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

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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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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 396 questions

1 Medium

1 of 396

Application +4 | -1

25^{\circ} \mathrm{C}

, the standard emf of cell having reactions involving two electron change is found to be

0.295 \mathrm{~V}

. The equilibrium constant of the reaction is

View answer and explanation

Correct option: C

10^{10}

Option C (

10^{10}

) is correct because it is calculated using the Nernst equation relating standard emf to the equilibrium constant.

2 Medium

2 of 396

Remembering +4 | -1

Which are used as secondary reference electrodes?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because Calomel electrode, Ag/AgCl electrode, and Hg/Hg2Cl2-KCl electrode are all commonly used as secondary reference electrodes.

3 Hard

3 of 396

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.

4 Medium

4 of 396

Application +4 | -1

A current of strength 2.5 A was passed through

\mathrm{CuS}{\mathrm{O}_4}

solution for 6 min 26 s. The amount of copper deposited is (At. Wt. of Cu = 63.5, 1F = 96500 C)

View answer and explanation

Correct option: A

0.3175 g

Option A (0.3175 g) is correct because it is the amount of copper deposited calculated using Faraday's laws of electrolysis.

5 Medium

5 of 396

Application +4 | -1

A current strength of 9.65 amperes is passed through excess fused

{AlCl}_3

for 5 hours. How many litres of chlorine will be liberated at

STP

({F}=96500 {C})

View answer and explanation

Correct option: D

20.16

Option A (2.016) is incorrect. Option D (20.16) is correct because it is calculated by converting the current and time to charge in Faradays and then using the stoichiometry of the electrolysis reaction to find the volume of chlorine gas liberated at STP.

6 Medium

6 of 396

Analyzing +4 | -1

The electrode potential (in millivolts) of

2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-}

in a solution buffered at

\mathrm{pH}=3

and which is also saturated with

0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S},

is (Given: for

\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}}

\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48},

E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V},

\log 2=0.3,

2.303 R T / F=0.06

)

View answer and explanation

Correct option: A

142 \mathrm{~mV}

Option A (142 mV) is correct because it involves calculating the cell potential using the Nernst equation, considering the pH and solubility product.

7 Easy

7 of 396

Application +4 | -1

An ion is reduced to the element when it absorbs

6 \times {10^{20}}

electrons. The number of equivalents of the ion is :

View answer and explanation

Correct option: C

0.001

Option C (0.001) is correct because 6 × 10²⁰ electrons is equivalent to 0.001 equivalents, using the relationship 6 × 10²³ electrons = 1 equivalent.

8 Hard

8 of 396

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+.

9 Medium

9 of 396

Application +4 | -1

The standard potentials for the half reactions are given below

\begin{aligned} & Z{n^{2+}}_{(\mathrm{aq})}+2 e \rightarrow Z n_{(s)}; E^0=-0.76 \mathrm{~V} \\ & {Fe}_{(\mathrm{aq})}^{2+}+2 e \rightarrow {Fe}_{(\mathrm{s})} ;E^0=-0.44 \mathrm{~V} \end{aligned}

E^0

for the cell reaction

{Fe}^{+2}+{Zn} \rightarrow {Zn}^{+2}+{Fe} \text { is }

View answer and explanation

Correct option: D

+0.32 \mathrm{~V}

Option D (+0.32 V) is correct because the cell potential is calculated by subtracting the anode's standard reduction potential from the cathode's standard reduction potential.

10 Medium

10 of 396

Understanding +4 | -1

Fuel cell, using hydrogen and oxygen as fuels,
A. has been used in spaceship
B. has as efficiency of

40 \%

to produce electricity
C. uses aluminium as catalysts
D. is eco-friendly
E. is actually a type of Galvanic cell only

View answer and explanation

Correct option: D

A, D, E only

Option D (A, D, E only) is correct because fuel cells are used in spaceships, are eco-friendly, and are a type of galvanic cell.

11 Medium

11 of 396

Understanding +4 | -1

Given:

E_{A{g^ + }/Ag}^\mathrm{o} = 0.82\,V,\,\,E_{C{o^{2 + }}/Co}^\mathrm{o} = - 0.28\,V,

E_{C{u^{2 + }}/Cu}^\mathrm{o} = + 0.34V,\,\,E_{Z{n^{2 + }}/Zn}^\mathrm{o} = - 0.76\,V

Which metal will corrode fastest?

View answer and explanation

Correct option: D

Zn

Option D (Zn) is correct because Zinc has the lowest reduction potential, making it the most easily oxidized and thus corrode fastest.

12 Medium

12 of 396

Application +4 | -1

By passage of same current for same time, in which of the following solutions maximum amount of copper can be deposited considering

1 \mathrm{M}

concentration of each solution?

View answer and explanation

Correct option: C

aq.

\mathrm{Cu}_2 \mathrm{C}{l}_2

Option C (aq.

\mathrm{Cu}_2 \mathrm{C}{l}_2

) is correct because it has the highest equivalent weight of copper, leading to maximum deposition for the same current and time.

13 Medium

13 of 396

Application +4 | -1

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

0.550 \mathrm{~A}

deposits

0.55 \mathrm{~g}

of a certain metal in 100 minutes. Calculate the atomic mass of the metal if eq. mass

=

mol. mass/3.

View answer and explanation

Correct option: C

48.25

Option D (48.25) is correct because it is the calculated atomic mass based on Faraday's law of electrolysis and the given equivalent mass relationship.

14 Medium

14 of 396

Understanding +4 | -1

Which process involves corrosion?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because brown deposits on iron, green deposits on battery terminals, and black deposits on silver coins are all examples of corrosion processes.

15 Medium

15 of 396

Understanding +4 | -1

When applied external opposite potential in case of Daniell cell is less than

1.1 \mathrm{~V}

, then

View answer and explanation

Correct option: B

electrons flow from zinc rod to copper rod

Option B (electrons flow from zinc rod to copper rod) is correct because when the external potential is less than 1.1V, the Daniell cell functions normally, with electrons flowing from the zinc anode to the copper cathode.

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16 Medium

16 of 396

Analyzing +4 | -1

Assertion (A): Gold chloride

\left({AuCl}_3\right)

solution cannot be stored in a vessel made of copper, iron, nickel, chromium, zinc or tin.

Reason (R): Gold is a very precious metal.

View answer and explanation

Correct option: B

Both (A) and (R) true but (R) is not the correct explanation of (A).

Option B is correct because both

Cl^-

and Ar have the same electronic configuration:

1s^22s^22p^63s^23p^6

17 Hard

17 of 396

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.

18 Medium

18 of 396

Application +4 | -1

An aqueous sodium acetate solution is electrolyzed using

\text{2 A}

current for

9650 \mathrm{~sec}.

Calculate volume of gas evolved at anode at S.T.P.

View answer and explanation

Correct option: C

6.72 \mathrm{~L}

Option C (

6.72 \mathrm{~L}

) is correct because the electrolysis of aqueous sodium acetate produces carbon dioxide and ethane at the anode, and the total volume of gas evolved is calculated using Faraday's laws of electrolysis.

19 Medium

19 of 396

Remembering +4 | -1

For an electrolyte

A_x B_y,

which of the following is in accordance with kohlraush law:

View answer and explanation

Correct option: B

x\left(\lambda_m^{\infty}\right)_{A^{y+}}+y\left(\lambda_m^{\infty}\right)_{B^{x-}}

Option B (

x(\lambda_m^{\infty})_{A^{y+}}+y(\lambda_m^{\infty})_{B^{x-}}

) is correct because it represents the sum of the ionic conductivities multiplied by their respective stoichiometric coefficients.

20 Easy

20 of 396

Understanding +4 | -1

Statement-1: Salt bridge maintains electrical neutrality in two half cell.

Statement-2: Salt bridge transfer ions of one solution into other solutions.

View answer and explanation

Correct option: C

Statement-1 is true but statement-2 is false

Option C is correct. Statement-1 is true as the salt bridge maintains electrical neutrality. Statement-2 is false because the salt bridge provides ions to each half-cell, not transferring ions from one solution to the other.

21 Easy

21 of 396

Application +4 | -1

Given the limiting molar conductivity as

\Lambda _\mathrm{m}^0\left( {\mathrm{HCl}} \right) = 425.9{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

\Lambda _\mathrm{m}^0\left( {\mathrm{NaCl}} \right) = 126.4{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

\Lambda _\mathrm{m}^0\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}} \right) = 91{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

The molar conductivity , at infinite dilution, of acetic acid (

\mathrm{in}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

) will be

View answer and explanation

Correct option: B

390.5

Option B (390.5) is correct because the molar conductivity of acetic acid at infinite dilution is calculated using Kohlrausch's law of independent migration of ions.

22 Hard

22 of 396

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.

23 Hard

23 of 396

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.

24 Medium

24 of 396

Analyzing +4 | -1

STATEMENT-1 : If SOP of substance is less than

-1.23 \mathrm{~V}

and over voltage

=0 \mathrm{~V}

, then it's oxidation in it's aqueous solution is not possible at

298 \mathrm{~K}

.

STATEMENT-2 : Standard reduction potential (SRP) of water is

+1.23 \mathrm{~V}

View answer and explanation

Correct option: C

If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE

Option C (If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE) is correct because the statement about SOP is correct, but the SRP of water is not +1.23V.

25 Hard

25 of 396

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.

26 Medium

26 of 396

Understanding +4 | -1

For a given cell reaction;

Cr + 3{H_2}O + OC{l^ - } \to C{r^{+3}}+3C{l^ - } + 6O{H^ - },

the species undergoing reduction is :

View answer and explanation

Correct option: C

OC{l^ - }

Option C (OCl-) is correct because in the given reaction, the oxidation number of Cl in OCl- decreases from +1 to -1, indicating reduction.

27 Medium

27 of 396

Application +4 | -1

Given the molar conductance of sodium butyrate, sodium chloride and hydrogen chloride as 83, 127 and

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

25^{\circ} \mathrm{C}

respectively. Calculate the molar conductance of butyric acid at infinite dilution.

View answer and explanation

Correct option: A

382

Option A (382) is correct because the molar conductance of butyric acid at infinite dilution is calculated using Kohlrausch's law.

28 Medium

28 of 396

Remembering +4 | -1

The equation representing kohlrausch law from the following is (

{V^ + } =

No. of cations,

{V^ - } =

No. of anions)

View answer and explanation

Correct option: B

\lambda _m^\mathrm{o} = {V^ + }\lambda _ + ^\mathrm{o} + {V^ - }\lambda _ - ^\mathrm{o}

Option B (

\lambda _m^\mathrm{o} = {V^ + }\lambda _ + ^\mathrm{o} + {V^ - }\lambda _ - ^\mathrm{o}

) is correct because it represents Kohlrausch's Law, stating that the limiting molar conductivity of an electrolyte is the sum of the contributions of the individual ions.

29 Hard

29 of 396

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.

30 Medium

30 of 396

Application +4 | -1

Time taken to completely (in hrs) decompose

36 {~g}

water by passing

3 {~A}

current is :

View answer and explanation

Correct option: A

35.8 {hrs}

Option A (35.8 hrs) is correct because the time taken to decompose 36 g of water by passing 3 A current is calculated using Faraday's laws of electrolysis.

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Key Takeaways

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