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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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121 Medium
121 of 230
Analyzing +4 | -1
Assertion: When aqueous sodium chloride solution is electrolysed, Oxygen gas is produced at the anode.

Reason: It is due to the overpotential for oxidation of water to oxygen.
View answer and explanation
Correct option: A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
The correct answer is A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) because the overpotential of water oxidation favors the production of hydrogen at the cathode instead of oxygen at the anode during the electrolysis of aqueous NaCl.
122 Medium
122 of 230
Application +4 | -1
The EMFs of the cell obtained by combining separately Zn\mathrm{Zn} and Cu\mathrm{Cu} electrodes of a Daniel cell with normal calomel electrodes are 1.083 V1.083 \mathrm{~V} and −0.018 V,-0.018 \mathrm{~V}, respectively, at 25∘C.25^{\circ} \mathrm{C}. If the potential of normal calomel electrode is −0.28 V,-0.28 \mathrm{~V}, the EMF of the Daniel cell is
View answer and explanation
Correct option: B
1.101 V1.101 \mathrm{~V}
Option B (1.101 V1.101 \mathrm{~V}) is correct because the EMF of the Daniel cell is the difference between the reduction potentials of the copper and zinc electrodes.
123 Medium
123 of 230
Application +4 | -1
The electrode potential of the following half-cell at 298 K298 \mathrm{~K}
X∣X+2(0.1M)∣∣Y+2(0.001M)∣YX\left|X^{+2}(0.1 M)\right|\left|Y^{+2}(0.001 M)\right| Y is ____ ×10−2 V\times 10^{-2} \mathrm{~V} (Nearest integer) given E0X+2/X=−1.58 VE0Y+2/Y=+0.74,2.303RTF=0.06 VE^0 X^{+2} / X=-1.58 \mathrm{~V} E^0 Y^{+2} / Y=+0.74, \frac{2.303 R T}{F}=0.06 \mathrm{~V}
View answer and explanation
Correct option: A
226
Option A (226) is correct because the cell potential is calculated using the Nernst equation with the given standard electrode potentials and concentrations.
124 Medium
124 of 230
Application +4 | -1
The resistance of 0.05 N0.05 \mathrm{~N} solution of an electrolyte was found to be 420 ohm420 \mathrm{~ohm} at 298 K298 \mathrm{~K}. Its conductance is
View answer and explanation
Correct option: A
2.4×10−3 ohm−12.4 \times 10^{-3} \mathrm{~ohm}^{-1}
Option A (2.4×10−3 ohm−12.4 \times 10^{-3} \mathrm{~ohm}^{-1}) is correct because conductance is the reciprocal of resistance.
125 Medium
125 of 230
Analyzing +4 | -1
Given, ECl2/Cl−o=1.36 V, ECr3+/Cro=−0.74 VE_{C{l_2}/C{l^ - }}^\mathrm{o} = 1.36\,V,\,E_{C{r^{3 + }}/Cr}^\mathrm{o} = - 0.74\,V
ECr2O72−/Cr3+o=1.33 V, EMnO4−/Mn2+o=1.51 VE_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^\mathrm{o} = 1.33\,V,\,E_{MnO_4^ - /M{n^{2 + }}}^\mathrm{o} = 1.51\,V
Among the following, the strongest reducing agent is
View answer and explanation
Correct option: A
CrCr
Option A (CrCr) is correct because the strongest reducing agent has the lowest (most negative) reduction potential. Among the given options, Cr has the lowest reduction potential (-0.74 V).
126 Medium
126 of 230
Understanding +4 | -1
Select the correct statement(s) for the weak electrolyte.
View answer and explanation
Correct option: D
At infinite dilution, electrolyte dissociate completely but at such low concentration the conductivity of solution is low and cannot be measured accurately.
Option D is correct because weak electrolytes dissociate completely only at infinite dilution, but the conductivity is too low to measure accurately.
127 Medium
127 of 230
Application +4 | -1
An object whose surface area is 80 cm280 \mathrm{~cm}^2 is to be plated with an even layer of gold 8.0×10−4 cm8.0 \times 10^{-4} \mathrm{~cm} thick. The density of gold is 19.7 g/cm3.19.7 \mathrm{~g} / \mathrm{cm}^3. The object is placed in a solution of Au(NO3)3\mathrm{Au}\left(\mathrm{NO}_3\right)_3 and a current of 2.4 A\text{2.4 A} is applied. The time required for the electroplating to be completed, assuming that the layer of gold builds up evenly, is (Au=197)(\mathrm{Au}=197)
View answer and explanation
Correct option: A
772 s772 \mathrm{~s}
Option A (772 s772 \mathrm{~s}) is correct because it is calculated using Faraday's laws of electrolysis and the given parameters.
128 Medium
128 of 230
Analyzing +4 | -1
E∘{E^ \circ } values of Mg(2+)M{g^{(2 + )}} / Mg is -2.37 V, of Zn2+\mathrm{Z}{\mathrm{n}^{2 + }} / Zn is -0.76 V and Fe2+\mathrm{F}{\mathrm{e}^{2 + }} / Fe is -0.44 V.
Which of the statements is correct?
View answer and explanation
Correct option: A
Zn will reduce Fe2+\mathrm{F}{\mathrm{e}^{2 + }}
Option A (Zn will reduce Fe²⁺) is correct because Zn has a more negative reduction potential than Fe, meaning it is a stronger reducing agent and can reduce Fe²⁺ to Fe.
129 Medium
129 of 230
Understanding +4 | -1
Hydrogen gas is not liberated when the following metal is added to dil. HCl
View answer and explanation
Correct option: A
Ag
Option A (Ag) is correct because silver is below hydrogen in the electrochemical series and therefore cannot displace hydrogen from dilute HCl.
130 Medium
130 of 230
Application +4 | -1
For given cell, Mg/Mg2+(0.01M)∥Ag+(0.01M)/Ag\mathrm{Mg} / \mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}, if Ecel ∘=3.17 V\mathrm{E}_{\text {cel }}^{\circ}=3.17 \mathrm{~V}, then Ecell \mathrm{E}_{\text {cell }} will be
View answer and explanation
Correct option: C
3.11 volt
Option C (3.11 volt) is correct because the cell potential is calculated using the Nernst equation, considering the concentrations of the ions.
131 Medium
131 of 230
Application +4 | -1
A certain current liberates 0.504 g of hydrogen in 2 hr. How many gram of copper can be liberated by the same current flowing for the same time in CuSO4\mathrm{CuS}{\mathrm{O}_4} solution?
View answer and explanation
Correct option: B
16
Option B (16) is correct because the mass of copper liberated is calculated using the electrochemical equivalence and Faraday's laws.
132 Medium
132 of 230
Understanding +4 | -1
Which process occurs in the electrolysis of aqueous solution of nickel chloride at nickel anode?
View answer and explanation
Correct option: D
Ni→Ni2++2e\mathrm{Ni} \to \mathrm{N}{\mathrm{i}^{2 + }} + 2e
Option D (Ni→Ni2++2e\mathrm{Ni} \to \mathrm{N}{\mathrm{i}^{2 + }} + 2e) is correct because at the nickel anode, nickel metal is oxidized to nickel ions.
133 Medium
133 of 230
Analyzing +4 | -1
Statement I: on dilution of the solution the electrode potential increases

Statement II: Effect of dilution on electrode potential depends on electrode whether it is metal or non-metal
View answer and explanation
Correct option: D
Statement I is false but Statement II is correct
Option D is correct because Statement I is false, and Statement II is correct regarding the effect of dilution on electrode potential depending on the electrode type.
134 Medium
134 of 230
Analyzing +4 | -1
Assertion (A): In weak electrolytes, the change in Λm\Lambda_{\mathrm{m}} with dilution is mainly due to increase in degree of dissociation
Reason (R): At a given concentration of weak electrolyte the degree of dissociation α=Λm0Λm\alpha=\frac{\Lambda_{\mathrm{m}}^0}{\Lambda_{\mathrm{m}}}
View answer and explanation
Correct option: D
A is correct but R is not correct
Option D is correct because the degree of dissociation is calculated as α=ΛmΛm0\alpha = \frac{{\Lambda_m}}{{\Lambda_m^0}} and not α=Λm0Λm\alpha=\frac{{\Lambda_{\mathrm{m}}^0}}{{\Lambda_{\mathrm{m}}}}.
135 Medium
135 of 230
Analyzing +4 | -1
Assertion: Sodium ions are discharged at the Hg\mathrm{Hg} electrode during electrolysis in preference to H+\mathrm{H}^{+}ions.

Reason: The nature of electrode also affects the order of discharge of cations.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because both the assertion and reason are true, and the reason correctly explains why sodium ions are discharged at the Hg electrode in preference to H+ ions.
136 Medium
136 of 230
Application +4 | -1
The conductivity of a 0.1NKCl0.1N{\kern 1pt} \mathrm{KCl} solution at 230C is 0.012 ohm−1cm−1\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt}{\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}} .The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be :
View answer and explanation
Correct option: B
0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}
Option B (0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}) is incorrect. The correct calculation using the conductivity and resistance yields 0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}.
137 Medium
137 of 230
Application +4 | -1
The quantity of charge required to completely reduce 0.5 mol0.5 \mathrm{~mol} of MnO4−\mathrm{MnO}_4^{-}to Mn2O3\mathrm{Mn}_2 \mathrm{O}_3 is
View answer and explanation
Correct option: C
2 F2 \mathrm{~F}
Option C (2 F) is correct because the reduction of 0.5 mol of MnO₄⁻ to Mn₂O₃ requires a change of 4 electrons per Mn atom, resulting in a total charge of 2 Faradays.
138 Medium
138 of 230
Application +4 | -1
The E∘{E^ \circ } of Fe2+/Fe\mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{Fe} and Sn2+/Sn\mathrm{S}{\mathrm{n}^{2 + }}/\mathrm{Sn} are -0.44 V and -0.14 V respectively. If cell reaction is

Fe+Sn2+→Fe2++Sn\mathrm{Fe} + \mathrm{S}{\mathrm{n}^{2 + }} \to \mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{Sn}

then emf of the cell is
View answer and explanation
Correct option: A
#ERROR!
Option A (+0.30 V) is correct because the cell emf is calculated by subtracting the anode's standard reduction potential from the cathode's.
139 Medium
139 of 230
Application +4 | -1
A 4.0 molar aqueous solution of NaCl\mathrm{NaCl} is prepared and 500 mL500 \mathrm{~mL} of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. If the cathode is a Hg\mathrm{Hg} electrode, the maximum weight (g)(\mathrm{g}) of amalgam (Na−Hg)(\mathrm{Na}-\mathrm{Hg}) formed from this solution is : (atomic mass: Na=23,Hg=200\mathrm{Na}=23, \mathrm{Hg}=200 )
View answer and explanation
Correct option: A
446
Option A (446) is correct because it is the mass of the Na-Hg amalgam formed based on the moles of Na+ reduced at the cathode.
140 Medium
140 of 230
Application +4 | -1
A solution is 1 molar in each of NaCl,CdCl2,ZnCl2{NaCl}, {CdCl}_2,{ZnCl}_2 and PbCl2{PbCl}_2. To this Sn{Sn} metal is added, which of the following is true?
[ Given E∘(Pb2+/Pb=−0.126 V)ESn2/Sn∘=−0.136 V,ECd2+/Cd∘=−0.40 VEZn2+/Zn∘=−0.763 V,ENa+/Na∘=−2.71 V]\left[\begin{array}{l} \text { Given } \quad {E}^{\circ}\left({Pb}^{2+} /{Pb}=-0.126 \mathrm{~V}\right) \\ {E}_{{Sn}^2 /{Sn}}^{\circ}=-0.136 \mathrm{~V},{E}_{{Cd}^{2+} / {Cd}}^{\circ}=-0.40 \mathrm{~V} \\ {E}_{{Zn}^{2+} / {Zn}}^{\circ}=-0.763 \mathrm{~V}, {E}_{{Na}^{+} /{Na}}^{\circ}=-2.71 \mathrm{~V} \end{array}\right]
View answer and explanation
Correct option: D
Sn{Sn} can reduce Pb2+{Pb}^{2+} to Pb{Pb}
Option D (Sn{Sn} can reduce Pb2+{Pb}^{2+} to Pb{Pb}) is correct because the standard reduction potential of Sn is lower than that of Pb, making Sn a stronger reducing agent.
141 Medium
141 of 230
Application +4 | -1
The equivalent conductivity of 0.1 M weak acid is 100 times less than that at infinite dilution. The degree of dissociation of weak electrolyte at 0.1 M is :
View answer and explanation
Correct option: C
0.01
Option A (333) is correct because it is calculated using Kohlrausch's law of independent migration of ions.
142 Medium
142 of 230
Application +4 | -1
A certain quantity electricity is passed through aqueous solutions of Al2(SO4)3A{l_2}{(S{O_4})_3} and CuSO4CuS{O_4} solutions connected in series. 0.09 g of AlAl is deposited on cathode during electrolysis. The amount of copper deposited on cathode in grams is (Atomic weight of Al=27Al = 27, Atomic weight of Cu=63.6Cu = 63.6)
View answer and explanation
Correct option: B
0.318
Option B (0.318) is correct because it is the amount of copper deposited calculated using Faraday's second law of electrolysis.
143 Medium
143 of 230
Application +4 | -1
The resistance of 2.5 N2.5 \mathrm{~N} solution of acetic acid is 230 ohm230 \mathrm{~ohm}, when measured in a cell having a cell constant of 1.15 cm−11.15 \mathrm{~cm}^{-1}. The equivalent conductance is ohm−1 cm2\mathrm{ohm}^{-1} \mathrm{~cm}^2 equiv −1{ }^{-1}
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equivalent conductance is calculated using the formula Λeq=K×1000normality\Lambda_{eq} = \frac{{K \times 1000}}{{\text{{normality}}}} where K is conductivity.
144 Medium
144 of 230
Application +4 | -1
The standard reduction potential of EBi33+/Bi\mathrm{E}_{\mathrm{Bi}^3}^{3+} / \mathrm{Bi} and ECu2+/Cu\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}} are 0.226 V0.226 \mathrm{~V} and 0.344 V0.344 \mathrm{~V} respectively A\mathrm{A} mixture of salts of Bi\mathrm{Bi} and Cu\mathrm{Cu} at unit concentration each is electrolysed at 25∘C25^{\circ} \mathrm{C}. To what value can [Cu2+]\left[\mathrm{Cu}^{2+}\right] be brought down before bismuth starts to deposit during electrolysis.
View answer and explanation
Correct option: A
10−410^{-4}
Option A (10−410^{-4}) is correct because it is the concentration of Cu2+\mathrm{Cu}^{2+} at which the reduction potential of copper equals that of bismuth, initiating bismuth deposition.
145 Medium
145 of 230
Application +4 | -1
During the electrolysis of 0.1 M−CuSO40.1 \mathrm{~M}-\mathrm{CuSO}_4 solution using copper electrodes, a depletion of Cu2+\mathrm{Cu}^{2+} occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the local concentration of Cu2+\mathrm{Cu}^{2+} near the anode and cathode are, respectively, 0.12 M0.12 \mathrm{~M} and 0.08 M,0.08 \mathrm{~M}, the back EMF developed at 298 K298 \mathrm{~K} is (log⁡1.5=0.18,2.303RT/F=0.06)(\log 1.5=0.18,2.303 R T / F=0.06)
View answer and explanation
Correct option: B
5.4 mV5.4 \mathrm{~mV}
Option B (5.4 mV5.4 \mathrm{~mV}) is correct because the back EMF is calculated using the Nernst equation with the given concentrations.
146 Medium
146 of 230
Application +4 | -1
Resistivity of a material is equal to resistance when cell
constant is
View answer and explanation
Correct option: A
1 cm−11 \mathrm{~cm}^{-1}
Option A (1 cm−11 \mathrm{~cm}^{-1}) is correct because resistivity is equal to resistance when the cell constant is 1.
147 Medium
147 of 230
Remembering +4 | -1
Which of the following statements is correct? Galvanic cell converts:
View answer and explanation
Correct option: A
Chemical energy into electrical energy
Option A (Chemical energy into electrical energy) is correct because a galvanic cell (also known as a voltaic cell) uses spontaneous chemical reactions to generate electricity.
148 Medium
148 of 230
Analyzing +4 | -1
Assertion (A) Ecell E_{\text {cell }} increases with increase in concentration of Ag+\mathrm{Ag}^{+}ions.

Reason (R) Ecell E_{\text {cell }} has positive value.
View answer and explanation
Correct option: B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion.
149 Medium
149 of 230
Application +4 | -1
12.3 g nitrobenzene is reduced into aniline by electrolysis, if current efficiency is 25% calculate charge required in coulombs
View answer and explanation
Correct option: C
24×965024 \times 9650 coulombs
The correct answer is C (24×965024 \times 9650 coulombs) because it is calculated by considering the number of moles of nitrobenzene, the number of electrons involved in the reduction, and the current efficiency.
150 Medium
150 of 230
Understanding +4 | -1
Assertion: In electrolytic refining, electrolytic cell consists of pure and impure Cu electrodes dipped in CuSO4(ag)\mathrm{CuSO}_4(\mathrm{ag}).

Reason: Pure cu is obtained at anode.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because in electrolytic refining, pure copper is obtained at the cathode, not the anode.
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