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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 61-90 of 396 questions

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61 Medium
61 of 396
Analyzing +4 | -1
STATEMENT-1 : We cannot add the electrode potential in order to get electrode potential of third electrode if no. of moles of electrons exchanged are not same.
STATEMENT-2 : Electrode potential is an extensive property.
View answer and explanation
Correct option: C
If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE
Option C is correct. Electrode potential is an intensive property, and we can only add the Gibbs free energy change to obtain the overall Gibbs free energy change for the third electrode.
62 Easy
62 of 396
Understanding +4 | -1
Which of the following reactions occurs at the cathode?
View answer and explanation
Correct option: D
Cu2++2e−→Cu\mathrm{Cu}^{2+}+2 {e}^{-} \rightarrow \mathrm{Cu}
Option D (Cu2++2e−→Cu\mathrm{Cu}^{2+}+2 {e}^{-} \rightarrow \mathrm{Cu}) is correct because reduction occurs at the cathode, and this equation represents the gain of electrons (reduction) by copper ions.
63 Medium
63 of 396
Remembering +4 | -1
The potential difference of two electrodes in a galvanic cell is known as
View answer and explanation
Correct option: A
EMF
The correct answer is A (EMF) because the potential difference between two electrodes in a galvanic cell is defined as the electromotive force (EMF).
64 Medium
64 of 396
Remembering +4 | -1
What is the density of solution of sulphuric acid used as on electrolyte in lead accumulator
View answer and explanation
Correct option: B
1.2  g  L−11.2\,\,{\text{g}}\,\,{{\text{L}}^{ - 1}}
Option B (1.2  g L−11.2\,\,{\text{g}}\,{{\text{L}}^{ - 1}}) is incorrect. The correct density is approximately 1.28 g/mL or 1280 g/L for a fully charged lead-acid battery electrolyte.
65 Easy
65 of 396
Remembering +4 | -1
A depolariser used in dry cell batteries is :
View answer and explanation
Correct option: B
Manganese dioxide
Option B (Manganese dioxide) is correct because manganese dioxide acts as a depolarizer in dry cell batteries.
66 Medium
66 of 396
Application +4 | -1
The standard potentials of OCl−/Cl−{OCl}^{-} / {Cl}^{-} and Cl−/Cl2{Cl}^{-} / {Cl}_2 are 0.94 V0.94 \mathrm{~V} and −1.36 V,-1.36 \mathrm{~V}, respectively. The E∘E^{\circ} value of OCl−/Cl2{OCl}^{-} / {Cl}_2 will be
View answer and explanation
Correct option: D
0.52 V0.52 \mathrm{~V}
Option D (0.52 V) is correct because it is the calculated standard potential using the given standard potentials and the appropriate stoichiometric coefficients.
67 Hard
67 of 396
Application +4 | -1
The resistance of 1N solution of acetic is 250Ω ,when measured in a cell having a cell constant of 1.15cm−11.15{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}} .The equivalent conduction (in ohm−1cm2equiv−1\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.
68 Hard
68 of 396
Application +4 | -1
An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO3\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
Pt(s),H2(g)∣H+(1M)∣∣Ag+(aq)∣Ag(s)\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 Ecell∘E_{\mathrm{cell}}^ \circ = 0.80 V, what is the percentage of Ag in the alloy?
[At 25∘C{25^ \circ }C , RT /F = 0.06]
View answer and explanation
Correct option: D
50
Option D (50) is correct because the percentage of Ag in the alloy is calculated using the Nernst equation and the given cell potential.
69 Easy
69 of 396
Analyzing +4 | -1
Two platinum electrodes were immersed in a solution of cupric sulphate and electric current passed through the solution. After sometime, it was found that the colour of copper sulphate disappeared with evolution of gas at the electrode. The colourless solution contains:
View answer and explanation
Correct option: D
Sulphuric acid
Option D (Sulphuric acid) is correct because during electrolysis of CuSO4CuSO_4, copper is deposited at the cathode and oxygen is evolved at the anode, leaving sulphuric acid in the solution.
70 Medium
70 of 396
Analyzing +4 | -1
Statement I: In case of Hg\mathrm{Hg} cathode Na\mathrm{Na} undergoes deposition
Statement II: Nature of electrode determines the deposition order
View answer and explanation
Correct option: A
If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A (If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the nature of the electrode influences the deposition order, and Na deposition is favored with a Hg cathode.
71 Easy
71 of 396
Understanding +4 | -1
Lithium is generally used as an electrode in high energy density batteries. This is because:
View answer and explanation
Correct option: B
Lithium has quite high negative reduction potential
Option B (Lithium has quite high negative reduction potential) is correct because its high negative reduction potential indicates it is easily oxidized, making it a good electrode material.
72 Medium
72 of 396
Application +4 | -1
The conductivity of N/50N/50 solution of KCl\mathrm{KCl} in a cell at 250C is 0.002765mhocm−10.002765{\kern 1pt} {\kern 1pt} {\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^{ - 1}} . If the resistance of a cell containing this solution is 400 ohm, the cell constant is :
View answer and explanation
Correct option: B
1.106cm−11.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}
Option B (1.106cm−11.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}) is correct because the cell constant is the product of the conductivity and the resistance.
73 Easy
73 of 396
Analyzing +4 | -1
Which one of the following solutions has highest conductance power?
View answer and explanation
Correct option: D
0.1MHCl0.1M{\kern 1pt} {\kern 1pt} \mathrm{HCl}
Option D (0.1M HCl) has the highest conductance power because HCl is a strong electrolyte and H+ ions have high mobility due to Grotthuss mechanism.
74 Easy
74 of 396
Remembering +4 | -1
If ϕ denotes standard reduction potential, which is true:
View answer and explanation
Correct option: A
Ecell∘=ϕR−ϕLE_{\mathrm{cell}}^ \circ = {\phi _R} - {\phi _L}
Option A is correct because the standard cell potential is the difference between the standard reduction potential of the right-hand electrode (cathode) and the left-hand electrode (anode).
75 Medium
75 of 396
Application +4 | -1
For the given reactions

Sn2++2e−→Sn\mathrm{S n}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{S n}

Sn4++2e−→Sn\mathrm{S n}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{S n}

The electrode potentials are; ESn2+/Sn0=−0.140 V\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^0=-0.140\text{ V} and ESn4+/Sn∘=0.010 V.\mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}}^{\circ}=0.010 \mathrm{~V}. The magnitude of standard electrode potential for Sn4+/Sn2+\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+} i.e. ESn4+/Sn2+\mathrm{E}_{\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}} is _____ ×10−2V.\times{10^{-{2}} \mathrm{V}}.
View answer and explanation
Correct option: C
16
Option C (16) is correct because the standard electrode potential for Sn4+/Sn2+ is calculated using Gibbs free energy changes for the given reactions.
76 Easy
76 of 396
Application +4 | -1
If ECu2+/Cu∘=+0.34  V{E}_{{Cu}^{2+} / {Cu}}^{\circ}=+0.34 \,{~V} then reduction potential of Cu2+/Cu{Cu}^{2+} / {Cu} electrode at pH=14{pH}=14 is (Ksp\left({K}_{{sp}}\right. of Cu(OH)2=1{Cu}({OH})_2=1 ×10−19\times 10^{-19} )
View answer and explanation
Correct option: B
−0.22  V-0.22 \,{~V}
Option B (-0.22 V) is correct because it is calculated using the Nernst equation, considering the solubility product of copper(II) hydroxide.
77 Hard
77 of 396
Application +4 | -1
Calculate the volume of H2{\mathrm{H}_2} gas at NTP obtained by passing 4 A through acidified H2O{\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.
78 Medium
78 of 396
Application +4 | -1
How much metal will be deposited when a current of 12 ampere with 75%75 \% efficiency is passed through the cell for 3 h3 \mathrm{~h}? (Given: Z=4×10−4{Z}=4 \times 10^{-4})
View answer and explanation
Correct option: B
38.8 g38.8 \mathrm{~g}
Option C (38.8 g) is correct because it is calculated using the formula for mass deposited in electrolysis, considering the efficiency of the process.
79 Medium
79 of 396
Application +4 | -1
For the galvanic cell

Cr(s)∣Cr3+(2.23×10−7M)∥Ag+(0.0264M)∣Ag(s)\mathrm{Cr}(\mathrm{s})\left|\mathrm{Cr}^{3+}\left(2.23 \times 10^{-7} \mathrm{M}\right) \| \mathrm{Ag}^{+}(0.0264 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s}) at 25∘C25^{\circ} \mathrm{C}

Calculate the emf of the cell.

Given: E∘(Ag+/Ag)=0.80 V\mathrm{E}^{\circ}\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)=0.80 \mathrm{~V}

E∘(Cr+3/Cr)=−0.74 V\mathrm{E}^{\circ}\left(\mathrm{Cr}^{+3} / \mathrm{Cr}\right)=-0.74 \mathrm{~V}
View answer and explanation
Correct option: A
1.5771.577
Option A (1.577) is correct because the emf of the cell is calculated using the Nernst equation with the given standard electrode potentials and concentrations.
80 Medium
80 of 396
Application +4 | -1
Using a particular conductivity cell, resistance offered is 60Ω60 \Omega. When the conductivity cell with area twice the earlier and having 75%75 \% more distance between the electrodes is used, resistance offered will be
View answer and explanation
Correct option: C
52.5Ω52.5 \Omega
Option C (52.5Ω52.5 \Omega) is correct because it correctly calculates the new resistance based on changes in length and area using the formula R=ρlAR = \rho \frac{l}{A}.
81 Medium
81 of 396
Understanding +4 | -1
On which factor, electrical conductivity of electrolytic cell doesn’t depend
View answer and explanation
Correct option: D
Nature of electrode
Option D is correct because the nature of the electrode material does not directly affect the electrical conductivity of the electrolytic cell; conductivity depends on electrolyte properties and temperature.
82 Medium
82 of 396
Analyzing +4 | -1
During electrolysis of sulphuric acid, which of the following processes is possible at anode?

A. 2H2O(l)⟶O2( g)+4H+(aq)+4e−Ecell ⊖=+1.23 V2 {H}_2 {O}({l}) \longrightarrow {O}_2({~g})+4 {H}^{+}({aq})+4 {e}^{-} \stackrel{\ominus}{{E}_{\text {cell }}}=+1.23 {~V}

B. 2SO42−(aq)⟶S2O82−(aq)+2e−Ecell ⊖=1.96 V2 {SO}_4^{2-}({aq}) \longrightarrow {S}_2 {O}_8^{2-}({aq})+2 {e}^{-} \stackrel{\ominus}{{E}_{\text {cell }}}= 1.96 {~V}

Choose the correct option based on following statements.

(i) Process A is preferred at higher concentration of sulphuric acid.

(ii) Process B is preferred at higher concentration of sulphuric acid.

(iii) Process A is preferred for dilute sulphuric acid.

(iv) Process B is preferred for dilute sulphuric acid.

(v) Both A and B are equally possible at higher concentration.
View answer and explanation
Correct option: B
(iii) and (ii)
Option B ((iii) and (ii)) is correct because process A is favored in dilute sulfuric acid, while process B is favored in concentrated sulfuric acid.
83 Medium
83 of 396
Application +4 | -1
The amount of charge that must be passed through a solution containing Cu2+\mathrm{Cu}^{2+} in order to deposit 1 g1 \mathrm{~g} atom of copper (63.55)(63.55) is
View answer and explanation
Correct option: D
193000 C193000 \mathrm{~C}
The correct answer is D (193000 C) because depositing 1 gram atom of copper (Cu2+) requires 2 Faradays of charge, which is 2 * 96500 Coulombs.
84 Medium
84 of 396
Application +4 | -1
A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has emf of 2.7 V. If the standard reduction potential of copper electrode is +0.34 V then that of the magnesium electrode is
View answer and explanation
Correct option: B
−2.36V- 2.36\mathrm{V}
Option B (-2.36V) is correct because the cell EMF is the difference between the reduction potentials of the cathode (copper) and the anode (magnesium).
85 Medium
85 of 396
Application +4 | -1
The Gibbs energy change (in KJ) for the given reaction at [Cu2+]=[Sn2+]=1M\left[ {C{u^{2 + }}} \right] = \left[ {S{n^{2 + }}} \right] = 1M and 298 K is Cu(s)+Sn2+(aq)→Cu2+(aq)+SN(s); (ESn2+/Sn0=−0.16V, ECu2+/Cu0=0.34V)Cu\left( s \right) + S{n^{2 + }}\left( {aq} \right) \to C{u^{2 + }}\left( {aq} \right) + SN\left( s \right);\,\left( {E_{S{n^{2 + }}/Sn}^0 = - 0.16V,\,E_{C{u^{2 + }}/Cu}^0 = 0.34V} \right)
Take, F=96500   C  mol−1F = 96500\,\,\,C\,\,mo{l^{ - 1}})
View answer and explanation
Correct option: A
96.5 KJ
Option A (96.5 KJ) is correct because it is the Gibbs free energy change calculated using the standard reduction potentials and Faraday's constant.
86 Medium
86 of 396
Understanding +4 | -1
When an acid cell is charged, then
View answer and explanation
Correct option: A
voltage of cell increases
Option A (voltage of cell increases) is correct because during charging, electrical energy is converted into chemical energy, increasing the cell's potential.
87 Medium
87 of 396
Understanding +4 | -1
When a lead storage battery is charged
View answer and explanation
Correct option: C
Sulphuric acid is regenerated
Option C (Sulphuric acid is regenerated) is correct because during the charging of a lead storage battery, lead sulfate is converted back to lead and lead dioxide, regenerating sulfuric acid.
88 Hard
88 of 396
Analyzing +4 | -1
A student made the following observations in the laboratory,
i) Clean copper metal did not react with 1 molar Pb(NO3)2\mathrm{Pb}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2} solution.
ii) Clean lead metal dissolved in a 1 molar AgNO3\mathrm{AgN}{\mathrm{O}_3} solution and crystals of Ag\mathrm{Ag} metal appeared.
iii) Clean silver metal did not react with 1 molar Cu(NO3)2\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
Pb,Cu,Ag\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.
89 Medium
89 of 396
Application +4 | -1
How much current is necessary to produce H2\mathrm{H}_2 gas at the rate of 1 cm31 \mathrm{~cm}^3 per second under STP.
View answer and explanation
Correct option: C
8.61amp8.61 \mathrm{amp}
Option C (8.61 amp) is correct because it is calculated using Faraday's laws of electrolysis, relating the volume of hydrogen gas produced to the current required.
90 Hard
90 of 396
Application +4 | -1
The standard emf for the cell reaction,

2Cu+(aq)→2Cu(s)+Cu+(aq)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
1.2×1061.2 \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.
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