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JEE Chemistry
Electrochemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

100 Q 5 Topics Take Test
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Difficulty: All Easy Medium Hard 61–70 of 100
Topics in JEE Chemistry
In a galvanic cell, oxidation occurs at which electrode?
A Cathode
B Anode
C Both anode and cathode
D Neither anode nor cathode
Correct Answer:  B. Anode
EXPLANATION

In a galvanic cell, oxidation occurs at the anode. The anode is the negative electrode where electrons are released and the species loses electrons.

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Q.62 Medium Electrochemistry
In the electrolysis of dilute H₂SO₄ with inert electrodes, the cathode reaction is:
A 2H⁺ + 2e⁻ → H₂
B SO₄²⁻ + 2e⁻ → S + 4O²⁻
C 2H₂O + 2e⁻ → H₂ + 2OH⁻
D H⁺ + e⁻ → H
Correct Answer:  A. 2H⁺ + 2e⁻ → H₂
EXPLANATION

In dilute H₂SO₄, H⁺ is preferentially reduced over water at the cathode (less negative reduction potential), producing H₂ gas.

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The conductivity of a solution containing 0.1 M NaCl is 1.29 S·m⁻¹. The molar conductivity is:
A 12.9 S·cm²·mol⁻¹
B 129 S·cm²·mol⁻¹
C 1290 S·cm²·mol⁻¹
D 0.129 S·cm²·mol⁻¹
Correct Answer:  A. 12.9 S·cm²·mol⁻¹
EXPLANATION

Λm = κ/C, where κ = 1.29 S·m⁻¹ = 0.0129 S·cm⁻¹ and C = 0.1 M. Λm = 0.0129/0.1 = 0.129 S·cm²·mol⁻¹ = 12.9 S·cm²·mol⁻¹.

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At 25°C, for a cell reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), if [Zn²⁺] = 1 M and [Ag⁺] = 0.1 M, and E°cell = 1.56 V, the Ecell is approximately:
A 1.62 V
B 1.50 V
C 1.44 V
D 1.68 V
Correct Answer:  A. 1.62 V
EXPLANATION

Using Nernst: Ecell = E° - (0.059/n)log(Q). Q = [Zn²⁺]/[Ag⁺]² = 1/(0.1)² = 100. Ecell = 1.56 - (0.059/2)log(100) = 1.56 + 0.059 ≈ 1.62 V.

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An electrochemical cell requires 193,700 C of charge to deposit 19.6 g of a metal X. The valency of metal X is:
A 2
B 3
C 1
D 4
Correct Answer:  B. 3
EXPLANATION

Charge = 193,700 C; moles of electrons = 193,700/96,500 = 2. If 19.6 g = ? mol; then valency n = (moles of e⁻)/(moles of metal). Assuming atomic mass from calculation gives valency = 3 (like Al).

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The standard reduction potential for Zn²⁺/Zn is -0.76 V and for Cu²⁺/Cu is +0.34 V. For the cell Zn-Cu, E°cell is:
A -1.10 V
B +1.10 V
C +0.42 V
D -0.42 V
Correct Answer:  B. +1.10 V
EXPLANATION

E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V. Cu²⁺ is reduced (cathode), Zn is oxidized (anode).

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Q.67 Medium Electrochemistry
Calculate the number of Faradays required to deposit 3.2 g of Cu (atomic mass 64) from CuSO₄ solution:
A 0.1 F
B 0.05 F
C 0.2 F
D 0.5 F
Correct Answer:  A. 0.1 F
EXPLANATION

Cu²⁺ + 2e⁻ → Cu. Moles of Cu = 3.2/64 = 0.05 mol. Charge = 0.05 × 2 = 0.1 F (since 1 F = 1 mole of electrons).

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Q.68 Medium Electrochemistry
In a concentration cell, the EMF depends on:
A Only temperature
B The ratio of concentrations of the two electrodes
C The nature of the electrodes
D The volume of solutions
Correct Answer:  B. The ratio of concentrations of the two electrodes
EXPLANATION

In a concentration cell, E°cell = 0, so EMF = (RT/nF)ln(C₁/C₂), depending only on the concentration ratio, temperature, and number of electrons transferred.

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Q.69 Medium Electrochemistry
The equivalent conductivity of a strong electrolyte at infinite dilution (Λ∞) for HCl is approximately:
A 426 S·cm²·mol⁻¹
B 149 S·cm²·mol⁻¹
C 198 S·cm²·mol⁻¹
D 65 S·cm²·mol⁻¹
Correct Answer:  A. 426 S·cm²·mol⁻¹
EXPLANATION

Λ∞ for HCl ≈ 426 S·cm²·mol⁻¹ (sum of Λ∞H⁺ ≈ 350 and Λ∞Cl⁻ ≈ 76). This is higher than monovalent salts due to high mobility of H⁺.

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Q.70 Medium Electrochemistry
Which of the following solutions has the highest conductivity?
A 0.1 M NaCl
B 0.1 M HCl
C 0.1 M CH₃COOH
D 0.1 M NH₃
Correct Answer:  B. 0.1 M HCl
EXPLANATION

HCl is a strong electrolyte with complete ionization. NaCl is also strong but HCl has higher molar conductivity. CH₃COOH and NH₃ are weak electrolytes with low ionization.

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