JEE Chemistry Electrochemistry MCQs Set 13

Answer: (a) II and III

Explanation:
1. In a galvanic cell the cathode is positive, so (I) is true and (II) is wrong.
2. Electrons flow from the anode to the cathode through the external circuit, so (III) is wrong.
3. Reduction occurs at the cathode: (IV) is true. The wrong statements are II and III.

Teacher's Note:
1. Electrons leave the anode (oxidation) and enter the cathode (reduction).
2. Conventional current flows in the opposite direction.

Answer: (c) 0.80 − (−0.25) = +1.05 V

Explanation:
1. Ag⁺/Ag is on the right, so it is the cathode; Ni is the anode.
2. E°cell = E°cathode − E°anode = 0.80 − (−0.25) = +1.05 V.

Teacher's Note:
1. Never add two reduction potentials to get the EMF.
2. Cathode minus anode, both as reduction potentials.

Answer: (c) Mg

Explanation:
1. Reducing agents are the reduced forms of the couples: Mg and Fe.
2. The couple with the lower E° has the stronger reduced form, so Mg (E° = −2.37 V) is the best reducing agent.

Teacher's Note:
1. Mg²⁺ and Fe³⁺ are oxidised forms, so they are oxidising agents.
2. Lower reduction potential means a stronger reductant.

Answer: (a) Br⁻ < Fe²⁺ < Al

Explanation:
1. The reducing power of the reduced species increases as E° of its couple decreases.
2. E°: Br₂/Br⁻ = +1.08 V, Fe³⁺/Fe²⁺ = +0.77 V, Al³⁺/Al = −1.66 V.
3. So the reducing power increases in the order Br⁻ < Fe²⁺ < Al.

Teacher's Note:
1. High E° means the oxidised form is strongly oxidising and the reduced form is weakly reducing.
2. List the couples in order of E° and reverse it.

Answer: (d) 0.772 V

Explanation:
1. Potentials cannot be subtracted directly; use free energies: ΔG° = −nFE°.
2. Fe³⁺ + 3e⁻ → Fe has ΔG₁ = −3F(−0.036); Fe²⁺ + 2e⁻ → Fe has ΔG₂ = −2F(−0.440).
3. Fe³⁺ + e⁻ → Fe²⁺ = reaction 1 − reaction 2, so −1·F·E° = ΔG₁ − ΔG₂ giving E° = 3(−0.036) − 2(−0.440) = −0.108 + 0.880 = +0.772 V.

Teacher's Note:
1. Potentials of half reactions with different numbers of electrons must be combined through ΔG°.
2. E° = (n₁E₁ − n₂E₂)/(n₁ − n₂).

Answer: (d) 2x₂ − x₁

Explanation:
1. Cu²⁺ + 2e⁻ → Cu: ΔG₂ = −2F x₂. Cu⁺ + e⁻ → Cu: ΔG₁ = −F x₁.
2. Cu²⁺ + e⁻ → Cu⁺ is the first minus the second: ΔG = ΔG₂ − ΔG₁ = −2Fx₂ + Fx₁ = −F(2x₂ − x₁).
3. So E° = 2x₂ − x₁.

Teacher's Note:
1. Add or subtract ΔG° values, never E° values directly, when the electron numbers differ.
2. Check with numbers: x₂ = 0.337, x₁ = 0.521 gives 0.153 V.

Answer: (c) C₂ > C₁

Explanation:
1. The cell reaction is Zn(s) + Zn²⁺(C₂) → Zn²⁺(C₁) + Zn(s), so Q = C₁/C₂.
2. E = −(0.0591/2) log(C₁/C₂) = (0.0591/2) log(C₂/C₁).
3. E > 0, that is ΔG < 0, requires C₂ > C₁.

Teacher's Note:
1. In a concentration cell the electrons flow from the dilute half to the concentrated half.
2. E° = 0, so the EMF comes only from the concentration ratio.

Answer: (c) \( 6.26\times10^{-3} \)

Explanation:
1. Reduction: Hg₂²⁺ + 2e⁻ → 2Hg (0.789 V). Oxidation: Hg → Hg²⁺ + 2e⁻ (−0.854 V). Sum: Hg₂²⁺ → Hg + Hg²⁺ with n = 2.
2. E° = 0.789 − 0.854 = −0.065 V.
3. \( \log K=\frac{2(-0.065)}{0.0591}=-2.20 \), so K = 6.3 × 10⁻³.

Teacher's Note:
1. Both half reactions involve 2 electrons, so the potentials can be combined directly.
2. K < 1 because E° is negative.

Answer: (b) 1, 3 and 2

Explanation:
1. Equal charge deposits equal equivalents. Equivalents = mass × valency/atomic weight.
2. A: 2.1 v_A/7 = 0.3 v_A; B: 2.7 v_B/27 = 0.1 v_B; C: 7.2 v_C/48 = 0.15 v_C.
3. Setting 0.3 v_A = 0.1 v_B = 0.15 v_C gives v_A : v_B : v_C = 1 : 3 : 2.

Teacher's Note:
1. Equivalent mass = atomic weight/valency.
2. Valencies must be small whole numbers; choose the smallest set that fits.

Answer: (c) \( 8.314\times10^{36} \)

Explanation:
1. \( \log K=\frac{nE^\circ}{0.0591}=\frac{2\times1.091}{0.0591}=36.92 \).
2. K = 10^{36.92} = 8.3 × 10³⁶.

Teacher's Note:
1. Use n = 2 for the Zn/Cu²⁺ reaction.
2. A large positive E° gives a very large K.

Answer: (c) 0.521 V

Explanation:
1. Cu²⁺ + 2e⁻ → Cu: ΔG° = −2F(0.337). Cu²⁺ + e⁻ → Cu⁺: ΔG° = −F(0.153).
2. Cu⁺ + e⁻ → Cu is the difference: ΔG° = −2F(0.337) + F(0.153) = −F(0.674 − 0.153).
3. E° = 0.674 − 0.153 = 0.521 V.

Teacher's Note:
1. Combine through ΔG° because the half reactions have different electron numbers.
2. Cu⁺ is unstable in water and disproportionates, consistent with E°(Cu⁺/Cu) > E°(Cu²⁺/Cu⁺).

Answer: (a) The potential decreases by 0.38 V with a decrease in oxidising power

Explanation:
1. \( E=E^\circ-\frac{0.0591}{5}\log\frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}=E^\circ+\frac{0.0591}{5}\times8\log[H^+] \).
2. For [H⁺] = 10⁻⁴: ΔE = (0.0591 × 8/5)(−4) = −0.378 V.
3. The potential falls by about 0.38 V, so permanganate is a weaker oxidising agent in less acidic solution.

Teacher's Note:
1. Oxidising power of MnO₄⁻ is greatest in strongly acidic medium.
2. The coefficient of [H⁺] in the Nernst term is 8, the number of H⁺ in the reaction.

Answer: (a) 0.34

Explanation:
1. The cell reaction Fe²⁺ + Ag⁺ → Fe³⁺ + Ag has E°cell = 0.7991 − 0.771 = 0.0281 V.
2. With [Fe²⁺] = [Fe³⁺], Q = 1/[Ag⁺] and E = E° − 0.0591 log(1/[Ag⁺]) = 0.0281 + 0.0591 log[Ag⁺].
3. Setting E = 0: log[Ag⁺] = −0.0281/0.0591 = −0.4755, so [Ag⁺] = 0.335 M ≈ 0.34 M.

Teacher's Note:
1. The Nernst equation is applied to the whole cell reaction with n = 1.
2. At E = 0 the cell is at equilibrium.

Answer: (a) 0.3667 V

Explanation:
1. \( E^\circ=\frac{0.0591}{n}\log K=\frac{0.0591}{2}\log(2.69\times10^{12}) \).
2. log K = 12.43, so E° = 0.02955 × 12.43 = 0.367 V.

Teacher's Note:
1. Use E° = (0.0591/n) log K at 298 K.
2. n = 2 for Sr → Sr²⁺.

Answer: (c) −0.041 V

Explanation:
1. Reaction: H₂(0.4 atm) + 2H⁺(0.01 M) → 2H⁺(0.1 M) + H₂(0.1 atm), n = 2, E° = 0.
2. \( Q=\frac{(0.1)^2(0.1)}{(0.01)^2(0.4)}=25 \).
3. E = −(0.0591/2) log 25 = −0.02955 × 1.398 = −0.041 V.

Teacher's Note:
1. pH = 1 means [H⁺] = 0.1 M and pH = 2 means 0.01 M.
2. In a concentration cell E° = 0 and the sign depends on Q.

Answer: (b) 27172 C

Explanation:
1. Volume of copper = 10 × 10 × 10⁻² = 1 cm³, mass = 8.94 g.
2. Moles of Cu = 8.94/63.5 = 0.1408 mol.
3. Cu²⁺ + 2e⁻ → Cu, so charge = 0.1408 × 2 × 96500 = 27172 C.

Teacher's Note:
1. Find the mass from volume × density, then moles, then charge.
2. Each mole of Cu needs 2 faradays.

Answer: (c) 66.7

Explanation:
1. Moles of electrons = 1.81 × 10²²/6.022 × 10²³ = 0.03006.
2. M²⁺ + 2e⁻ → M, so moles of M = 0.01503.
3. Atomic weight = 1/0.01503 = 66.5 ≈ 66.7.

Teacher's Note:
1. Convert electrons to moles by dividing by N_A.
2. Divide by 2 because each M²⁺ needs two electrons.

Answer: (b) p₁ > p₂

Explanation:
1. Anode: H₂(p₁) → 2H⁺ + 2e⁻. Cathode: 2H⁺ + 2e⁻ → H₂(p₂). Overall H₂(p₁) → H₂(p₂), so Q = p₂/p₁ and E° = 0.
2. E = −(0.0591/2) log(p₂/p₁).
3. E > 0 requires p₂/p₁ < 1, that is p₁ > p₂.

Teacher's Note:
1. Gas flows from high pressure to low pressure in a hydrogen concentration cell.
2. E° = 0 for identical electrodes.

Answer: (b) 0.2281 cm⁻¹

Explanation:
1. Cell constant = κ × R = 2.768 × 10⁻³ × 82.4 = 0.2281 cm⁻¹.

Teacher's Note:
1. κ = (1/R)(l/A), so l/A = κR.
2. The unit of cell constant is cm⁻¹.

Answer: (c) Pt | H₂(g) | HCl(sol) || Cl⁻ | AgCl(s) | Ag

Explanation:
1. Oxidation: ½H₂ → H⁺ + e⁻ at a platinum hydrogen electrode (anode, left), in HCl solution.
2. Reduction: AgCl + e⁻ → Ag + Cl⁻ at the silver-silver chloride electrode (cathode, right), written Cl⁻ | AgCl(s) | Ag.
3. Only option (c) shows these two half cells with a salt bridge between them.

Teacher's Note:
1. The cathode must involve AgCl and Cl⁻, not Ag⁺ from AgNO₃.
2. Write the oxidation half on the left and the reduction half on the right.