JEE Chemistry Electrochemistry MCQs Set 08

Practice MCQs for JEE Chemistry Electrochemistry

Access targeted multiple-choice questions for Electrochemistry designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Electrochemistry Questions and Solutions

Access the complete set of multiple-choice questions for Electrochemistry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Which of the following displacements does not occur ?
(a) Zn + 2H⁺ → Zn²⁺ + H₂ ↑
(b) Fe + 2Ag⁺ → Fe²⁺ + 2Ag ↓
(c) Cu + Fe²⁺ → Cu²⁺ + Fe ↓
(d) Zn + Pb²⁺ → Zn²⁺ + Pb ↓
Show Answer & Explanation

Answer: (c) Cu + Fe²⁺ → Cu²⁺ + Fe ↓

Explanation:
1. Cu has a higher reduction potential than Fe (0.34 V against −0.44 V).
2. Cu cannot reduce Fe²⁺, so reaction (c) does not occur. The others are all feasible.

Teacher's Note:
1. The metal on the left must be above the ion's metal in the series.
2. Copper is below iron.

Question 2: E°(Ni²⁺/Ni) = −0.25 V and E°(Au³⁺/Au) = 1.50 V. The emf of the voltaic cell Ni | Ni²⁺ (1.0 M) || Au³⁺ (1.0 M) | Au is :
(a) 1.25 V
(b) −1.75 V
(c) 1.75 V
(d) 4.0 V
Show Answer & Explanation

Answer: (c) 1.75 V

Explanation:
1. E°cell = E°cathode − E°anode = 1.50 − (−0.25).
2. = 1.75 V.

Teacher's Note:
1. Gold is the cathode and nickel is the anode.
2. Use the same unit (volts) throughout.

Question 3: The emf of the cell in which the following reaction occurs, Zn(s) + Ni²⁺ (a = 0.1) ⇌ Zn²⁺ (a = 1.0) + Ni(s), is found to be 0.5105 V at 298 K. The standard emf of the cell is :
(a) −0.5105 V
(b) 0.5400 V
(c) 0.4810 V
(d) 0.5696 V
Show Answer & Explanation

Answer: (b) 0.5400 V

Explanation:
1. E = E° − (0.0591/2) log([Zn²⁺]/[Ni²⁺]) = E° − 0.02955 × log(1.0/0.1) = E° − 0.02955.
2. So E° = 0.5105 + 0.02955 = 0.540 V.

Teacher's Note:
1. Reaction quotient Q = product / reactant.
2. Q is greater than 1 here, so E is less than E°.

Question 4: The emf of the cell Tl(s) | Tl⁺ (0.0001 M) || Cu²⁺ (0.01 M) | Cu(s) is 0.83 V. The emf of this cell will be increased by :
(a) increasing the concentration of Cu²⁺ ions
(b) decreasing the concentration of Tl⁺
(c) increasing the concentration of both
(d) (a) and (b) both
Show Answer & Explanation

Answer: (d) (a) and (b) both

Explanation:
1. The cell reaction is 2Tl + Cu²⁺ → 2Tl⁺ + Cu, so E = E° − (0.0591/2) log([Tl⁺]²/[Cu²⁺]).
2. E increases when [Cu²⁺] increases or [Tl⁺] decreases, so both (a) and (b) raise the emf.

Teacher's Note:
1. Increasing a reactant concentration or decreasing a product concentration increases E.
2. Increasing both concentrations has mixed effects.

Question 5: The potential of a hydrogen electrode (P(H₂) = 1 atm, C(H⁺) = 0.1 M) at 25 °C will be :
(a) 0.00 V
(b) −0.059 V
(c) 0.118 V
(d) 0.059 V
Show Answer & Explanation

Answer: (b) −0.059 V

Explanation:
1. E = −0.0591 × pH, and pH = 1 for 0.1 M H⁺.
2. E = −0.059 V as a reduction potential.

Teacher's Note:
1. A hydrogen electrode has E = 0 at pH 0 and becomes more negative as pH increases.
2. Each unit of pH changes it by 0.059 V.

Question 6: Consider the reaction Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂. The emf of the cell when [Cl⁻] = [Br₂] = [Br⁻] = 0.01 M and Cl₂ gas is at 1 atm pressure will be (E° for the above reaction = 0.29 V) :
(a) 0.54 V
(b) 0.35 V
(c) 0.24 V
(d) −0.29 V
Show Answer & Explanation

Answer: (b) 0.35 V

Explanation:
1. Q = [Cl⁻]²[Br₂] / ([Br⁻]² p(Cl₂)) = 0.01 (taking the given concentrations).
2. E = 0.29 − (0.0591/2) log 0.01 = 0.29 + 0.059 = 0.349 V, which is about 0.35 V.

Teacher's Note:
1. A small Q (products below reactants) increases the emf.
2. Use n = 2 for this reaction.

Question 7: The standard emf for the cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu is 1.10 V at 25 °C. The emf for the cell reaction when 0.1 M Cu²⁺ and 0.1 M Zn²⁺ solutions are used at 25 °C is :
(a) 1.10 V
(b) 0.110 V
(c) −1.10 V
(d) −0.110 V
Show Answer & Explanation

Answer: (a) 1.10 V

Explanation:
1. E = E° − (0.0591/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.02955 × log(0.1/0.1).
2. The log term is zero, so E = 1.10 V.

Teacher's Note:
1. Equal ion concentrations give Q = 1.
2. Then the emf equals the standard emf.

Question 8: E° for F₂ + 2e⁻ → 2F⁻ is 2.8 V. E° for ½F₂ + e⁻ → F⁻ is :
(a) 2.8 V
(b) 1.4 V
(c) −2.8 V
(d) −1.4 V
Show Answer & Explanation

Answer: (a) 2.8 V

Explanation:
1. Electrode potential is an intensive property and does not depend on the quantity of species in the half reaction.
2. Halving the equation does not change E°.

Teacher's Note:
1. ΔG° changes with the equation, but E° does not.
2. Do not divide or multiply E° by coefficients.

Question 9: How much will the oxidation potential of a Zn/Zn²⁺ electrode change if the solution of Zn²⁺ is diluted 10 times ?
(a) increases by 0.03 V
(b) decreases by 0.03 V
(c) increases by 0.059 V
(d) decreases by 0.059 V
Show Answer & Explanation

Answer: (a) increases by 0.03 V

Explanation:
1. For Zn → Zn²⁺ + 2e⁻ the oxidation potential is E = E° − (0.0591/2) log [Zn²⁺].
2. Diluting 10 times decreases log [Zn²⁺] by 1, so E increases by 0.0591/2 = 0.03 V.

Teacher's Note:
1. Oxidation potential has the opposite sign to the reduction potential.
2. The change is 0.0591/2, which is about 0.03 V.

Question 10: How much will the potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralised to pH = 7 ?
(a) increases by 0.059 V
(b) decreases by 0.059 V
(c) increases by 0.41 V
(d) decreases by 0.41 V
Show Answer & Explanation

Answer: (d) decreases by 0.41 V

Explanation:
1. E(red) = −0.0591 × pH.
2. Going from pH 0 to pH 7 changes E from 0 to −0.41 V, a decrease of 0.41 V.

Teacher's Note:
1. Each pH unit changes the potential by 0.059 V.
2. Seven units give about 0.41 V.

Question 11: Which of the following will increase the voltage of the cell with the cell reaction Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s) ?
(a) Increase in the size of silver rod
(b) Increase in the concentration of Sn²⁺ ions
(c) Increase in the concentration of Ag⁺ ions
(d) Decrease in the concentration of Ag⁺ ions
Show Answer & Explanation

Answer: (c) Increase in the concentration of Ag⁺ ions

Explanation:
1. E = E° − (0.0591/2) log([Sn²⁺]/[Ag⁺]²).
2. Raising [Ag⁺] (a reactant) lowers Q and increases E. The size of an electrode does not matter.

Teacher's Note:
1. Electrode size does not enter the Nernst equation.
2. Reactant concentration up means voltage up.

Question 12: E° for the reaction Fe + Zn²⁺ → Zn + Fe²⁺ is −0.35 V. The given cell reaction is :
(a) feasible
(b) not feasible
(c) in equilibrium
(d) can't say anything
Show Answer & Explanation

Answer: (b) not feasible

Explanation:
1. A reaction is feasible only if E°cell is positive (ΔG° negative).
2. E° = −0.35 V is negative, so the reaction is not feasible under standard conditions.

Teacher's Note:
1. Sign of E° decides spontaneity.
2. Here Zn would reduce Fe²⁺ instead.

Question 13: For a reaction A(s) + 2B⁺ → A²⁺ + 2B(s), Kc has been found to be 10¹². The E°cell is :
(a) 0.354 V
(b) 0.708 V
(c) 0.0098 V
(d) 1.36 V
Show Answer & Explanation

Answer: (a) 0.354 V

Explanation:
1. E°cell = (0.0591/n) log K = (0.0591/2) × 12.
2. = 0.3546 V.

Teacher's Note:
1. Two electrons are transferred in this reaction.
2. log 10¹² = 12.

Question 14: The standard electrode potential (E°) for OCl⁻/Cl⁻ and Cl⁻/½Cl₂ respectively are 0.94 V and −1.36 V. The E° value of OCl⁻/½Cl₂ will be :
(a) −2.20 V
(b) −0.42 V
(c) 0.52 V
(d) 1.04 V
Show Answer & Explanation

Answer: (c) 0.52 V

Explanation:
1. OCl⁻ + 2e⁻ → Cl⁻ gives ΔG₁ = −2F(0.94) and Cl⁻ → ½Cl₂ + e⁻ gives ΔG₂ = −F(−1.36).
2. The sum, OCl⁻ + e⁻ → ½Cl₂, has ΔG = −F(1.88 − 1.36) = −0.52F, so E° = 0.52 V.

Teacher's Note:
1. Add free energies, not potentials, when the numbers of electrons differ.
2. ΔG° = −nFE°.

Question 15: The standard reduction potentials of Cu²⁺/Cu and Cu²⁺/Cu⁺ are 0.337 V and 0.153 V respectively. The standard electrode potential of the Cu⁺/Cu half cell is :
(a) 0.184 V
(b) 0.827 V
(c) 0.521 V
(d) 0.490 V
Show Answer & Explanation

Answer: (c) 0.521 V

Explanation:
1. Cu²⁺ + 2e⁻ → Cu has ΔG = −2F(0.337), and Cu²⁺ + e⁻ → Cu⁺ has ΔG = −F(0.153).
2. Cu⁺ + e⁻ → Cu: −F E° = −2F(0.337) + F(0.153), so E° = 0.674 − 0.153 = 0.521 V.

Teacher's Note:
1. Combine half reactions through ΔG°, not E°.
2. The Cu⁺ ion disproportionates since 0.521 V is greater than 0.153 V.

Question 16: The emf of the cell Ni | Ni²⁺ (1.0 M) || Au³⁺ (0.1 M) | Au is given as : [E°(Ni²⁺/Ni) = −0.25 V, E°(Au³⁺/Au) = 1.50 V]
(a) 1.25 V
(b) −1.75 V
(c) 1.75 V
(d) 1.73 V
Show Answer & Explanation

Answer: (d) 1.73 V

Explanation:
1. The cell reaction is 3Ni + 2Au³⁺ → 3Ni²⁺ + 2Au with n = 6 and E° = 1.75 V.
2. E = 1.75 − (0.0591/6) log(1³/0.1²) = 1.75 − 0.00985 × 2 = 1.73 V.

Teacher's Note:
1. Balance the electrons first to find n.
2. The concentration term is small here.

Question 17: When an electric current is passed through acidified water, 112 mL of hydrogen gas at STP collects at the cathode in 965 s. The current passed, in ampere, is :
(a) 1.0
(b) 0.5
(c) 0.1
(d) 2.0
Show Answer & Explanation

Answer: (a) 1.0

Explanation:
1. 112 mL H₂ = 0.005 mol, which needs 0.010 mol of electrons = 0.010 × 96500 = 965 C.
2. I = Q/t = 965 / 965 = 1.0 A.

Teacher's Note:
1. One mole of H₂ needs 2 F of charge.
2. Use molar volume 22.4 L at STP.

Question 18: Two electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride, are connected in series. The ratio of iron deposited at the cathodes in the two cells when electricity is passed through the cells will be :
(a) 3 : 1
(b) 2 : 1
(c) 1 : 1
(d) 3 : 2
Show Answer & Explanation

Answer: (d) 3 : 2

Explanation:
1. Fe²⁺ + 2e⁻ → Fe needs 2 F per mole, and Fe³⁺ + 3e⁻ → Fe needs 3 F per mole.
2. For the same charge, moles of Fe are in the ratio 1/2 : 1/3 = 3 : 2.

Teacher's Note:
1. Same charge, so the amount deposited is inversely proportional to the charge on the ion.
2. Compare equivalent amounts.

Question 19: A current of 9.65 A flowing for 10 minutes deposits 3.0 g of a metal. The equivalent weight of the metal is :
(a) 10
(b) 30
(c) 50
(d) 96.5
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. Charge = 9.65 × 600 = 5790 C = 0.06 F.
2. Equivalent weight = 3.0 / 0.06 = 50.

Teacher's Note:
1. One faraday deposits one gram equivalent.
2. Find the number of faradays first.

Question 20: The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate [Ni(NO₃)₂] and chromium nitrate [Cr(NO₃)₃] respectively. If 0.3 g of nickel was deposited in the first cell, the amount of chromium deposited in the other cell is (atomic weight of Ni = 59, atomic weight of Cr = 52) :
(a) 0.1 g
(b) 0.17 g
(c) 0.3 g
(d) 0.6 g
Show Answer & Explanation

Answer: (b) 0.17 g

Explanation:
1. Equivalent weights: Ni = 59/2 = 29.5 and Cr = 52/3 = 17.33.
2. By Faraday's second law, W(Cr) = 0.3 × 17.33 / 29.5 = 0.176 g, which is about 0.17 g.

Teacher's Note:
1. Same charge means the masses are in the ratio of equivalent weights.
2. The valency of Cr is 3.

Question 21: How many coulombs of electricity are required for the oxidation of 1 mol of H₂O to O₂ ?
(a) 9.65 × 10⁴ C
(b) 4.825 × 10⁵ C
(c) 1.93 × 10⁵ C
(d) 1.93 × 10⁴ C
Show Answer & Explanation

Answer: (c) 1.93 × 10⁵ C

Explanation:
1. H₂O → ½O₂ + 2H⁺ + 2e⁻, so 1 mol of water releases 2 mol of electrons.
2. Q = 2 × 96500 = 1.93 × 10⁵ C.

Teacher's Note:
1. Balance the oxidation half reaction to find the electrons per mole.
2. 1 F = 96500 C.

Question 22: 10800 C of electricity passed through an electrolyte deposited 2.977 g of a metal with atomic mass 106.4 g mol⁻¹. The charge on the metal cation is :
(a) +4
(b) +3
(c) +2
(d) +1
Show Answer & Explanation

Answer: (a) +4

Explanation:
1. Moles of metal = 2.977/106.4 = 0.02798 mol, and charge passed = 10800/96500 = 0.1119 F.
2. Charge on ion = 0.1119/0.02798 = 4, so the cation is M⁴⁺.

Teacher's Note:
1. Charge per ion = faradays divided by moles deposited.
2. Round to the nearest whole number.

Question 23: On passing electricity through dilute H₂SO₄ solution, the amounts of the substances liberated at the cathode and anode are in the ratio :
(a) 1 : 8
(b) 8 : 1
(c) 16 : 1
(d) 1 : 16
Show Answer & Explanation

Answer: (a) 1 : 8

Explanation:
1. The cathode gives H₂ and the anode gives O₂.
2. By Faraday's second law, the masses are in the ratio of the equivalent weights, 1 : 8.

Teacher's Note:
1. Equivalent weight of hydrogen is 1 and of oxygen is 8.
2. The mole ratio of H₂ to O₂ would be 2 : 1.

Question 24: During electrolysis of fused calcium hydride, hydrogen is produced at :
(a) cathode
(b) anode
(c) hydrogen is not liberated at all
(d) H₂ produced reacts with oxygen to form water
Show Answer & Explanation

Answer: (b) anode

Explanation:
1. Fused CaH₂ contains Ca²⁺ and H⁻ ions.
2. H⁻ is oxidised at the anode (2H⁻ → H₂ + 2e⁻), so hydrogen forms at the anode.

Teacher's Note:
1. Hydride ion is a negative ion and moves to the anode.
2. Calcium is deposited at the cathode.

Question 25: A silver cup is plated with silver by passing 965 A current for one second. The mass of Ag deposited is (at. wt. of Ag = 107.87) :
(a) 9.89 g
(b) 107.87 g
(c) 1.0787 g
(d) 100.2 g
Show Answer & Explanation

Answer: (c) 1.0787 g

Explanation:
1. Charge = 965 × 1 = 965 C = 0.01 F.
2. Mass of Ag = 0.01 × 107.87 = 1.0787 g.

Teacher's Note:
1. One faraday deposits one mole of silver.
2. 965 C is one hundredth of a faraday.

Question 26: When electricity is passed through a solution of AlCl₃, 13.5 g of Al is deposited. The number of faradays must be :
(a) 5.0
(b) 1.0
(c) 1.5
(d) 3.0
Show Answer & Explanation

Answer: (c) 1.5

Explanation:
1. 13.5 g Al = 0.5 mol, and Al³⁺ + 3e⁻ → Al.
2. Faradays = 0.5 × 3 = 1.5.

Teacher's Note:
1. Each mole of Al needs 3 F.
2. Use the molar mass 27 g/mol.

Question 27: One faraday of electricity will liberate one mole of the metal from the solution of :
(a) auric chloride
(b) silver nitrate
(c) calcium chloride
(d) copper sulphate
Show Answer & Explanation

Answer: (b) silver nitrate

Explanation:
1. One faraday is the charge of one mole of electrons.
2. Only a univalent metal ion is reduced to one mole of metal by one mole of electrons, which is Ag⁺ in silver nitrate. (Ca does not deposit from an aqueous solution.)

Teacher's Note:
1. Moles of metal = faradays / charge on ion.
2. Au³⁺ needs 3 F and Cu²⁺ needs 2 F.

Question 28: When 96500 C of electricity is passed through molten barium chloride, the amount of barium deposited will be :
(a) 0.5 mol
(b) 1.0 mol
(c) 1.5 mol
(d) 2.0 mol
Show Answer & Explanation

Answer: (a) 0.5 mol

Explanation:
1. Ba²⁺ + 2e⁻ → Ba, so 1 F deposits half a mole of Ba.
2. 96500 C = 1 F, giving 0.5 mol.

Teacher's Note:
1. Ba²⁺ carries a charge of 2+.
2. In aqueous solution barium would not be deposited, so the molten salt is used.

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