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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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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 230 questions

1 Medium

1 of 230

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 230

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 Medium

3 of 230

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.

4 Medium

4 of 230

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.

5 Medium

5 of 230

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.

6 Medium

6 of 230

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.

7 Medium

7 of 230

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.

8 Medium

8 of 230

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.

9 Medium

9 of 230

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.

10 Medium

10 of 230

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.

11 Medium

11 of 230

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.

12 Medium

12 of 230

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.

13 Medium

13 of 230

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

14 Medium

14 of 230

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.

15 Medium

15 of 230

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.

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16 Medium

16 of 230

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.

17 Medium

17 of 230

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.

18 Medium

18 of 230

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.

19 Medium

19 of 230

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.

20 Medium

20 of 230

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.

21 Medium

21 of 230

Application +4 | -1

The equivalent conductivity of

0.05\,\,N

solution of a monobasic acid

{\text{15}}{\text{.8}}\,\,{\text{mho}}\,\,\,{\text{c}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{ - 1}}.

If equivalent conductivity of the acid at infinite dilution is

350\,\,{\text{mho}}\,\,\,{\text{c}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{ - 1}}

at the same temperature. What is the degree of dissociation and dissociation constant of acid?

View answer and explanation

Correct option: A

0.04514,2.06 \times {10^{ - 4}}

Option A (0.04514, 2.06 x 10⁻⁴) is correct because it is calculated by first finding the degree of dissociation using the ratio of equivalent conductivities and then using that to find the dissociation constant.

22 Medium

22 of 230

Application +4 | -1

{18^ \circ }\mathrm{C}

the conductance of

{\mathrm{H}^ + }

and

\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }

at infinite dilution are 315 and 35

{\rm{mho {\kern 1pt} {\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{e}{\mathrm{q}^{ - 1}}

respectively. The equivalent conductivity of

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

at infinite dilution is ……

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

View answer and explanation

Correct option: A

350

Option A (350) is correct because it is the sum of the ionic conductances of H+ and CH3COO- at infinite dilution.

23 Medium

23 of 230

Application +4 | -1

{E}_{{A\ell}^{+3} / {A} \ell}=-1.66 {~V} \& {~K}_{{sp}}

{A} \ell({OH})_3=1.0 \times 10^{-33}

Reduction potential of the above couple at

{pH}=14

View answer and explanation

Correct option: A

-2.31 {~V}

Option A (-2.31 V) is correct because the reduction potential at pH=14 is calculated using the Nernst equation, considering the solubility product of Al(OH)₃.

24 Medium

24 of 230

Remembering +4 | -1

The composition of rust is

View answer and explanation

Correct option: A

F{e_2}{O_3} \cdot x{H_2}O

Option A (

F{e_2}{O_3} \cdot x{H_2}O

) is correct because rust is hydrated iron(III) oxide, where 'x' represents a variable number of water molecules.

25 Medium

25 of 230

Remembering +4 | -1

According to Faraday’s first law of electrolysis mass of substances liberated is equal to

View answer and explanation

Correct option: B

Option B (eQ) is correct because Faraday's first law of electrolysis states that the mass of a substance liberated is directly proportional to the charge (Q) passed, and 'e' represents the electrochemical equivalent.

26 Medium

26 of 230

Application +4 | -1

Iron powder is added to

1.0 \mathrm{~M}

solution of

{CdCl}_2

298 \mathrm{~K}

. The reaction occurring is

{Cd}^{2+}(a q)+{Fe}(s) \rightarrow {Cd}(s)+{Fe}^{2+}(a q)

. If the standard potential of a cell producing this reaction is

0.037 \mathrm{~V}

, the concentrations of

{Cd}^{2+}

and

{Fe}^{2+}

ions in the above reaction at equilibrium respectively will be (Hint:-

\left.10^{1.3}=20\right)

View answer and explanation

Correct option: A

0.05 \mathrm{~M}, 0.95 \mathrm{~M}

Option A (0.05 M, 0.95 M) is correct because it is calculated using the Nernst equation at equilibrium and the given standard potential.

27 Medium

27 of 230

Remembering +4 | -1

Hydrogen peroxide is obtained by electrolysis of

View answer and explanation

Correct option: B

Sulphurie acid

Correct Answer + Core Reason: Option B (Electrolysis of

H_2SO_4

) is correct because the electrolysis of sulfuric acid produces hydrogen peroxide.

28 Medium

28 of 230

Remembering +4 | -1

Cell constant of a conductivity cell is usually derived by using a solution of :

View answer and explanation

Correct option: A

\mathrm{KCl}

Option A (

\mathrm{KCl}

) is correct because KCl solutions have well-established conductivity values at various concentrations, making them suitable for calibrating conductivity cells.

29 Medium

29 of 230

Analyzing +4 | -1

Copper can be deposited from acidified copper sulphate and alkaline cuprous cyanide. If the same current is passed for a definite time:

View answer and explanation

Correct option: B

The mass of copper deposited from alkaline cuprous cyanide will be higher

Option B is correct because the mass of copper deposited is proportional to the equivalent weight, which is higher for Cu+ in alkaline cuprous cyanide than for Cu2+ in acidified copper sulfate.

30 Medium

30 of 230

Application +4 | -1

One Faraday of electricity is passed through molten

{Al}_2 {O}_3

, aqueous solution of

{CuSO}_4

and molten

{NaCl}

taken in three different electrolytic cells connected in series. The mole ratio of

{Al}, {Cu}

and

{Na}

deposited at the respective cathode is

View answer and explanation

Correct option: A

2: 3: 6

The correct answer is 2:3:6 because one Faraday deposits 1/3 mole of Al, 1/2 mole of Cu, and 1 mole of Na, leading to the ratio 2:3:6.

Revision notes and key takeaways

Review ELECTRO CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 230 single-correct-answer MCQs on ELECTRO CHEMISTRY.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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