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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 151-180 of 396 questions

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151 Medium
151 of 396
Application +4 | -1
The specific conductivity of 0.1 N KCl solution is 0.0129 Ω−1cm−1{\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^{ - 1}} . The resistance of the solution in the cell 100Ω. The cell constant of the cell will be
View answer and explanation
Correct option: B
1.29
Option B (1.29) is correct because the cell constant is calculated by multiplying the specific conductivity by the resistance.
152 Medium
152 of 396
Remembering +4 | -1
Which of the following metal has the highest conductivity at room temperature ?
View answer and explanation
Correct option: C
Ag
Option C (Ag) is correct because silver (Ag) has the highest electrical conductivity at room temperature compared to the other metals listed.
153 Medium
153 of 396
Application +4 | -1
For the cell reaction

Pb+Sn2+→Pb2++Sn{Pb}+{Sn}^{2+} \rightarrow {Pb}^{2+}+{Sn}

Given that, Pb→Pb2+,E∘=0.13 V{Pb} \rightarrow {Pb}^{2+}, {E}^{\circ}=0.13 \mathrm{~V}

Sn2++2e−→Sn;E∘=−0.14 V{Sn}^{2+}+2 {e}^{-} \rightarrow {Sn} ; {E}^{\circ}=-0.14 \mathrm{~V}

What would be the ratio of cation concentration for which E=0?{E}=0?
View answer and explanation
Correct option: B
1/21/2
The correct answer is 1/2 because at E=0, the ratio of cation concentrations is determined by the Nernst equation and the standard cell potential.
154 Medium
154 of 396
Understanding +4 | -1
Assertion : Galvanised iron does not rust.

Reason : Zinc has a more negative electrode potential than iron.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
Option A is correct because galvanized iron does not rust due to the zinc coating, and zinc has a more negative electrode potential than iron, making it a sacrificial anode that corrodes instead of the iron.
155 Medium
155 of 396
Application +4 | -1
Calculate the time required to deposite 2×10−3 cm2 \times 10^{-3} \mathrm{~cm} thick layer of silver (density 1.05 g cm−31.05 \mathrm{~g} \mathrm{~cm}^{-3} ) on the surface of area 100 cm−3100 \mathrm{~cm}^{-3} by passing a current of 10 A10 \mathrm{~A} through silver nitrate solution?
View answer and explanation
Correct option: C
18.7 s18.7 \mathrm{~s}
Option C (18.7 s) is correct because it is calculated using Faraday's law of electrolysis, relating the mass of silver deposited to the current, time, and electrochemical equivalent.
156 Medium
156 of 396
Application +4 | -1
The cell constant of a given cell is 0.47 cm−10.47 {~cm}^{-1}. Theresistance of a solution placed in this cell is measured to be 31.6 ohm31.6 \, {ohm}. The conductivity of the solution (in Scm−1{S} {cm}^{-1} where S{S} has usual meaning) is
View answer and explanation
Correct option: C
0.0150.015
Option C is correct because the conductivity is calculated using the cell constant and the resistance of the solution.
157 Easy
157 of 396
Remembering +4 | -1
In metallic conductor the current is conducted by flow of
View answer and explanation
Correct option: C
Electrons
Option C (Electrons) is correct because in metallic conductors, the current is conducted by the flow of electrons.
158 Medium
158 of 396
Understanding +4 | -1
Which substance is obtained in the solution on electrolysis of aqueous CuSO4\mathrm{CuS}{\mathrm{O}_4} solution using graphite electrodes?
View answer and explanation
Correct option: A
H2O{\mathrm{H}_2}\mathrm{O}
Option B is correct. Electrolysis of aqueous CuSO4 using graphite electrodes results in the formation of H2SO4 in the solution as Cu2+ ions are reduced to Cu at the cathode and water is oxidized to O2 and H+ at the anode.
159 Medium
159 of 396
Analyzing +4 | -1
For the reaction 2Fe3++Fe→3Fe2+, E∘=1.21 V,2F{e^{3 + }} + Fe \to 3F{e^{2 + }},\,E^\circ = 1.21\,V, hence
View answer and explanation
Correct option: D
All of the above
The correct answer is D (All of the above) because the positive E° indicates a spontaneous reaction, Fe acts as the anode, and Fe would directly react with Fe3+Fe^{3+} if used as the cathode.
160 Medium
160 of 396
Application +4 | -1
The charge present on 42 g42 \mathrm{~g} of nitride ions in coulomb is
View answer and explanation
Correct option: D
8.6×1058.6 \times 10^5
Option D (8.6×1058.6 \times 10^5) is correct because it accurately calculates the total charge of the nitride ions based on the number of moles and Faraday's constant.
161 Medium
161 of 396
Understanding +4 | -1
Identify the factor from the following that does not
affect electrolytic conductance of a solution.
View answer and explanation
Correct option: D
The nature of the electrode used
Option D (The nature of the electrode used) is correct because the electrolytic conductance of a solution does not depend on the nature of the electrode used.
162 Medium
162 of 396
Application +4 | -1
Number of moles of MnO4−\mathrm{MnO}_4^{-} getting reduced to MnO2\mathrm{MnO}_2 under neutral condition by 4.517×10244.517 \times 10^{24} electrons is:
View answer and explanation
Correct option: C
2.5 moles
Option C (2.5 moles) is correct because the number of moles of MnO4- reduced is calculated based on the number of electrons transferred in the reduction reaction.
163 Medium
163 of 396
Analyzing +4 | -1
250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO30.1\,M\,AgN{O_3} and 0.1 M AgCl.0.1\,M\,AgCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electro deposited will be:
(EAg+/Ag0=0.80 V,EAu+/Au0 =1.69 V)\left( {E_{A{g^ + }/Ag}^0 = 0.80\,V,E_{A{u^ + }/Au}^0\, = 1.69\,V} \right)
View answer and explanation
Correct option: A
Only gold
Option A (Only gold) is correct because gold has a higher standard reduction potential than silver, so it will be deposited first.
164 Easy
164 of 396
Understanding +4 | -1
The cell reaction is spontaneous, when
View answer and explanation
Correct option: C
ΔG∘\Delta {G^ \circ } is negative
Option C (ΔG∘\Delta {G^ \circ } is negative) is correct because a spontaneous cell reaction is characterized by a negative change in Gibbs free energy.
165 Hard
165 of 396
Application +4 | -1
The resistance of a conductivity cell was found to be 700Ω700 \Omega with 0.01 M KCl0.01 \mathrm{~M} \mathrm{~KCl} and 800Ω800 \Omega with 0.01 M AgNO30.01 \mathrm{~M} \mathrm{~AgNO}_3.

Molar conductance of 0.01 M KCl=150 ohm−1 cm2 mol−10.01 \mathrm{~M} \mathrm{~KCl}=150 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}.

Thus, molar conductance of AgNO3\mathrm{AgNO}_3 is
View answer and explanation
Correct option: A
131 ohm−1 cm2 mol−1131 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
Option A (131 ohm−1 cm2 mol−1131 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}) is correct because it is calculated by finding the ratio of molar conductances using the given resistances and molarity.
166 Medium
166 of 396
Application +4 | -1
The molar conductivity of HCl,NaCl\mathrm{HCl},\mathrm{NaCl} and CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa} are 425, 188, 96 Scm2mol−1S{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt}\mathrm{mo}{\mathrm{l}^{ - 1}} at 298 K. The molar conductivity of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} at the same temperature is …. Scm2mol−1S{\kern 1pt}{\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt}\mathrm{mo}{\mathrm{l}^{ - 1}}
View answer and explanation
Correct option: A
333
Option D (300) is correct because charge is the product of current and time, and 5 amperes for 1 minute equals 300 coulombs.
167 Medium
167 of 396
Analyzing +4 | -1
At 298 K298 \mathrm{~K}, the standard oxidation potential of A∣A+2, B∣B+2,C∣C+1,D∣D+3\mathrm{A}\left|\mathrm{A}^{+2}, \mathrm{~B}\right| \mathrm{B}^{+2}, \mathrm{C}\left|\mathrm{C}^{+1} , \mathrm{D}\right| \mathrm{D}^{+3} are 0.28 V,−0.56 V0.28 \mathrm{~V},-0.56 \mathrm{~V}, 0.15 V,−2.0 V0.15 \mathrm{~V},-2.0 \mathrm{~V} respectively.

On the basis of standard oxidation potential, predict which of the following reaction can not occur:
View answer and explanation
Correct option: C
A+2\mathrm{A}^{+2} will be reduced by D\mathrm{D}
The correct answer is that A+2 will be reduced by D because the cell potential for the reaction is negative, indicating a non-spontaneous reaction.
168 Medium
168 of 396
Application +4 | -1
Reduction potential of hydrogen electrode at pH{pH} of 3 and pH2=1 atm{p}_{{H}_2}=1 {~atm} is
View answer and explanation
Correct option: B
−0.177 V-0.177 {~V}
Option B (−0.177 V-0.177 {~V}) is correct because it's calculated using the Nernst equation with the given pH and hydrogen pressure.
169 Easy
169 of 396
Understanding +4 | -1
Which of the following is correct?
View answer and explanation
Correct option: D
Zinc displaces tin from its solution
Option D (Zinc displaces tin from its solution) is correct because zinc is more reactive than tin and can displace it from its solution, as indicated by the electrochemical series.
170 Medium
170 of 396
Application +4 | -1
Equilibrium constant for the following redox reaction at 25∘C25^{\circ} \mathrm{C} is

Zn2+(aq)+Mg(s)⟶Zn(s)+Mg2+(aq)\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{Mg}^{2+}(\mathrm{aq})

( Given EZn2+/Zn0=−0.76 V,EMg2+/Mg0=−2.36 V)\left(\text { Given } \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^0=-2.36 \mathrm{~V}\right)
View answer and explanation
Correct option: C
1.72×10541.72 \times 10^{54}
The correct answer is C (1.72 × 1054) because it is calculated using the Nernst equation and the relationship between standard cell potential and equilibrium constant.
171 Easy
171 of 396
Remembering +4 | -1
The standard H electrode is written as :
View answer and explanation
Correct option: C
PtH2(g)(1atm)/H+(a=1)\mathrm{Pt}{\mathrm{H}_2}\left( \mathrm{g} \right)\left( {1\mathrm{atm}} \right)/{\mathrm{H}^ + }\left( {a = 1} \right)
Option C is correct because it accurately represents the standard hydrogen electrode (SHE) with platinum electrode, hydrogen gas at 1 atm, and H+ ions at unit activity.
172 Medium
172 of 396
Application +4 | -1
The molar conductivities of KCl, NaCl and KNO3\mathrm{KN}{\mathrm{O}_3} are 152, 128 and 111 Scm2mol(−1)S{\kern 1pt} {\kern 1pt} c{m^2}{\kern 1pt} mo{l^{( - 1)}} respectively. What is the molar conductivity of NaNO3\mathrm{NaN}{\mathrm{O}_3} ?
View answer and explanation
Correct option: B
87 Scm2mol(−1)S{\kern 1pt} {\kern 1pt} c{m^2}{\kern 1pt} mo{l^{( - 1)}}
Option B (87 Scm2mol(−1)S{\kern 1pt} {\kern 1pt} c{m^2}{\kern 1pt} mo{l^{( - 1)}}) is correct because it is calculated using Kohlrausch's Law of independent migration of ions.
173 Medium
173 of 396
Understanding +4 | -1
KCl(aq)\mathrm{KCl}\left( {aq} \right) cannot be used as a salt bridge for the cell Cu(s)∣CuSO4(aq)∣∣AgNO3(aq)∣Ag(s)\mathrm{Cu}\left( s \right)\left| {\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right)} \right|\left| {\mathrm{AgN}{\mathrm{O}_3}\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right) because :
View answer and explanation
Correct option: C
AgCl\mathrm{AgCl} is precipitated
Option C (AgCl is precipitated) is correct because Ag+ ions from AgNO3 react with Cl- ions from KCl to form insoluble AgCl, disrupting the salt bridge function.
174 Easy
174 of 396
Application +4 | -1
The molar conductivity at infinite dilution of AgNO3,NaCl\mathrm{AgN}{\mathrm{O}_3},\mathrm{NaCl} and NaNO3\mathrm{NaN}{\mathrm{O}_3} are 116.5, 110.3 and 105.2 mhocm2mol−1{\rm{mho{\kern 1pt} {\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}} respectively. The molar conductivity of AgCl is :
View answer and explanation
Correct option: A
121.6
Option A (121.6) is correct because it is calculated by applying Kohlrausch's law of independent migration of ions.
175 Easy
175 of 396
Application +4 | -1
Given the limiting molar conductivity as

Λm0(HCl)=425.9Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{HCl}} \right) = 425.9{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

Λm0(NaCl)=126.4Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{NaCl}} \right) = 126.4{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

Λm0(CH3COONa)=91Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}} \right) = 91{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

The molar conductivity , at infinite dilution, of acetic acid (inΩ−1cm2mol−1\mathrm{in}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}) will be
View answer and explanation
Correct option: B
390.5
Option B (390.5) is correct because it is calculated by applying Kohlrausch's law to find the molar conductivity of acetic acid at infinite dilution.
176 Medium
176 of 396
Understanding +4 | -1
Electrolysis of H2SO4{H}_2 {SO}_4 (conc.) gives the following product at anode
View answer and explanation
Correct option: D
H2 S2O8{H}_2 {~S}_2 {O}_8
The correct answer is D (H2 S2O8{H}_2 {~S}_2 {O}_8) because during the electrolysis of concentrated H2SO4{H}_2 {SO}_4, peroxydisulfuric acid (H2 S2O8{H}_2 {~S}_2 {O}_8) is formed at the anode.
177 Medium
177 of 396
Understanding +4 | -1
Assertion: Molar conductance of 0.1 M NH4OH0.1 \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH} solution is less than that of 0.001 M NH4OH0.001 \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH} solution.

Reason: Dilution increases the degree of ionization of NH4OH\mathrm{NH}_4 \mathrm{OH}.
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 molar conductance decreases with increasing concentration due to decreased ionization, and dilution increases the degree of ionization for weak electrolytes.
178 Hard
178 of 396
Application +4 | -1
The emf of the cell Zn∣Zn2+(0.01M)∣∣Fe2+(0.001M)∣Fe\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\left( {0.01\mathrm{M}} \right)} \right|\left| {\mathrm{F}{\mathrm{e}^{2 + }}\left( {0.001\mathrm{M}} \right)} \right|\mathrm{Fe} at 298 K is 0.2905. The value of equilibrium constant for the cell reaction is
View answer and explanation
Correct option: A
10100.320.0298{10^{{{10}^{\frac{{0.32}}{{0.0298}}}}}}
Option A (100.320.0298{10^{{\frac{{0.32}}{{0.0298}}}}}) is correct because the equilibrium constant is calculated using the Nernst equation, relating the cell potential to the equilibrium constant.
179 Easy
179 of 396
Application +4 | -1
Given the limiting molar conductivity as

Λm0(HCl)=425.9Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{HCl}} \right) = 425.9{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

Λm0(NaCl)=126.4Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{NaCl}} \right) = 126.4{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

Λm0(CH3COONa)=91Ω−1cm2mol−1\Lambda _\mathrm{m}^0\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}} \right) = 91{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

The molar conductivity , at infinite dilution, of acetic acid (inΩ−1cm2mol−1\mathrm{in}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}) will be
View answer and explanation
Correct option: B
390.5
Option B (390.5) is correct because it is calculated by applying Kohlrausch's law to find the molar conductivity of acetic acid at infinite dilution.
180 Medium
180 of 396
Application +4 | -1
The standard reduction potentials for Zn2+/Zn,Ni2+/Ni\mathrm{Z}{\mathrm{n}^{2 + }}/\mathrm{Zn},\mathrm{N}{\mathrm{i}^{2 + }}/\mathrm{Ni} and Fe2+/Fe\mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{Fe} are −0.76,−0.23- 0.76, - 0.23 and−0.44V- 0.44\mathrm{V} respectively. The reactionX+Y2+→X2++YX + {Y^{2 + }} \to {X^{2 + }} + Y will be spontaneous when :
View answer and explanation
Correct option: C
X=Zn,Y=NiX = \mathrm{Zn},Y = \mathrm{Ni}
Option C (X=Zn,Y=NiX = \mathrm{Zn}, Y = \mathrm{Ni}) is correct because the cell potential for this combination is positive, indicating a spontaneous reaction.
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