JEE Chemistry Electrochemistry MCQs Set 09

Answer: (b) 18 kg

Explanation:
1. 40 kg Ca = 1000 mol, needing 2000 F.
2. Al³⁺ + 3e⁻ → Al, so 2000 F gives 2000/3 mol = 666.7 mol, which is 18 kg.

Teacher's Note:
1. Same charge means the same number of gram equivalents.
2. Equivalent weights are Ca = 20 and Al = 9.

Answer: (b) 1 : 1/2 : 1/3

Explanation:
1. Moles deposited = F / charge on the ion.
2. For 1 F: Ag = 1, Cu = 1/2 and Fe = 1/3 mol, giving 1 : 1/2 : 1/3.

Teacher's Note:
1. Higher charge means fewer moles per faraday.
2. Equal equivalents are deposited, not equal moles.

Answer: (b) a decrease in pH

Explanation:
1. Cu²⁺ is reduced at the cathode and water is oxidised at the anode, 2H₂O → O₂ + 4H⁺ + 4e⁻.
2. H⁺ accumulates (as H₂SO₄), so the pH decreases.

Teacher's Note:
1. Anode reaction produces H⁺.
2. The solution becomes acidic and the blue colour fades.

Answer: (c) NaCl(aq)

Explanation:
1. In aqueous NaCl, water is reduced at the cathode to give H₂ and Cl⁻ is oxidised at the anode to give Cl₂.
2. CuSO₄ and CuCl₂ deposit copper, and pure water gives H₂ and O₂.

Teacher's Note:
1. Na⁺ is not reduced from aqueous solution.
2. Concentrated NaCl gives Cl₂ at the anode.

Answer: (c) sulphuric acid is regenerated

Explanation:
1. Charging reverses the discharge reaction: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄.
2. PbSO₄ is converted back to Pb and PbO₂ and sulphuric acid is regenerated.

Teacher's Note:
1. Acid is used up in discharging and made again in charging.
2. The density of the acid is a measure of the charge.

Answer: (c) 1.212 V

Explanation:
1. Fe is oxidised at the anode and Fe³⁺ is reduced at the cathode.
2. E°cell = 0.771 − (−0.441) = 1.212 V.

Teacher's Note:
1. The cathode is the Fe³⁺/Fe²⁺ couple with the higher potential.
2. Subtract the anode potential from the cathode potential.

Answer: (d) A + B⁺ → A⁺ + B

Explanation:
1. In the cell notation the left electrode (A) is the anode where oxidation occurs, and the right electrode (B) is the cathode where B⁺ is reduced.
2. A positive emf means the reaction is spontaneous as written: A + B⁺ → A⁺ + B.

Teacher's Note:
1. Anode is written on the left and the cathode on the right.
2. A positive emf confirms this arrangement.

Answer: (c) H₂, Br₂ and NaOH

Explanation:
1. At the cathode water is reduced to H₂ (Na⁺ is not reduced from water) leaving OH⁻, which forms NaOH.
2. At the anode Br⁻ is oxidised to Br₂ in preference to water.

Teacher's Note:
1. Na⁺ never discharges from aqueous solution.
2. The overall products are H₂, Br₂ and NaOH.

Answer: (c) 4.0 × 10¹⁵

Explanation:
1. log K = nE°/0.0591 = 2 × 0.46 / 0.0591 = 15.6.
2. K = 10^15.6, which is about 4.0 × 10¹⁵.

Teacher's Note:
1. Two electrons are transferred.
2. Use log K = nE° / 0.0591 at 298 K.

Answer: (b) infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

Explanation:
1. Kohlrausch's law of independent migration of ions applies at infinite dilution.
2. Each ion contributes a fixed amount to the equivalent conductance, irrespective of the other ion.

Teacher's Note:
1. Independent migration holds only when inter-ionic attraction is absent.
2. That is the case only at infinite dilution.

Answer: (b) 1.25 × 10⁻⁵

Explanation:
1. α = 8/400 = 0.02.
2. Ka = C α² / (1 − α) = (1/32)(0.02)² / 0.98 ≈ 1.28 × 10⁻⁵, which is about 1.25 × 10⁻⁵.

Teacher's Note:
1. Find α from the ratio of conductances.
2. For a weak acid, Ka is approximately C α².

Answer: (b) increase in ionic mobility of ions

Explanation:
1. A strong electrolyte is already completely ionised, so the number of ions per mole cannot increase.
2. On dilution the inter-ionic attraction decreases, so the mobility of the ions increases.

Teacher's Note:
1. For weak electrolytes the increase is due to greater ionisation.
2. For strong electrolytes it is due to greater ionic mobility.

Answer: (c) (b) and (d)

Explanation:
1. emf = SOP(anode) + SRP(cathode), which is (b), and since SRP(cathode) = −SOP(cathode), it equals SOP(anode) − SOP(cathode), which is (d).
2. The relations (a) and (c) give wrong signs.

Teacher's Note:
1. Reduction potential is the negative of oxidation potential for the same half cell.
2. Convert one form into the other to check.

Answer: (a) Λ°(Al³⁺) + Λ°(SO₄²⁻)

Explanation:
1. Equivalent conductance at infinite dilution is the sum of the equivalent conductances of the cation and the anion.
2. Since both values are already per equivalent, Λ°eq = Λ°(Al³⁺) + Λ°(SO₄²⁻).

Teacher's Note:
1. Per equivalent means the valence is already included.
2. The fractions 1/3 and 1/2 apply only to molar conductances.

Answer: (b) −89.0 kJ

Explanation:
1. The reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag has n = 2.
2. ΔG° = −nFE° = −2 × 96500 × 0.46 = −88780 J ≈ −89.0 kJ.

Teacher's Note:
1. Remember the factor n = 2.
2. Check the unit, kJ against J.

Answer: (b) −(RT/F) ln (0.1)

Explanation:
1. This is a concentration cell with E° = 0 and n = 2.
2. E = (RT/2F) ln(0.1/0.001) = (RT/2F) ln 100 = (RT/F) ln 10 = −(RT/F) ln 0.1.

Teacher's Note:
1. The more concentrated half cell is the cathode.
2. ln 100 = 2 ln 10.

Answer: (c) Z > X > Y

Explanation:
1. Reducing power increases as the standard reduction potential decreases.
2. Z (−3.0 V) > X (−1.2 V) > Y (+0.5 V).

Teacher's Note:
1. Rank from the most negative E° to the most positive.
2. The metal with the lowest E° is the best reducing agent.

Answer: (d) ΔG° > 0 ; Keq < 1

Explanation:
1. ΔG° = −nFE°, so a negative E° gives a positive ΔG°.
2. Since ΔG° = −RT ln K, a positive ΔG° means K < 1.

Teacher's Note:
1. Negative E° means non-spontaneous in the forward direction.
2. Equilibrium then favours the reactants.

Answer: (d) I⁻ will be oxidised to I₂

Explanation:
1. E°(Fe³⁺/Fe²⁺) is greater than E°(I₂/I⁻), so Fe³⁺ is the stronger oxidising agent.
2. The favourable reaction is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, in which I⁻ is oxidised to I₂.

Teacher's Note:
1. The couple with the higher E° acts as the oxidising agent.
2. Iodine cannot oxidise Fe²⁺ here.

Answer: (b) Λ°m(NH₄Cl) + Λ°m(NaOH) − Λ°m(NaCl)

Explanation:
1. NH₄OH contains NH₄⁺ and OH⁻.
2. Λ°(NH₄OH) = Λ°(NH₄Cl) + Λ°(NaOH) − Λ°(NaCl), because the Na⁺ and Cl⁻ contributions cancel.

Teacher's Note:
1. Choose strong electrolytes that supply each required ion.
2. Cancel the unwanted ions by subtraction.

Answer: (c) 0.59 V

Explanation:
1. The oxidation potential is E(ox) = +0.0591 × pH.
2. For pH = 10, E(ox) = 0.591 V ≈ 0.59 V.

Teacher's Note:
1. Oxidation potential has the opposite sign to reduction potential.
2. The reduction potential would be −0.59 V.

Answer: (b) 1.10 V

Explanation:
1. Ag₂O is reduced at the cathode and Zn is oxidised at the anode.
2. E°cell = 0.34 − (−0.76) = 1.10 V.

Teacher's Note:
1. Cathode minus anode reduction potentials.
2. Button cells give about 1.1 V to 1.5 V.

Answer: (d) fuel cell

Explanation:
1. A fuel cell continuously converts the free energy of a fuel and oxygen directly into electricity.
2. The other devices are not direct fuel energy converters.

Teacher's Note:
1. Fuel cells have a higher efficiency than burning the fuel.
2. The H₂-O₂ fuel cell was used in space missions.

Answer: (d) 0.38 V

Explanation:
1. E = E° + (0.0591/2) log [Cu²⁺] = E° + 0.02955 × (−2) = E° − 0.059.
2. So E° = 0.32 + 0.059 = 0.379 V ≈ 0.38 V.

Teacher's Note:
1. A dilute solution gives a lower potential than the standard value.
2. Add back the correction to get E°.

Answer: (a) 10⁻¹⁴ atm

Explanation:
1. For 2H⁺ + 2e⁻ → H₂, E = −(0.0591/2) log (P/[H⁺]²).
2. E = 0 requires P = [H⁺]² = (10⁻⁷)² = 10⁻¹⁴ atm.

Teacher's Note:
1. [H⁺] in pure water is 10⁻⁷ M.
2. E° of H⁺/H₂ is zero.

Answer: (d) 11.52 S cm²/mol

Explanation:
1. Λm = κ × 1000 / C = 5.76 × 10⁻³ × 1000 / 0.5.
2. = 11.52 S cm² mol⁻¹.

Teacher's Note:
1. Concentration in mol per litre equals mol/dm³.
2. Use κ in S cm⁻¹ with the factor 1000.

Answer: (d) 110 minutes

Explanation:
1. 2Cl⁻ → Cl₂ + 2e⁻, so 0.10 mol Cl₂ needs 0.20 F = 19300 C.
2. t = 19300 / 3 = 6433 s ≈ 107 minutes, which is nearest to 110 minutes.

Teacher's Note:
1. Charge = moles of electrons × 96500.
2. Convert seconds to minutes at the end.

Answer: (b) E₁ > E₂

Explanation:
1. E₁ = E° − (0.059/2) log(0.01/1) = E° + 0.059.
2. E₂ = E° − (0.059/2) log(1/0.01) = E° − 0.059, so E₁ > E₂.

Teacher's Note:
1. Lower product concentration and higher reactant concentration give a larger emf.
2. Compare the values of Q in the two cases.