JEE Chemistry Electrochemistry MCQs Set 12

Answer: (a) Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)

Explanation:
1. Oxidation (anode) is Cu → Cu²⁺ + 2e⁻ and reduction (cathode) is Ag⁺ + e⁻ → Ag.
2. In cell notation the anode is written on the left and the cathode on the right, with a double line for the salt bridge: Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s).

Teacher's Note:
1. Left electrode = oxidation, right electrode = reduction.
2. Each electrode is written as metal | ion.

Answer: (c) The velocities of K⁺ and NO₃⁻ are nearly the same

Explanation:
1. A salt bridge must not set up its own junction potential.
2. K⁺ and NO₃⁻ have almost equal ionic mobilities, so the liquid junction potential is negligible. KNO₃ also does not react with the ions in the half cells.

Teacher's Note:
1. Other suitable salts: KCl (but not with Ag⁺ because AgCl precipitates) and NH₄NO₃.
2. High solubility alone is not the reason.

Answer: (a) Anode is positive

Explanation:
1. In a galvanic cell oxidation occurs at the anode, which is the negative electrode because it releases electrons.
2. The cathode is positive and reduction occurs there.
3. The electrode with the lower reduction potential is oxidised, so (b), (c) and (d) are correct. Only (a) is wrong.

Teacher's Note:
1. Polarity is reversed in an electrolytic cell: anode positive, cathode negative.
2. In both types, oxidation is at the anode and reduction at the cathode.

Answer: (d) −0.30 V

Explanation:
1. Cathode (reduction): Fe²⁺/Fe, E° = −0.44 V. Anode (oxidation): Sn/Sn²⁺, E° = −0.14 V.
2. E°cell = E°cathode − E°anode = −0.44 − (−0.14) = −0.30 V.
3. The negative value shows that the reaction as written is not spontaneous.

Teacher's Note:
1. E°cell = E°(reduction half) − E°(oxidation half), both taken as reduction potentials.
2. A negative E°cell means the reverse reaction is spontaneous.

Answer: (c) 2.7 V

Explanation:
1. Cu²⁺/Cu has the higher reduction potential, so it is the cathode and Mg is the anode.
2. E°cell = 0.33 − (−2.37) = 2.70 V.

Teacher's Note:
1. The cell EMF is always positive when the electrodes are connected in the spontaneous direction.
2. Subtract the anode potential from the cathode potential.

Answer: (c) If the copper electrode is the anode, E°cell is +0.46 V

Explanation:
1. Copper as anode and silver as cathode: E° = 0.80 − 0.34 = +0.46 V, so (c) is right.
2. Copper as cathode and silver as anode gives 0.34 − 0.80 = −0.46 V; the values in (a) and (b) are wrong, and silver as cathode gives +0.46, not −0.34 V.

Teacher's Note:
1. E°cell = E°cathode − E°anode.
2. Silver, the stronger oxidising agent, is the cathode in the spontaneous cell.

Answer: (d) No reaction will take place

Explanation:
1. E°(Cu²⁺/Cu) = +0.34 V and E°(Fe²⁺/Fe) = −0.44 V.
2. Copper is lower than iron in the activity series, so it cannot reduce Fe²⁺. E° for Cu + Fe²⁺ → Cu²⁺ + Fe is −0.44 − 0.34 = −0.78 V, so no reaction occurs.

Teacher's Note:
1. A metal displaces another metal from solution only if its reduction potential is lower.
2. Fe displaces Cu, not the reverse.

Answer: (a) Zn(s)

Explanation:
1. The species with the most negative reduction potential is the easiest to oxidise.
2. Zn/Zn²⁺ has the lowest value (−0.762 V), so Zn is the strongest reducing agent.

Teacher's Note:
1. Strong reducing agent means low (negative) E° of the couple it belongs to.
2. Compare the potentials of the couples, not of the individual species.

Answer: (c) Ag, Hg, Cu

Explanation:
1. The ion with the highest reduction potential is discharged first: Ag (0.80), then Hg (0.79), then Cu (0.34).
2. Mg²⁺ (−2.37 V) is much harder to reduce than water, so in an aqueous solution hydrogen is evolved instead and Mg is not deposited.

Teacher's Note:
1. Higher E° means easier reduction and earlier deposition.
2. Very active metals (K, Na, Ca, Mg, Al) are not deposited from aqueous solutions.

Answer: (a) Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

Explanation:
1. Zn + 2Ag⁺ → Zn²⁺ + 2Ag: E°cell = E°ox(Zn) + E°red(Ag) = 0.76 + 0.80 = +1.56 V, positive, so spontaneous.
2. The reverse reaction has E° = −1.56 V. Options (c) and (d) are not balanced redox reactions.

Teacher's Note:
1. Add the oxidation potential of the species oxidised and the reduction potential of the species reduced.
2. The reaction with positive E°cell occurs spontaneously.

Answer: (b) −212.30 kJ

Explanation:
1. The cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu involves n = 2 electrons.
2. ΔG° = −nFE° = −2 × 96500 × 1.10 = −212300 J = −212.3 kJ.

Teacher's Note:
1. The maximum electrical work obtainable equals −ΔG°.
2. Use ΔG° = −nFE° with E° in volts to get joules.

Answer: (d) \( 1\times10^{10} \)

Explanation:
1. \( \log K=\frac{nE^\circ}{0.0591}=\frac{2\times0.295}{0.0591}=9.98\approx10 \).
2. So K = 10¹⁰.

Teacher's Note:
1. At 25 °C: log K = nE°/0.0591.
2. A positive E° means K > 1.

Answer: (b) 15.9 g

Explanation:
1. Equivalents of H₂ = 0.50/1.008 ≈ 0.496.
2. The same charge liberates the same number of equivalents of Cu. Equivalent mass of Cu = 63.5/2 = 31.75 g.
3. Mass of Cu = 0.496 × 31.75 = 15.75 g ≈ 15.9 g.

Teacher's Note:
1. Faraday's second law: equal charge liberates equal equivalents of different substances.
2. Equivalent mass of Cu²⁺ is half the atomic mass.

Answer: (c) 1.50

Explanation:
1. Moles of Al = 13.5/27 = 0.5.
2. Al³⁺ + 3e⁻ → Al, so each mole needs 3 F.
3. Charge = 0.5 × 3 = 1.5 F.

Teacher's Note:
1. Faradays = moles of metal × charge on the ion.
2. Number of faradays equals the moles of electrons.

Answer: (a) 108 g, 29.5 g, 17.3 g

Explanation:
1. One faraday deposits one gram equivalent of each metal.
2. Ag: 108/1 = 108 g; Ni: 59/2 = 29.5 g; Cr: 52/3 = 17.3 g.

Teacher's Note:
1. Equivalent mass = atomic mass/charge on the ion.
2. Cells connected in series carry the same charge.

Answer: (d) Sulphuric acid is consumed

Explanation:
1. Discharge reactions: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ (anode) and PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O (cathode).
2. Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Sulphuric acid is consumed and PbSO₄ is formed.

Teacher's Note:
1. The density of the acid falls during discharge and is used to judge the state of charge.
2. Charging reverses these reactions.

Answer: (a) 630 ohm⁻¹ cm² mol⁻¹

Explanation:
1. \( \Lambda_m=\frac{\kappa\times1000}{C}=\frac{6.3\times10^{-2}\times1000}{0.1}=630\ \text{ohm}^{-1}\text{cm}^2\text{mol}^{-1} \).

Teacher's Note:
1. Λm = 1000κ/C with κ in ohm⁻¹ cm⁻¹ and C in mol L⁻¹.
2. The factor 1000 converts litres to cm³.

Answer: (d) 0.1 M difluoroacetic acid

Explanation:
1. All are weak acids at the same concentration, so conductivity depends on the degree of ionisation.
2. Electron-withdrawing fluorine atoms stabilise the carboxylate ion, so difluoroacetic acid is the strongest acid and is the most ionised.

Teacher's Note:
1. Order of acidity: CHF₂COOH > CH₂FCOOH > CH₂ClCOOH > CH₃COOH.
2. More ions means higher conductivity.

Answer: (c) 388.8

Explanation:
1. Apply Kohlrausch's law: λ(ClCH₂COOH) = λ(ClCH₂COONa) + λ(HCl) − λ(NaCl).
2. = 224 + 203 − 38.2 = 388.8 ohm⁻¹ cm² geq⁻¹.

Teacher's Note:
1. Kohlrausch's law: the limiting molar conductivity of an electrolyte is the sum of the ionic contributions.
2. Arrange the data so that Na⁺ and Cl⁻ cancel.

Answer: (c) 203

Explanation:
1. At infinite dilution the ions migrate independently: λ°(BaCl₂) = λ°(½Ba²⁺) + λ°(Cl⁻).
2. = 127 + 76 = 203 ohm⁻¹ cm² eq⁻¹.

Teacher's Note:
1. Use equivalent conductances of the ions per equivalent: add them directly.
2. Do not divide by the number of ions.