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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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Practice format and chapter quiz

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 31-60 of 230 questions
31 Medium
31 of 230
Understanding +4 | -1
The increase in equivalent conductivity of a weak electrolyte solution with dilution is attributed to :
View answer and explanation
Correct option: C
Both (a) and (b)
Option C (Both (a) and (b)) is correct because the increase in equivalent conductivity of a weak electrolyte upon dilution is due to both an increase in the degree of dissociation and an increase in ionic mobility.
32 Medium
32 of 230
Application +4 | -1
Equivalence conductance at infinite dilution of NH4Cl,NaOH{NH}_4 {Cl}, {NaOH} and NaCl{NaCl} are 129.8,217.4129.8, 217.4 and 108.9 Ω−1 cm2 mol−1,108.9 \, \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, respectively. If the equivalent conductance of 0.01 N0.01 \mathrm{~N} solution of NH4OH{NH}_4 {OH} is 9.532 Ω−1 cm2 mol−1,9.532 \, \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, then the degree of dissociation of NH4OH{NH}_4 {OH} at this temperature is
View answer and explanation
Correct option: C
4.0%4.0 \%
Option C (4.0%4.0 \% ) is correct because it is the degree of dissociation of NH4OH{NH}_4 {OH} calculated using Kohlrausch's law.
33 Medium
33 of 230
Application +4 | -1
The reduction potential of a hydrogen electrode having pH\mathrm{pH} value 'x{x}' and y{y} atm pressure of H2{H}_2-gas at 25∘C25^{\circ} \mathrm{C} is correctly given as
View answer and explanation
Correct option: C
−0.0295[2x+log⁡y]-0.0295[2 x+\log y]
Option C (−0.0295[2x+log⁡y]-0.0295[2 x+\log y]) is correct because it correctly applies the Nernst equation for the hydrogen electrode, considering both pH and pressure.
34 Medium
34 of 230
Analyzing +4 | -1
Statement I: When iron rod is immersed in NaCl\mathrm{NaCl} solution such that half of its length is corroded and other half part when exposed to air is not corroded.

Statement II: Metal portion with less access to O2\mathrm{O}_2 corrodes.
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
The correct answer is A because both statements are correct, and Statement II correctly explains why the iron rod corrodes more where it has less access to oxygen.
35 Medium
35 of 230
Application +4 | -1
A current of 9.65 flowing for 10 minute deposits 3.0 g of a metal. The equivalent weight of the metal is
View answer and explanation
Correct option: C
50
Option C (50) is correct because it is the calculated equivalent weight of the metal using Faraday's laws of electrolysis.
36 Medium
36 of 230
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.
37 Medium
37 of 230
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).
38 Medium
38 of 230
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.
39 Medium
39 of 230
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.
40 Medium
40 of 230
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.
41 Medium
41 of 230
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.
42 Medium
42 of 230
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.
43 Medium
43 of 230
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.
44 Medium
44 of 230
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.
45 Medium
45 of 230
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}.
46 Medium
46 of 230
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.
47 Medium
47 of 230
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.
48 Medium
48 of 230
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.
49 Medium
49 of 230
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).
50 Medium
50 of 230
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.
51 Medium
51 of 230
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.
52 Medium
52 of 230
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.
53 Medium
53 of 230
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.
54 Medium
54 of 230
Understanding +4 | -1
Chlorine gas is passed into a solution containing KF, KI, and KBr and CHCl3\mathrm{CHC}{\mathrm{l}_3} is added. The initial colour in CHCl3\mathrm{CHC}{\mathrm{l}_3} layer is :
View answer and explanation
Correct option: A
Violet due to formation of I2{\mathrm{I}_2}
Option A (Violet due to formation of I2{\mathrm{I}_2}) is correct because chlorine will oxidize iodide ions to iodine, which dissolves in CHCl3\mathrm{CHC}{\mathrm{l}_3} giving it a violet color.
55 Medium
55 of 230
Analyzing +4 | -1
How does the electrical conductivity of 20 ml20 \mathrm{~ml} of 0.2 M−MgSO40.2 \mathrm{~M}-\mathrm{MgSO}_4 change when 0.5 M−Ba(OH)20.5 \mathrm{~M}-\mathrm{Ba}(\mathrm{OH})_2 solution is gradually added in it, to excess?
View answer and explanation
Correct option: D
Decreases and then increases
Option D is correct because the conductivity initially decreases as ions are removed by precipitation, then increases as excess Ba(OH)₂ adds more ions.
56 Medium
56 of 230
Analyzing +4 | -1
STATEMENT-1 : When 2 faraday of electricity is passed through 0.1 M H2SO4(aq),11.20.1 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4(\mathrm{aq}), 11.2 litre O2\mathrm{O}_2 evolved at STP.

STATEMENT-2 : Molecular weight of oxygen is 32 .
View answer and explanation
Correct option: B
If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
Option B is correct. Statement-1 is true because 2 Faradays of electricity will produce 11.2 L of O2 at STP. Statement-2 is also true, but it doesn't explain Statement-1.
57 Medium
57 of 230
Understanding +4 | -1
The products formed when an aqueous solution of NaBr{NaBr} is electrolysed in a cell having inert electrodes are:
View answer and explanation
Correct option: C
H2,Br2{H}_2, {Br}_2 and NaOH{NaOH}
Option C (H2,Br2{H}_2, {Br}_2 and NaOH{NaOH}) is correct because during the electrolysis of aqueous NaBr, hydrogen gas and hydroxide ions are produced at the cathode, while bromine gas is produced at the anode.
58 Medium
58 of 230
Application +4 | -1
During electrolysis of water the volume of O2{\mathrm{O}_2} liberated is 2.24 dm3\mathrm{d}{\mathrm{m}^3}. The volume of hydrogen liberated, under same conditions will be
View answer and explanation
Correct option: C
4.48dm34.48{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{d}{\mathrm{m}^3}
Option C (4.48dm34.48{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{d}{\mathrm{m}^3}) is correct because during electrolysis of water, the volume of hydrogen liberated is twice the volume of oxygen liberated.
59 Medium
59 of 230
Application +4 | -1
By how much would the oxidizing power of MnO4−/Mn2+\mathrm{MnO}_4{ }^{-} / \mathrm{Mn}^{2+} couple change if the H+\mathrm{H}^{+} ions concentration is decreased 100 times at 25∘C?25^{\circ} \mathrm{C}?
View answer and explanation
Correct option: B
Decreases by 189 mV189 \mathrm{~mV}
Option B (Decreases by 189 mV189 \mathrm{~mV}) is correct because decreasing the H+ concentration decreases the oxidizing power, as shown by the Nernst equation.
60 Medium
60 of 230
Application +4 | -1
In the electrolysis of water, 1 F of electrical energy would evolve
View answer and explanation
Correct option: C
8 g of oxygen
The correct answer is 8 g of oxygen because 4F of charge liberate 32g of oxygen, therefore 1F will liberate 8g of oxygen.
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