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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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121 Medium
121 of 396
Application +4 | -1
An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g0.108 \mathrm{~g} more at the end of the electrolysis. The volume of oxygen evolved at STP is
View answer and explanation
Correct option: C
5.6 mL5.6 \mathrm{~mL}
Option C (5.6 mL5.6 \mathrm{~mL}) is correct because the volume of oxygen evolved is calculated using Faraday's laws of electrolysis, relating the mass of silver deposited to the volume of oxygen produced.
122 Hard
122 of 396
Application +4 | -1
Salts of metals X,Y{X}, {Y} and Z{Z} are electrolysed under identical condition using same quantity of electricity. It was observed that 4.2 g4.2 {~g} of X,5.4 g{X}, 5.4 {~g} of Y{Y} and 19.2 g19.2 {~g} of Z{Z} were deposited at respective cathode. If the atomic weights of X,Y,Z{X}, {Y}, {Z} are 7 , 27 and 64 respectively, then their ratio of valencies is
View answer and explanation
Correct option: B
1:3:21: 3: 2
Option B (1:3:21: 3: 2) is correct because it represents the ratio of valencies calculated using Faraday's laws of electrolysis.
123 Easy
123 of 396
Analyzing +4 | -1
Electrolysis of aq. Solution of LiCl\mathrm{LiCl} shows :
View answer and explanation
Correct option: C
pH>7\mathrm{pH} > 7
Option C (pH>7\mathrm{pH} > 7) is correct because during the electrolysis of aqueous LiCl, water is reduced at the cathode, producing hydroxide ions and increasing the pH.
124 Easy
124 of 396
Application +4 | -1
Maximum number of mole of oxygen gas that can be obtained by the electrolytic decomposition of 90 g of water will be:
View answer and explanation
Correct option: B
2.5
Option B (2.5) is correct because the electrolytic decomposition of 90g of water (5 moles) produces 2.5 moles of oxygen gas, based on the stoichiometry of the reaction.
125 Medium
125 of 396
Understanding +4 | -1
In the electrochemical cell,
A(s)+B2+(aq)⟶A2+(aq)+B(s){A}({s})+{B}^{2+}({aq}) \longrightarrow {A}^{2+}({aq})+{B}({s})

Increase in the concentration of A2+\mathrm{A}^{2+}
View answer and explanation
Correct option: A
Decreases EMF of the cell
Option A (Decreases EMF of the cell) is correct because according to the Nernst equation, increasing the concentration of product (A2+A^{2+}) decreases the cell potential.
126 Hard
126 of 396
Application +4 | -1
What pressure (bar) of H2\mathrm{H}_2 would be required to make emf of hydrogen electrode zero in pure water at 25∘C25^{\circ} \mathrm{C}?
View answer and explanation
Correct option: A
10−1410^{-14}
Option A (10−1410^{-14}) is correct because it is derived from the Nernst equation, setting the cell potential to zero and solving for the hydrogen pressure.
127 Hard
127 of 396
Application +4 | -1
Given the ionic equivalent conductivities for the following ions:

λ∘eq K+=73.5 cm2ohm−1eq−1\lambda^{\circ}{ }_{\text {eq }} \mathrm{K}^{+}=73.5 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

λ∘eq Al3+=149 cm2ohm−1eq−1\lambda^{\circ}{ }_{\text {eq }} \mathrm{Al}^{3+}=149 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

λ∘eq SO42−=85.8 cm2ohm−1eq−1\lambda^{\circ}{ }_{\text {eq }} \mathrm{SO}_4{ }^{2-}=85.8 \mathrm{~cm}^2 \mathrm{ohm}^{-1} \mathrm{eq}^{-1}

The Λeq ∘\Lambda_{\text {eq }}^{\circ} for potash alum (K2SO4⋅Al2(SO4)3\left(\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\right.. 24H2O24 \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.
128 Hard
128 of 396
Analyzing +4 | -1
Cu+2HCl→CuCl2+H2(g)\mathrm{Cu} + 2\mathrm{HCl} \to \mathrm{CuC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

[ECu2+/Cu∘=+0.34V]\left[ {E_{C{u^{2 + }}/Cu}^ \circ = + 0.34V} \right]

Zn+2HCl→ZnCl2+H2(g)\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

[EZn2+/Zn∘=−0.76V]\left[ {E_{Z{n^{2 + }}/Zn}^ \circ = - 0.76V} \right]

Ag+2HCl→AgCl+12H2(g)\mathrm{Ag} + 2\mathrm{HCl} \to \mathrm{AgCl} + \frac{1}{2}{\mathrm{H}_2}\left( \mathrm{g} \right)

[EAg+/Ag∘=+0.80V]\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 Ecell∘E_{cell}^\circ value, while the other reactions have negative values.
129 Medium
129 of 396
Remembering +4 | -1
The standard electrode potential is measured by
View answer and explanation
Correct option: B
Voltmeter
Option B (Voltmeter) is correct because a voltmeter is used to measure the standard electrode potential.
130 Medium
130 of 396
Application +4 | -1
The limiting molar conductivity of calcium acetate is
View answer and explanation
Correct option: B
200.8200.8
Option B (200.8200.8) is correct because it is calculated by summing the limiting molar conductivities of the constituent ions.
131 Medium
131 of 396
Understanding +4 | -1
Select the appropriate statement for Daniell cell containing Cu2+\mathrm{Cu}^{2+} and Zn2+\mathrm{Zn}^{2+} ions.
View answer and explanation
Correct option: C
Ecell E_{\text {cell }} increases with decrease in the concentration of Zn2+\mathrm{Zn}^{2+} ions
Option C is correct because, according to the Nernst equation, the cell potential (EcellE_{\text{cell}}) increases when the concentration of Zn2+\mathrm{Zn}^{2+} ions decreases.
132 Medium
132 of 396
Application +4 | -1
0.5 F0.5 \mathrm{~F} charge is passed through 1 L,1 M  CuCl2(aq)1 \mathrm{~L}, 1 \mathrm{~M}\,\, {CuCl}{ }_2(\mathrm{aq}), using inert electrodes. Final concentration of Cu2+{Cu}^{2+} in the solution will be :-
View answer and explanation
Correct option: B
0.75 M0.75 \mathrm{~M}
Option B (0.75 M0.75 \mathrm{~M}) is correct because 0.25 moles of Cu2+Cu^{2+} are deposited, leaving 0.75 moles in the 1L solution.
133 Medium
133 of 396
Understanding +4 | -1
Equivalent conductivity of Fe2(SO4)3\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 is related to molar conductivity by the expression:
View answer and explanation
Correct option: D
Λeq=Λm/6\Lambda_{\mathrm{eq}}=\Lambda_{\mathrm{m}} / 6
Option D is correct because the relationship between equivalent conductivity and molar conductivity is given by Λeq=Λm/n\Lambda_{eq} = \Lambda_m / n, where n is the valence factor. For Fe2(SO4)3Fe_2(SO_4)_3, n = 6.
134 Hard
134 of 396
Analyzing +4 | -1
Find the standard cell potential involving the cell reaction:

In2++Cu+2→In+3+Cu+, at 298 K{In}^{2+}+{Cu}^{+2} \rightarrow {In}^{+3}+{Cu}^{+} \text {, at } 298 {~K}

Given:

ECu+2/Cu+0=X1 V,EIn3+/In+0=X2 VEIn2+/In+0=X3 V{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
X1+X3−2X2X_1+X_3-2 X_2
The correct answer is C (X1+X3−2X2X_1+X_3-2 X_2) because it correctly applies Hess's law to calculate the standard cell potential from the given reduction potentials.
135 Hard
135 of 396
Application +4 | -1
0.04 NN solution of a weak acid has conductivity 4.23×10−4mhocm−14.23 \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 ……mhocm2eq.−1{\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.
136 Easy
136 of 396
Remembering +4 | -1
Which metal is most readily corroded in moist air?
View answer and explanation
Correct option: B
Iron
Option B (Iron) is correct because iron is known to corrode readily in moist air, forming rust, which is a hydrated iron oxide.
137 Medium
137 of 396
Application +4 | -1
Total charge required for the oxidation of two moles Mn3O4{Mn}_3 {O}_4 into MnO42−{MnO}_4^{2-} in presence of alkaline medium is
View answer and explanation
Correct option: C
20 F20 \mathrm{~F}
Option C (20 F20 \mathrm{~F}) is correct because the total charge required is calculated based on the change in oxidation state of Mn during the reaction.
138 Easy
138 of 396
Remembering +4 | -1
Galvanised iron sheets are coated with
View answer and explanation
Correct option: C
Zinc
Option C (Zinc) is correct because galvanized iron sheets are coated with zinc to prevent rusting.
139 Medium
139 of 396
Remembering +4 | -1
The most durable metal plating on iron to protect against corrosion is
View answer and explanation
Correct option: D
Zinc plating
Option D (Zinc plating) is correct because zinc has a higher oxidation potential than iron, providing sacrificial protection against corrosion.
140 Easy
140 of 396
Application +4 | -1
What weight of copper will be deposited by passing 2 faraday of electricity through a solution of Cu(II) salt?
View answer and explanation
Correct option: B
63.5 g
Option B (63.5 g) is correct because 2 Faradays of electricity will deposit 1 mole of Cu, which is equal to its atomic weight.
141 Hard
141 of 396
Analyzing +4 | -1
Assertion (A) At the end of electrolysis using platinum electrodes, an aqueous solution of CuSO4\mathrm{CuSO}_4 turns colourless.

Reason (R) During the electrolysis CuSO4\mathrm{CuSO}_4 changes to Cu(OH)2\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.
142 Medium
142 of 396
Application +4 | -1
Three faradays of electricity is passed through three electrolytic cells connected in series containing Ag+,Ca2+{Ag}^{+}, {Ca}^{2+} and Al3+{Al}^{3+} ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is
View answer and explanation
Correct option: C
6:3:26: 3: 2
Option C (6:3:2) is correct because, according to Faraday's second law, the gram equivalents are equal, and the molar ratio is found by dividing by the n-factor.
143 Medium
143 of 396
Application +4 | -1
Also the [H+{\mathrm{H}^ + }] for problem 9 using the same data is :
View answer and explanation
Correct option: A
0.00133 M
Option A (0.00133 M) is correct because it is the calculated hydrogen ion concentration based on the given data and the assumption of a weak acid.
144 Medium
144 of 396
Application +4 | -1
At 18∘C{18^ \circ }\mathrm{C} the conductance of H+{\mathrm{H}^ + } and CH3COO−\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - } at infinite dilution are 315 and 35 mhocm2eq−1{\rm{mho {\kern 1pt} {\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{e}{\mathrm{q}^{ - 1}} respectively. The equivalent conductivity of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} at infinite dilution is ……mhocm2eq.−1{\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.
145 Easy
145 of 396
Application +4 | -1
The value of electrode potential (10−4 M)H+∣H2(1 atm)∣Pt\left(10^{-4} \, \mathrm{M}\right) {H}^{+}\left|{H}_2(1 \mathrm{~atm})\right| {Pt} at 298 K298 \mathrm{~K} would be
View answer and explanation
Correct option: A
−0.236 V-0.236 \mathrm{~V}
Option A (-0.236 V) is correct because it is calculated using the Nernst equation for the given hydrogen electrode conditions.
146 Medium
146 of 396
Application +4 | -1
Assume that during electrolysis of AgNO3AgN{O_3}, only H2O{H_2}O is electrolysed and O2{O_2} is formed as: 2H2O→4H++O2+4e−2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - } . O2{O_2} formed at NTP due to passage of 2A of current for 965 s is
View answer and explanation
Correct option: A
0.112 L
Option A (0.112 L) is correct because it correctly calculates the volume of oxygen formed at NTP using Faraday's laws of electrolysis.
147 Medium
147 of 396
Application +4 | -1
During electrolysis of an aqueous solution of Cu2+\mathrm{C}{\mathrm{u}^{2 + }} sulphate, 0.635 g of copper was deposited at cathode. The amount of electricity consumed in coulomb is :
View answer and explanation
Correct option: A
1930
Option A (1930) is correct because it is calculated using Faraday's law of electrolysis, relating the mass of copper deposited to the amount of electricity consumed.
148 Medium
148 of 396
Analyzing +4 | -1
Assertion: More negative the electrode potential greater is the power to act as oxidising agent.

Reason: As the electrode potential becomes more negative there is greater tendency to undergo oxidation.
View answer and explanation
Correct option: D
Assertion is wrong, but reason is correct statement.
Option D (Assertion is wrong, but reason is correct statement) is correct because a more negative electrode potential indicates a greater tendency to undergo oxidation, thus a stronger reducing agent, not oxidizing agent.
149 Medium
149 of 396
Understanding +4 | -1
Mark the false statement?
View answer and explanation
Correct option: C
Nernst equation for single electrode potential is
E=Eo−RTnFlog⁡aMn+E=E^o-\frac{R T}{n F} \log _a M^{n+}
Option C is the correct representation of the Nernst equation for single electrode potential, relating it to the standard electrode potential and the activity of the metal ion.
150 Easy
150 of 396
Analyzing +4 | -1
The oxidation potential values of A, B, C and D are -0.03, +0.108 V,-0.07 V and +0.1 V respectively. The non-spontaneous cell reaction takes place between
View answer and explanation
Correct option: A
A and B
Option A (A and B) is correct because for a non-spontaneous cell reaction, the cell potential must be negative. This occurs when the electrode with the lower oxidation potential acts as the anode, which is the case with A and B.
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