For the following cell with hydrogen electrodes at two different pressures ${p_1}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {p_2}$ $\mathop {{\rm{Pt}}\left( {{{\rm{H}}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{{\rm{H}}^ + }\left( {aq} \right)}\limits_{1{\rm{ }}M} } \right|\mathop {Pt\left( {{{\rm{H}}_2}} \right)}\limits_{{p_2}} $ emf is given by

$\frac{{RT}}{F}{\log _e}\frac{{{p_1}}}{{{p_2}}}$

$\frac{{RT}}{F}{\log _e}\frac{{{p_2}}}{{{p_1}}}$

$\frac{{RT}}{{2F}}{\log _e}\frac{{{p_2}}}{{{p_1}}}$

When a copper wire is immersed in a solution of ${\rm{AgN}}{{\rm{O}}_3}$ , the colour of the solution becomes blue because copper :

Is oxidised to ${\rm{C}}{{\rm{u}}^{2 + }}$

Forms a soluble complex with ${\rm{AgN}}{{\rm{O}}_3}$

Is reduced to ${\rm{C}}{{\rm{u}}^{2 - }}$

Splits up into atomic form and dissolves

A student made the following observations in the laboratory, i) Clean copper metal did not react with 1 molar ${\rm{Pb}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}$ solution. ii) Clean lead metal dissolved in a 1 molar ${\rm{AgN}}{{\rm{O}}_3}$ solution and crystals of ${\rm{Ag}}$ metal appeared. iii) Clean silver metal did not react with 1 molar ${\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}$ solution. The order of decreasing reducing character of the three metals is :

${\rm{Pb}},{\rm{Cu}},{\rm{Ag}}$

${\rm{Cu}},{\rm{Pb}},{\rm{Ag}}$

${\rm{Cu}},{\rm{Ag}},{\rm{Pb}}$

${\rm{Pb}},{\rm{Ag}},{\rm{Cu}}$

The emf of the cell ${\rm{Zn}}\left| {{\rm{Z}}{{\rm{n}}^{2 + }}\left( {0.01{\rm{M}}} \right)} \right|\left| {{\rm{F}}{{\rm{e}}^{2 + }}\left( {0.001{\rm{M}}} \right)} \right|{\rm{Fe}}$ at 298 K is 0.2905. The value of equilibrium constant for the cell reaction is

${e^{\frac{{0.32}}{{0.0295}}}}$

${10^{\frac{{0.32}}{{0.0591}}}}$

${10^{\frac{{0.26}}{{0.0295}}}}$

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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Hard

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 63 questions

1 Hard

1 of 63

Application +4 | -1

For the following cell with hydrogen electrodes at two different pressures

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

emf is given by

View answer and explanation

Correct option: B

\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}

Option B (

\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}

) is correct because it represents the Nernst equation applied to a concentration cell with hydrogen electrodes at different pressures.

2 Hard

2 of 63

Analyzing +4 | -1

Assertion :

E^{\circ}

for

{Mn}^{3+} / {Mn}^{2+}

is more positive than

{Cr}^3 / {Cr}^{2+}

.

Reason : The third ionization energy of

{Mn}

is larger than that of

{Cr}

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

Option A is correct because both the assertion and the reason are correct, and the reason is the correct explanation of the assertion. The higher third ionization energy of Mn leads to a more positive E° for Mn3+/Mn2+.

3 Hard

3 of 63

Understanding +4 | -1

When a copper wire is immersed in a solution of

\mathrm{AgN}{\mathrm{O}_3}

, the colour of the solution becomes blue because copper :

View answer and explanation

Correct option: B

Is oxidised to

\mathrm{C}{\mathrm{u}^{2 + }}

Option B (Is oxidised to

\mathrm{C}{\mathrm{u}^{2 + }}

) is correct because copper is more reactive than silver and gets oxidized to Cu2+ ions, which impart a blue color to the solution.

4 Hard

4 of 63

Application +4 | -1

Calculate the standard cell potential (in V) of the cell in which following reaction takes place

F{e^{2 + }}(aq) + A{g^ + }(aq) \to F{e^{3 + }}(aq) + Ag(s)

Given that,

\begin{array}{l} E{^\circ _{A{g^ + }/Ag}} = x\mathrm{V}\\ E{^\circ _{F{e^{2 + }}/Fe}} = y\mathrm{V}\\ E{^\circ _{F{e^{3 + }}/Fe}} = z\mathrm{V} \end{array}

View answer and explanation

Correct option: A

x + 2y - 3z

Option A (

x + 2y - 3z

) is correct because it correctly combines the given standard reduction potentials using the Gibbs free energy relationship.

5 Hard

5 of 63

Analyzing +4 | -1

100 {~mL}

0.05 {MCuSO}_4(

aq) solution was electrolyzed using inert electrodes by passing current till the

{pH}

of the resulting solution was 2 . The solution after electrolysis was neutralized and then treated with excess

{KI}

and formed

{I}_2

titrated with

0.04 M {Na}_2{~S}_2 {O}_3

. Calculate the required volume (in

{mL}

) of

{Na}_2 {~S}_2 {O}_3

View answer and explanation

Correct option: A

112.5 {~mL}

Option A (112.5 mL) is correct because it involves stoichiometric calculations based on electrolysis and titration reactions.

6 Hard

6 of 63

Analyzing +4 | -1

Given that,

\begin{aligned} & {Co}^{3+}+{e}^{-} \rightarrow {Co}^{2+} \quad \quad\quad \quad{E}^0=+1.82 \mathrm{~V} \\ & 2 {H}_2 {O} \rightarrow {O}_2+4 {H}^{+}+4 {e}^{-} \quad {E}^0=-1.23 \mathrm{~V} \end{aligned}

Select correct statement based on above data in standard conditions :

View answer and explanation

Correct option: A

{Co}^{3+}

is unstable in aqueous medium

The correct answer is A (Co3+ is unstable in aqueous medium) because the positive reduction potential indicates that Co3+ is readily reduced to Co2+ in water.

7 Hard

7 of 63

Application +4 | -1

The potential of the cell below was

1.05 \mathrm{~V}.

What is the

\mathrm{pH} ? \, E_{\mathrm{Ag}^+ / \mathrm{Ag}}^{\circ}=0.799 \mathrm{~V}

\mathrm{Pt}\left|\mathrm{H}_2(1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Ag}^{+}(0.1 \mathrm{~M}) |\mathrm{Ag}

View answer and explanation

Correct option: B

5.25

Option B (5.25) is correct because it is calculated using the Nernst equation, considering the given cell potential and standard reduction potential.

8 Hard

8 of 63

Application +4 | -1

The emf of the cell

\mathrm{Ni}\left| {\mathrm{N}{\mathrm{i}^{2 + }}\left( {1.0\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{u}^{3 + }}\left( {1.0\mathrm{M}} \right)} \right|\mathrm{Au}

is [

{\mathrm{E}^ \circ }\left( {\mathrm{N}{\mathrm{i}^{2 + }}/\mathrm{Ni}} \right) =

-0.25 V and

{E^ \circ }\left( {\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}} \right) =

+1.5 V]

View answer and explanation

Correct option: D

1.75 V

Option D (1.75 V) is correct because the EMF of the cell is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.

9 Hard

9 of 63

Understanding +4 | -1

{\mathrm{H}_2}

cannot be displaced by

View answer and explanation

Correct option: D

\mathrm{A}{\mathrm{g}^ + }

Option D (

\mathrm{A}{\mathrm{g}^ + }

) is correct because silver is below hydrogen in the electrochemical series and thus cannot displace it.

10 Hard

10 of 63

Application +4 | -1

Calculate the volume of hydrogen at NTP obtained by passing a current of 0.4 ampere through acidified water for 30 minute :

View answer and explanation

Correct option: A

0.0836 litre

Option A (0.0836 litre) is correct because it is calculated using Faraday's law of electrolysis and the given current and time.

11 Hard

11 of 63

Application +4 | -1

Given the following molar conductivities at

25^{\circ} \mathrm{C}, \mathrm{HCl}, 426 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} ; \mathrm{NaCl}, 126 \Omega^{-1}

\mathrm{cm}^2 \mathrm{~mol}^{-1} ; \mathrm{NaC}

(sodium crotonate),

83 \Omega^{-1}

\mathrm{cm}^2 \mathrm{~mol}^{-1}

. What is the dissociation constant of crotonic acid, if the conductivity of a

0.001 \mathrm{M}

crotonic acid solution is

3.83 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}

View answer and explanation

Correct option: B

1.11 \times 10^{-5}

Option B (

1.11 \times 10^{-5}

) is correct because the dissociation constant is calculated using molar conductivities and the degree of dissociation.

12 Hard

12 of 63

Application +4 | -1

Standard electrode potentials of two half cells are given as

Z{n^{2 + }}/Zn = - 0.76\,\,V

A{g^ + }/Ag = 0.80\,V

The EMF of the cell

Zn\left| {Z{n^{2 + }}\left( {1M} \right)} \right|\left| {A{g^ + }\left( {1M} \right)} \right|Ag

will be:

View answer and explanation

Correct option: A

1.56 V

The correct answer is 1.56 V. The EMF of the cell is calculated by summing the standard reduction potentials of the half-cells, considering the correct signs.

13 Hard

13 of 63

Application +4 | -1

The resistance of 1N solution of acetic is 250Ω ,when measured in a cell having a cell constant of

1.15{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}}

.The equivalent conduction (in

\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{equi}{\mathrm{v}^{ - 1}}

) of 1N acetic acid is

View answer and explanation

Correct option: B

4.6

Option B (4.6) is correct because it is calculated using the formula for equivalent conductance and the given values for resistance and cell constant.

14 Hard

14 of 63

Application +4 | -1

An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute

\mathrm{HN}{\mathrm{O}_3}

and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set up

\mathrm{Pt}\left( s \right),{\mathrm{H}_2}\left( g \right)\left| {{\mathrm{H}^ + }\left( {1\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{g}^ + }\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right)

Was 0.62 V. If

E_{\mathrm{cell}}^ \circ

= 0.80 V, what is the percentage of Ag in the alloy?
[At

{25^ \circ }C

, RT /F = 0.06]

View answer and explanation

Correct option: D

Option D (50) is correct because the percentage of Ag in the alloy is calculated using the Nernst equation and the given cell potential.

15 Hard

15 of 63

Application +4 | -1

Calculate the volume of

{\mathrm{H}_2}

gas at NTP obtained by passing 4 A through acidified

{\mathrm{H}_2}\mathrm{O}

for 30 min is

View answer and explanation

Correct option: D

0.836 L

Option D (0.836 L) is correct because it is calculated using Faraday's laws of electrolysis and the ideal gas law at NTP.

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16 Hard

16 of 63

Analyzing +4 | -1

A student made the following observations in the laboratory,
i) Clean copper metal did not react with 1 molar

\mathrm{Pb}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}

solution.
ii) Clean lead metal dissolved in a 1 molar

\mathrm{AgN}{\mathrm{O}_3}

solution and crystals of

\mathrm{Ag}

metal appeared.
iii) Clean silver metal did not react with 1 molar

\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}

solution.
The order of decreasing reducing character of the three metals is :

View answer and explanation

Correct option: C

\mathrm{Pb},\mathrm{Cu},\mathrm{Ag}

Option C (Pb, Cu, Ag) is correct because lead dissolved in silver nitrate, copper did not react with lead nitrate, and silver did not react with copper nitrate, indicating the decreasing order of reducing power.

17 Hard

17 of 63

Application +4 | -1

The standard emf for the cell reaction,

2\mathrm{C}{\mathrm{u}^ + }\left( {aq} \right) \to 2\mathrm{Cu}\left( s \right) + \mathrm{C}{\mathrm{u}^ + }\left( {aq} \right)

is +0.36 V at 298 K. The equilibrium constant of the reaction is

View answer and explanation

Correct option: D

1.2 \times {10^6}

The correct answer is D (1.2 × 10⁶) because it is calculated using the relationship between standard emf and equilibrium constant: log K = nE°/0.059.

18 Hard

18 of 63

Application +4 | -1

A current of 12 A is passed through an electrolytic cell containing aqueous

\mathrm{NiS}{\mathrm{O}_4}

solution. Both Ni and

{\mathrm{H}_2}

gas are formed at the cathode. The current efficiency is 60%. What is the mass of nickel deposited on the cathode per hour?

View answer and explanation

Correct option: A

7.883 g

The correct answer is A (7.883 g) because it is calculated using Faraday's law of electrolysis, considering the current efficiency.

19 Hard

19 of 63

Application +4 | -1

How many electrons flow when a current of

5 \mathrm{~A}

is passed through a solution for

200 \mathrm{~s}?

View answer and explanation

Correct option: A

6.24 \times 10^{21}

Option A (

6.24 \times 10^{21}

) is correct because it represents the number of electrons calculated using the formula Q/e, where Q is the total charge and e is the charge of an electron.

20 Hard

20 of 63

Application +4 | -1

The limiting molar conductivities at

25^{\circ} \mathrm{C}

\mathrm{HCl}, \mathrm{NaCl}

and

\mathrm{NaA}

are

426 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mole}^{-1};

126 \Omega^{-1} \mathrm{~cm}^2\text{ mole}^{-1}

and

83 \Omega^{-1} \mathrm{~cm}^2 \text{ mole} ^{-1}

respectively. The conductivity of

0.001 \mathrm{~M}

acid

\text{(HA)}

solution is

3.83 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}.

The

\mathrm{pH}

of the acid solution is

\left[ {{\text{consider}},\,\,\left( {1 - \alpha } \right) \approx 1} \right]

View answer and explanation

Correct option: A

Option A (4) is correct because the pH is calculated from the hydrogen ion concentration, which is determined using limiting molar conductivities and the degree of dissociation.

21 Hard

21 of 63

Application +4 | -1

Salts of metals

{X}, {Y}

and

{Z}

are electrolysed under identical condition using same quantity of electricity. It was observed that

4.2 {~g}

{X}, 5.4 {~g}

{Y}

and

19.2 {~g}

{Z}

were deposited at respective cathode. If the atomic weights of

{X}, {Y}, {Z}

are 7 , 27 and 64 respectively, then their ratio of valencies is

View answer and explanation

Correct option: B

1: 3: 2

Option B (

1: 3: 2

) is correct because it represents the ratio of valencies calculated using Faraday's laws of electrolysis.

22 Hard

22 of 63

Application +4 | -1

What pressure (bar) of

\mathrm{H}_2

would be required to make emf of hydrogen electrode zero in pure water at

25^{\circ} \mathrm{C}

View answer and explanation

Correct option: A

10^{-14}

Option A (

10^{-14}

) is correct because it is derived from the Nernst equation, setting the cell potential to zero and solving for the hydrogen pressure.

23 Hard

23 of 63

Application +4 | -1

Given the ionic equivalent conductivities for the following ions:

\lambda^{\circ}{ }_{\text {eq }} \mathrm{K}^{+}=73.5 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

\lambda^{\circ}{ }_{\text {eq }} \mathrm{Al}^{3+}=149 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

\lambda^{\circ}{ }_{\text {eq }} \mathrm{SO}_4{ }^{2-}=85.8 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

The

\Lambda_{\text {eq }}^{\circ}

for potash alum

\left(\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\right.

24 \mathrm{H}_2 \mathrm{O}

) is

View answer and explanation

Correct option: A

215.92

The correct answer is A (215.92) because it is calculated by summing the equivalent conductivities of the ions present in potash alum.

24 Hard

24 of 63

Analyzing +4 | -1

\mathrm{Cu} + 2\mathrm{HCl} \to \mathrm{CuC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{C{u^{2 + }}/Cu}^ \circ = + 0.34V} \right]

\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{Z{n^{2 + }}/Zn}^ \circ = - 0.76V} \right]

\mathrm{Ag} + 2\mathrm{HCl} \to \mathrm{AgCl} + \frac{1}{2}{\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{A{g^ + }/Ag}^ \circ = + 0.80V} \right]

Which of the following reaction is feasible ?

View answer and explanation

Correct option: A

(ii)

Option A (ii) is correct because the reaction with zinc is feasible, as indicated by a positive

E_{cell}^\circ

value, while the other reactions have negative values.

25 Hard

25 of 63

Analyzing +4 | -1

Find the standard cell potential involving the cell reaction:

{In}^{2+}+{Cu}^{+2} \rightarrow {In}^{+3}+{Cu}^{+} \text {, at } 298 {~K}

Given:

{E}_{{Cu}^{+2} / {Cu}^{+}}^0={X}_1 {~V}, {E}_{{In}^{3+} / {In}^{+}}^0={X}_2 {~V} {E}_{{In}^{2+} / {In}^{+}}^0={X}_3 {~V}

View answer and explanation

Correct option: C

X_1+X_3-2 X_2

The correct answer is C (

X_1+X_3-2 X_2

) because it correctly applies Hess's law to calculate the standard cell potential from the given reduction potentials.

26 Hard

26 of 63

Application +4 | -1

0.04

N

solution of a weak acid has conductivity

4.23 \times {10^{ - 4}}{\rm{mho {\kern 1pt}{\kern 1pt}c}}{\mathrm{m}^{ - 1}}

. If the degree of dissociation of acid at this dilution is 0.0612, then equivalent conductivity at infinite dilution is ……

{\rm{mho {\kern 1pt}{\kern 1pt}c}}{\mathrm{m}^2}\mathrm{eq}{.^{ - 1}}

∶

View answer and explanation

Correct option: A

172.8

Option A (172.8) is correct because it is the calculated equivalent conductivity at infinite dilution using the given conductivity, concentration, and degree of dissociation.

27 Hard

27 of 63

Analyzing +4 | -1

Assertion (A) At the end of electrolysis using platinum electrodes, an aqueous solution of

\mathrm{CuSO}_4

turns colourless.

Reason (R) During the electrolysis

\mathrm{CuSO}_4

changes to

\mathrm{Cu}(\mathrm{OH})_2

during electrolysis.

View answer and explanation

Correct option: C

Assertion is true but Reason is false.

Option C (Assertion is true but Reason is false) is correct because the CuSO4 solution turns colorless due to Cu deposition at the cathode, not due to the formation of Cu(OH)2.

28 Hard

28 of 63

Application +4 | -1

The resistance of a conductivity cell was found to be

700 \Omega

with

0.01 \mathrm{~M} \mathrm{~KCl}

and

800 \Omega

with

0.01 \mathrm{~M} \mathrm{~AgNO}_3

.

Molar conductance of

0.01 \mathrm{~M} \mathrm{~KCl}=150 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

.

Thus, molar conductance of

\mathrm{AgNO}_3

View answer and explanation

Correct option: A

131 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Option A (

131 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

) is correct because it is calculated by finding the ratio of molar conductances using the given resistances and molarity.

29 Hard

29 of 63

Application +4 | -1

The emf of the cell

\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\left( {0.01\mathrm{M}} \right)} \right|\left| {\mathrm{F}{\mathrm{e}^{2 + }}\left( {0.001\mathrm{M}} \right)} \right|\mathrm{Fe}

at 298 K is 0.2905. The value of equilibrium constant for the cell reaction is

View answer and explanation

Correct option: A

{10^{{{10}^{\frac{{0.32}}{{0.0298}}}}}}

Option A (

{10^{{\frac{{0.32}}{{0.0298}}}}}}

) is correct because the equilibrium constant is calculated using the Nernst equation, relating the cell potential to the equilibrium constant.

30 Hard

30 of 63

Application +4 | -1

To a beaker containing

0.1 \mathrm{~M}-{HCl},

a little pure solid

{AgCl}

is added. Both a silver and a hydrogen electrode

\left(P_{{H}_2}=1.0 \mathrm{~bar}\right)

are then placed in the solution. What is the approximate value of EMF measured between the electrodes at

25^{\circ} \mathrm{C}?

(\text{Given: } E_{{Ag}^+ |{Ag}}^{\circ}=0.80 \mathrm{~V} \text{ and } K_{\mathrm{sp}}({AgCl})=10^{-10})

[2.303RT/F =0.06]

View answer and explanation

Correct option: B

0.32 \mathrm{~V}

Option B (0.32 V) is correct because it is calculated using the Nernst equation, considering the standard reduction potential of Ag+/Ag and the Ksp of AgCl.

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