O\, +\, CO_{2}\, +\, O_{2}\, \rightarrow\, [CuCO_{3}.Cu(OH)_{2}]$

Hence, options A and C are correct

How many coulombs are required for the oxidation of 1 mol of $H_2O \ to \ O_2$ ?

0%
$93000 C$
0%
$1.93\, \times\, 10^5 C$
0%
$9.65\, \times\, 10^4 C$
0%
$19.3\, \times\, 10^2 C$

Explanation

$H_{2}O\, \rightarrow\, 2H^{\oplus}\, +\, 2e^-\, +\, \displaystyle \frac{1}{2}\, O_2$

$2\, F\, \equiv\, 1\, mol\, of\, H_2O\, =\, \displaystyle \frac{1}{2}\, mol\, of\, O_2$

$\therefore\, 2\, \times\, 96500\, C\, =\, 1.93\, \times\, 10^5\, C$

Hence, option B is correct.

$E^{\ominus}\, for\, Cr^{3+}\, +\, 3e^-\, \rightarrow\, Cr\,\space and \space Cr^{3+}\, +\, e^-\, \rightarrow\, Cr^{2+}$ are

$-0.74 V$ and

$-0.40 V$ , respectively.

$E^{\ominus}$ for the reaction is:

$Cr^{+2}\, +\, 2e^-\, \rightarrow\, Cr$

0%
$-0.91 V$
0%
$+0.91 V$
0%
- $1.14 V$
0%
$+0.34 V$

Explanation

$Cr^{3+}\, +\, 3e^-\, \rightarrow\, Cr;$

$Cr^{3+}\, +\, e^-\, \rightarrow\, Cr^{2+};$

$Cr^{2+}\, +\, 2e^-\, \rightarrow\, Cr;$

$\displaystyle E^{\ominus}_{3}\, =\, \dfrac{n_1 E^{\ominus}_1\, -\, n_2 E^{\ominus}_2}{n_3}$

$\displaystyle =\, \dfrac{[3\, \times\, (-0.74)]\, -\, [(-1)\, \times\, (-40)]}{2}\, =\, -0.91\, V$

Electrolysis can be used to determine atomic masses. A current of

$0.550$ A deposits

$0.55$ g of a certain metal in

$100$ minutes. Calculate the atomic mass of the metal is

$n = 3$ :

0%
$100$
0%
$45.0$
0%
$48.25$
0%
$144.75$

Explanation

$I=0.55$ A

$W=0.55$ g

$t=100$ min

$=6000$ sec

$n=3$

We know that,

According to Faraday's

$1$ st law,

$W=Z\times I\times t$

$\Rightarrow 0.55=\dfrac{Atomic mass}{n\times 96500}\times 0.55\times 6000$

$\Rightarrow 1=\dfrac{At. mass}{3\times 965}\times 60$

$\Rightarrow$ At mass

$=104.2$ u

$\cong 100u$ .

1117815_1159550_ans_3f3dce4981eb4b8ab455692dca18f0af.jpg

A magnet is pulled away from a conducting coil in a circuit. The direction of the induced emf in the coil can be found from...........

0%
Faraday's law of electromagnetic induction
0%
Joule's law
0%
Lenz's law
0%
Fleming's left hand rule

Which of the following statement is correct-

0%
$E_{cell}$ & $\Delta_r G$ of cell reaction are extensive properties
0%
$E_{cell}$ & $\Delta_r G$ of cell reaction are intensive properties
0%
$E_{cell}$ is an intensive property while $\Delta_rG$ of cell reaction is an extensive property
0%
$E_{cell}$ is an intensive property while $\Delta_rG$ of cell reaction is an intensive property

Same quantity of current is passed through molten NaCl and molten

$Al_{2}O_{3}$ . If 4.6 g of Na was deposited in one cell, the mass of Al deposited in other cell is:-

0%
0.9 g
0%
1.8 g
0%
2.7 g
0%
3.6 g

Charge produced in butane

$-O_{2}$ Fuel cell if

$2\ mol$ butane is consumed, will be

0%
$23 F$
0%
$49 F$
0%
$26 F$
0%
$52 F$

The standard reduction potentials of

$Cu^{2+}\, |\, Cu$ and

$Cu^{2+}\, |\, Cu^{\bigoplus}$ are 0.337 V and 0.153 V, respectively. The standard electrode potential of

$Cu^{\bigoplus}\, |\, Cu$ half cell is:

0%
0.184 V
0%
0.827 V
0%
0.521 V
0%
0.490 V

Explanation

$Cu^{2+}\, +\, 2e^-\, \rightarrow\, Cu\, \quad\, ...(i) E^{\circleddash}_1$

$Cu^{2+}\, +\, e^-\, \rightarrow\, Cu^{\bigoplus}\, \quad\, ...(ii) E^{\circleddash}_2$

Net equation

$(Cu^{\bigoplus}\, +\, e^-\, \rightarrow\, Cu)\, \quad\, ...(iii)$

$E^{\circleddash}_3\, =\, \displaystyle \dfrac{n_1E^{\circleddash}_1\, -\, n_2E^{\circleddash}_2}{n_3}\, =\, \displaystyle \dfrac{2\, \times\, 0.337\, -\, 1\, \times\, 0.153}{1}\, =\, 0.521\, V$

Hence, option C is correct.

In an electrolytic cell current flows:

0%
from cathode to anode in outer circuit
0%
from anode to cathode outside the cell
0%
from cathode to anode inside the cell
0%
none of the above

On the basis of the electrochemical theory of aqueous corrosion, the reaction occurring at the cathode is :

0%
$O_{2(g)}+4H_{(aq)}^{+}+4e^{-}\rightarrow 2H_{2}O_{(l)}$
0%
$Fe_{(s)}\rightarrow Fe_{(aq)}^{2+}+2e^{-}$
0%
$Fe_{(aq)}^{2+}\rightarrow Fe_{(aq)}^{3+}+e^{-}$
0%
$H_{2(g)}+2OH_{(aq)}^{-}\rightarrow 2H_{2}O_{(l)}+2e^{-}$

Explanation

Answer is Option

$A$

At cathode :

$2{{H}^{+}}_{(aq)}+2{{e}^{-}}\to 2H$

$2H+\frac{1}{2}{{O}_{2}}\to {{H}_{2}}O$

$\overline{2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\to {{H}_{2}}O}$

Write true or false for the following statements :
A solution of an ionic compound in water contains ions , which move to the opposite electrodes when electricity is passed through the solution.

0%
True
0%
False

Explanation

Let's take an example of

$NaCl_{aq}$ . Now, when we pass electricity to the solution, the

$NaCl$ atoms split up into

$Na^+$ and

$Cl^-$ .the Na ions are attracted to the negative electrode (opposite charges attract) i.e. the cathode and Cl ions are attracted to the positive electrode i.e. the anodes. Similarly even water gets split into

$H^+$ and

${O_2}^{-}$ .

$H$ ions move to cathode and

$O$ ions move to anode

Write true or false for the following statements :
At ordinary temperature, the surface of metals such as magnesium, aluminium, zinc, lead, etc., are covered with a thin layer of oxide which protects the metals from the further oxidation.

0%
True
0%
False

Calcium oxide reacts vigorously with water to produce:

0%
Calcium hydroxide releasing a large amount of heat.
0%
Calcium hydroxide absorbing a large amount of heat.
0%
Calcium oxide and hydrogen with a release of large amount of heat.
0%
Calcium oxide any hydrogen with a absorption of large amount of heat

Explanation

$\text{When calcium oxide-}$

$CaO$

$\text{is added to water, it reacts vigorously with water to produce slaked lime}$

$\text{(calcium hydroxide) and releases a large amount of heat.}$

$\text{The reaction that takes place is:}$

$CaO + H_2O \rightarrow Ca(OH)_2$

Write true or false for the following statements:
Due to corrosion iron gets a brown coating, copper gets a green coating and silver gets a black coating.

0%
True
0%
False

Explanation

When a metal is attacked by substances around it such as moisture, acids, etc. it is said to corrode and this process is called corrosion.
After corrosion, the colour of:
Iron changes to

$\rightarrow$ Reddish brown
Silver changes to

$\rightarrow$ black
Copper changes to

$\rightarrow$ green
Hence, the given statement is true.

Calculate

$\Delta G$ for the following cell reaction.

$Zn_{(s)}+ Ag_2O_{(s)} + H_2O_{(l)} \rightarrow Zn^{2+} _{(aq)} + 2Ag_{(s)} +2OH{^-}{_{(aq)}}$

${ E }_{ { { Ag }^{ + } }/{ Ag } }^{ 0 }=+0.80V$ and

${ E }_{ { { Zn }^{ +2 } }/{ Zn } }^{ 0 }=-0.76V$

0%
$-305\ {kJ}/{mol}$
0%
$-301\ {kJ}/{mol}$
0%
$305\ {kJ}/{mol}$
0%
$301\ {kJ}/{mol}$

Explanation

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

$E^o_{cell} = 0.80 - ( - 0.76 )$

$E^o_{cell} = 1.56 V$

Now,

$\Delta G^o = - n F E^o_{cell}$

$\Delta G^o = - 2\times96500 \times 1.56$

$\Delta G^o = - 301080\ J / mol = - 301\ kJ/mol$

What is not Avogadro's number ?

0%
The number of atoms in 11.2 L of $\displaystyle { O }_{ 2 }$ at STP
0%
The number of atoms in 1 mole of $\displaystyle He$ at STP
0%
The number of electrons in 96,500 coulombs
0%
The number of $\displaystyle { SO }_{ 4 }^{ 2- }$ ions in 1 L of 0.5 M sulphuric acid
0%
None of the above

When

$0.1$ mol

$MnO_4^{2-}$ is oxidised the quantity of electricity required to completely

$MnO_4^{2-}$ to

$MnO_4{^-}$ is:

0%
$96500\ C$
0%
$2\times 96500\ C$
0%
$9650\ C$
0%
$96.50\ C$

Explanation

The oxidation reaction is as shown below.

$MnO_4^{2-} + e^-\rightarrow MnO_4^-$

Oxidation of one mole of

$MnO_4^{2-}$ requires 1 mole of electrons.

Hence, the oxidation of 0.1 mole of electrons will require 0.1 mole of electrons.
This corresponds to

$96500 \times 0.1= 9650 C$ .

1C electricity deposits:

0%
10.8 g of Ag
0%
electrochemical equivalent of Ag
0%
half of electrochemical equivalent of Ag
0%
96500 g of Ag

Explanation

From Faraday's first law,

$w=Z\times Q$

When

$Q=1\ C$

$Z = \dfrac {108}{1} = 108$

$w=Z$

Where

$Z =$ electrochemical equivalent of Ag.

Hence, option B is correct.

An electrode connected to the positive terminal of a battery is called anode.

0%
True
0%
False

Anode in the galvanic cell is :

0%
Negative electrode
0%
Positive electrode
0%
Neutral electrode
0%
None of the above

Current in an electrolyte is carried by

0%
only electrons
0%
only - ve ions
0%
only + ve ions
0%
both + ve and - ve ions

Electric energy is produced by the :

0%
separation of charges
0%
separation of protons
0%
separation of holes
0%
separation of plates

When the lead accumulator is charged, it acts as:

0%
an electrolytic cell
0%
a galvanic cell
0%
a Daniel cell
0%
none of the above

According to the electrochemical concept, corrosion occurs at :

0%
anodic area of metal
0%
cathodic area of metal
0%
both cathodic and anodic areas of metal
0%
neither cathodic area nor anodic areas of metal

Explanation

Corrosion can be defined as the deterioration of materials by chemical processes.

Of these, the most important by far is electrochemical corrosion of metals in which the oxidation process

$M \rightarrow M^+ + e^-$ is facilitated by the presence of a suitable electron acceptor. This is possible because metals are conductive and so, the electrons can flow through the metal from the anodic to the cathodic regions.
Metallic corrosion occurs when metal atoms are oxidized and subsequently leave the metal lattice as ions. The oxidation reaction of the metal is referred to as the anodic half-reaction. The areas on a metal surface where oxidation occurs are referred to as anodes.

Option A is correct.

As dilution of the solution increases then the conductivity of solution ___________.

0%
increases
0%
decreases
0%
remain unchanged
0%
none of the above

A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if:

0%
both the length and radius of wire are halved
0%
both length and radius of wire are doubled
0%
the radius of wire is doubled
0%
the length of the wire is doubled

Explanation

$R=\displaystyle\frac{\rho\ell}{\pi r^2}$

When

$\ell$ is

$\displaystyle\frac{\ell}{2}$ and radius is

$\displaystyle\frac{r}{2}$ ,

$\therefore\;R'=\displaystyle\frac{\rho\ell4}{\pi2r^2}=\displaystyle\frac{2\rho\ell}{\pi r^2}$

$R'=2R$ . Therefore, heat is doubled according to

$H=I^2RT$ .

Hence, option B is correct.

Splitting up of the water molecule by an electric current is an example of electrolysis.

0%
True
0%
False

For reducing one mole of

$Cr_{2} O_{7}^{2-}$ to

$Cr^{3+}$ , the charge required is:

0%
$3\times 96500\ coulomb$
0%
$6\times 96500\ coulomb$
0%
$0.3\ Faradays$
0%
$0.6\ Faradays$

Explanation

$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7 H_{2} O$
For reducing one mole of

$Cr_{2}O_{7}^{2-}$ , charge required

$= 6\times 96500\ coulomb$ .

When the power delivered by a

$100$ volt battery is

$40$ watts, the equivalent resistance of the circuit is:

0%
$100\;ohms$
0%
$250\;ohms$
0%
$300\;ohms$
0%
$350\;ohms$

Explanation

$P=\displaystyle\frac{V^2}{R}\,\Rightarrow\;R=\displaystyle\frac{V^2}{P}=\frac {100\times100}{40}=250\: ohms$

Hence, option B is correct.

The internal resistance of a lead acid battery can be reduced by

0%
Using strips of wood or celluloid as separators between electrodes
0%
Using grids of hard lead-antinomy alloy
0%
Using a specific manner of assembly of electrodes with wood or celluloid separators in between
0%
Both option A and B

The charge required for the reduction of

$1\ mol$ of

${MnO_{4}}^{-}$ to

$MnO_{2}$ is:

0%
$1\ F$
0%
$3\ F$
0%
$5\ F$
0%
$6\ F$

Explanation

$MnO_{4}^{-} + 4H^{+} + 3e^- \rightarrow MnO_{2} + 2H_{2} O$

Oxidation number of Mn in

${MnO_4}^-$ = +7
Oxidation number of Mn in

$MnO_2$ = +4

So,

$3\ F$ of charge will be required to reduce

$1\ mole$ of

$MnO_{4}^{-}$ .

Hence, option B is correct.

Three copper wires of lengths and cross sectional areas are

$(\ell,\,A),\,(2\,\ell,\,A/2)$ and

$(\ell/2,\,2A)$ . Resistance is minimum in:

0%
wire of cross-sectional area $A/2$
0%
wire of cross-sectional area A
0%
cross-sectional area $2A$
0%
same in all the three cases

Explanation

$R\propto\displaystyle\frac{\ell}{A}$ ;

So, the resistance of the wire will be minimum when the area of cross-section is maximum and length is minimum.

Hence, option C is correct.

The

$ZnCl$ used in a dry cell is which helps to maintain the moistness of paste contained in the cell between anode and cathode is

0%
Highly hydroscopic
0%
Weak hydroscopic
0%
Strongly hydrophobic
0%
Weak hydrophobic

Faraday law of electrolysis indirectly shows

0%
Quantisation of charge
0%
Quantisation of angular momentum
0%
Quantisation of current
0%
Quantisation of viscosity

Explanation

Faraday's law says that the amount of substance liberated or deposited at an electrode is proportional to the quantity of charge passed through it. Thus, mass

$\propto$ charge. Since mass is quantised hence indirectly charge is quantised.

When a corroded metal object is rubbed by a sand paper then:

0%
outer metal surface is removed
0%
metal becomes thin
0%
outer corroded layer will be removed
0%
all of the above

The standard reduction potential for

$Mg^{2+}/Mg$ is

$-2.37V$ and for

$Cu^{2+}/Cu$ is

$0.337$ . The

$E^o_{cell}$ for the following reaction is

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

0%
$+2.03V$
0%
$-2.03V$
0%
$-2.7V$
0%
$+2.7V$

Explanation

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$=E^o_{(Cu^{2+}/Cu)}-E^o_{(Mg^{2+}/Mg)}=0.337-(-2.37)V$

$=2.7V$

Electrolytes which allow large current to pass through them are known as :

0%
weak electrolytes
0%
strong electrolytes
0%
non-electrolytes
0%
none of the above

_____________ ionizes in a fused or in aqueous solution state and furnish positive ions other than

$H^+$ and negative ions other than

$OH^-$ .

0%
Mineral acids
0%
Bases
0%
Ionic salts
0%
Compounds

Explanation

Ionic salts ionizes in a fused or in aqueous solution state and furnish positive ions other than

$H^+$ and negative ions other than

$OH^−$ .
For example, aq NaCl ionises to form

$\displaystyle Na^+$ ions and chloride

$\displaystyle Cl^-$ ions.

$\displaystyle NaCl \rightleftharpoons Na^+ + Cl^-$

A metal is released in the electrolysis of a salt. It gets deposited on the :

0%
sides of the container
0%
half of positive terminal and half of negative terminal
0%
negative terminal
0%
positive terminal

Explanation

The metal which is released in the electrolysis is of simple salt, so it would generally be deposited on the cathode.

For example, suppose we take sodium chloride (NaCl) then on breaking this into simpler form we get,

$NaCl\rightarrow Na^++Cl^-$

This can be said that as Na acquires positive charge, it shall move to negative electrode (or negative terminal) i.e. cathode.

Hence , option C is correct .

______ ionizes in fused or in aqueous solution state and furnish ions in solution.

0%
Mineral acids
0%
Bases
0%
Ionic salts
0%
Compounds

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.