CBSE Class 12 Chemistry Electrochemistry MCQs Set 12

Answer: (c) electrolysis

Explanation:
1. The decomposition of an electrolyte (chemical change) by passing electric current is called electrolysis.

Teacher's Note:
1. Electrolysis uses electrical energy for a non-spontaneous reaction; a galvanic cell does the reverse.

Answer: (b) 3 F

Explanation:
1. Oxidation state of Mn changes from +7 in MnO₄⁻ to +4 in MnO₂.
2. MnO₄⁻ + 3e⁻ → MnO₂, so 3 F is required.

Teacher's Note:
1. MnO₄⁻ needs 5 F to become Mn²⁺ but only 3 F to become MnO₂; always check the product.

Answer: (b) 251.5

Explanation:
1. Λ°_m(NH₄OH) = Λ°_m(NH₄Cl) + Λ°_m(NaOH) − Λ°_m(NaCl)
2. = 129.8 + 248.1 − 126.4 = 251.5 ohm⁻¹ cm² mol⁻¹

Teacher's Note:
1. Choose the combination so that the unwanted ions (Na⁺ and Cl⁻) cancel out.

Answer: (c) 50 g

Explanation:
1. Q = I × t = 9.65 × 10 × 60 = 5790 C
2. Equivalent wt. = \( \frac{W \times 96500}{Q} \) = \( \frac{3.0 \times 96500}{5790} \) = 50 g

Teacher's Note:
1. Always convert minutes to seconds before calculating charge in coulombs.

Answer: (a) free energy change

Explanation:
1. The maximum electrical work done by a galvanic cell equals the decrease in Gibbs free energy: ΔG = −nFE_cell.

Teacher's Note:
1. The negative sign shows that work is done by the system when ΔG is negative.

Answer: (a) an electrolytic cell

Explanation:
1. During charging, external electrical energy drives the non-spontaneous reverse reaction, 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄, so it works as an electrolytic cell.

Teacher's Note:
1. A secondary cell is galvanic while discharging and electrolytic while charging.

Answer: (b) cathode to anode

Explanation:
1. Electrons flow from anode to cathode in the external circuit.
2. Conventional current flows in the opposite direction, i.e. from cathode to anode (Cu rod to Zn rod in a Daniell cell).

Teacher's Note:
1. Electron flow and current flow are always opposite; read the question carefully to see which one is asked.

Answer: (c) Hg

Explanation:
1. Metals below hydrogen in the electrochemical series (positive E°) such as Hg cannot reduce water to H₂.
2. Na reacts with cold water, Mg with hot water and Fe with steam to give H₂.

Teacher's Note:
1. A metal with positive reduction potential (Cu, Hg, Ag, Au) cannot liberate hydrogen from water or dilute acids.

Answer: (b) a decrease in pH

Explanation:
1. Cathode: Cu²⁺ + 2e⁻ → Cu; Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻.
2. H⁺ ions (H₂SO₄) accumulate in the solution, so the pH decreases.

Teacher's Note:
1. With copper electrodes instead, the anode dissolves and the solution composition does not change.

Answer: (a) Al³⁺

Explanation:
1. A metal can displace only those metal ions whose reduction potential is higher than its own.
2. E°(Al³⁺/Al) = −1.66 V is lower than E°(Zn²⁺/Zn) = −0.76 V, so Zn cannot displace Al³⁺; Cu²⁺, Fe²⁺ and Ag⁺ are displaced.

Teacher's Note:
1. A metal higher in the activity series (more negative E°) displaces one lower down, never the reverse.

Answer: (b) Ni-Cd cell

Explanation:
1. A secondary battery can be recharged by passing current in the opposite direction.
2. The Ni-Cd cell is rechargeable; the Leclanche (dry) cell, mercury cell and Daniell cell are primary cells.

Teacher's Note:
1. The lead storage battery and the Ni-Cd cell are the two secondary cells in the syllabus.

Answer: (c) molar conductance

Explanation:
1. On increasing concentration, the number of ions per unit volume increases, so conductance and conductivity (specific conductance) increase.
2. Molar conductance decreases because interionic attractions increase (and for weak electrolytes the degree of dissociation decreases).

Teacher's Note:
1. Conductivity and molar conductivity change in opposite directions with dilution; this is a favourite exam trap.

Answer: (a) A–R, B–Q, C–S, D–P

Explanation:
1. Conductance G = 1/R: unit S (siemens). Conductivity κ = G × l/A: unit S cm⁻¹.
2. Molar conductivity Λ_m = κ × 1000/C: unit S cm² mol⁻¹. Cell constant l/A: unit m⁻¹ (or cm⁻¹).

Teacher's Note:
1. Derive units from the formula rather than memorising them; e.g. cell constant = length/area gives m⁻¹.

Answer: (d) A–Q, B–S, C–R, D–P

Explanation:
1. Mn: +7 → +2 needs 5 F; Cu²⁺ → Cu needs 2 F; Al³⁺ → Al needs 3 F; Na⁺ → Na needs 1 F.

Teacher's Note:
1. Number of faradays per mole of product = change in oxidation number of the element deposited.

Answer: (b) C₂H₅OH

Explanation:
1. C₂H₅OH is a covalent compound and does not ionise in solution, so it does not conduct electricity.
2. CH₃COONa, NaCl and KOH are ionic and give ions in solution or in the molten state.

Teacher's Note:
1. Conduction through an electrolyte needs free ions; covalent molecules like ethanol or sugar give none.

Answer: (c) 3

Explanation:
1. Charge on L.H.S. = −1 + 4 = +3
2. Charge on R.H.S. = 0
3. Hence, 3 e⁻ should be added on L.H.S. to balance the charges.

Teacher's Note:
1. Cross-check with oxidation numbers: N goes from +5 in NO₃⁻ to +2 in NO, a gain of 3 electrons.

Answer: (c) 1.5

Explanation:
1. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
2. Thus, 5 moles of electrons reduce 1 mole of MnO₄⁻.
3. ∴ 7.5 moles of electrons will reduce MnO₄⁻ = \( \frac{1}{5} \) × 7.5 = 1.5 moles

Teacher's Note:
1. Mn changes from +7 to +2, so each MnO₄⁻ needs 5 electrons – divide the electrons available by 5.

Answer: (b) H₂ is evolved at cathode

Explanation:
1. NaCl (aq) → Na⁺ (aq) + Cl⁻ (aq)
2. H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
3. As H⁺ ions have lower discharge potential than Na⁺ ion, H⁺ ions are discharged at the cathode.

Teacher's Note:
1. Na is obtained only from molten NaCl; in water, water is reduced in preference to Na⁺ (E° of Na⁺/Na = −2.71 V).

Answer: (a) 6 F

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
2. Thus, 1 mole of Cr₂O₇²⁻ requires for reduction charge = 6 F

Teacher's Note:
1. There are two Cr atoms, each going from +6 to +3 – students often forget to multiply 3 by 2.

Answer: (b) 540 g

Explanation:
1. Ag⁺ + e⁻ → Ag; 1 F deposits 1 mol (108 g) of Ag.
2. 5 F deposit 5 × 108 = 540 g of Ag.

Teacher's Note:
1. For a monovalent ion, number of faradays = number of moles deposited.

Answer: (d) 63.5 g

Explanation:
1. Cu²⁺ + 2e⁻ → Cu
2. Thus, 2 F charge deposit 1 mol of Cu, i.e., 63.5 g.

Teacher's Note:
1. Write the electrode reaction first; the number of electrons tells you how many faradays deposit one mole.

Answer: (c) 9650 C

Explanation:
1. The oxidation reaction is MnO₄²⁻ → MnO₄⁻ + e⁻
2. 1 mol of MnO₄²⁻ for oxidation requires electricity = 1 F = 96500 C
3. ∴ 0.1 mol of MnO₄²⁻ will require electricity = 9650 C

Teacher's Note:
1. Mn goes only from +6 to +7 here, a change of one electron – do not confuse with MnO₄⁻ → Mn²⁺ (5 electrons).

Answer: (b) 110 minutes

Explanation:
1. 2Cl⁻ → Cl₂ (g) + 2e⁻
2. To produce 1 mol of Cl₂, charge required = 2 F = 2 × 96500 C
3. ∴ To produce 0.1 mol of Cl₂, charge required = 2 × 96500 × 0.1 = 19300 C
4. Q = I × t ∴ t = \( \frac{Q}{I} \) = \( \frac{19300}{3} \) s = \( \frac{19300}{3 \times 60} \) min = 107.22 min ≈ 110 min

Teacher's Note:
1. Convert seconds to minutes only at the end, and remember that each Cl₂ molecule needs 2 electrons.

Answer: (c) Decrease and increase

Explanation:
1. On dilution, the number of ions per unit volume decreases, so specific conductance (κ) decreases.
2. Molar conductance (Λ_m = κ × V) increases because the volume containing 1 mole increases much more than κ decreases.

Teacher's Note:
1. κ is about one cm³ of solution, Λ_m is about one mole of electrolyte – that is why they move in opposite directions on dilution.

Answer: (a) CsCl

Explanation:
1. CsCl is a strong electrolyte while NH₄OH and CH₃COOH are weak electrolytes.
2. Only for strong electrolytes is Λ_m vs √C a straight line (Λ_m = Λ°_m − A√C).

Teacher's Note:
1. Weak electrolytes give a curve that rises steeply at low concentration, so their Λ°_m cannot be found by extrapolation.

Answer: (b) Increase in ionic mobility of ions

Explanation:
1. A strong electrolyte is completely ionized at all concentrations. Hence, number of ions remains the same.
2. However, on dilution, interionic forces decrease and hence ionic mobility increases. Therefore, equivalent conductance increases.

Teacher's Note:
1. For weak electrolytes the main cause is the increase in degree of dissociation; for strong electrolytes it is the increase in ionic mobility.

Answer: (d) λ_c = λ_∞ − (B) √C

Explanation:
1. According to Debye-Huckel theory, for a strong electrolyte (like NaCl),
2. λ_c = λ_∞ − (B) √C

Teacher's Note:
1. Conductance at any concentration is less than at infinite dilution, so the sign must be minus, and the variable is √C, not C.

Answer: (b) 11.52 S cm²/mol

Explanation:
1. Λ_m = \( \frac{\kappa \times 1000}{\text{Molarity}} \)
2. = \( \frac{(5.76 \times 10^{-3}\ \text{S cm}^{-1})(1000\ \text{cm}^3\ \text{L}^{-1})}{0.5\ \text{mol L}^{-1}} \)
3. = \( \frac{5.76}{0.5} \) = 11.52 S cm² mol⁻¹

Teacher's Note:
1. The factor 1000 converts L to cm³ because κ is in S cm⁻¹; 1 dm³ = 1 L.

Answer: (d) remains constant for a cell

Explanation:
1. Cell constant = l/a, where l is the distance between the electrodes and a their area of cross-section.
2. It depends only on the geometry of the cell, so it remains constant for a given cell.

Teacher's Note:
1. Cell constant (unit cm⁻¹) belongs to the cell, not to the solution put in it.

Answer: (c) 128 S cm² mol⁻¹

Explanation:
1. Λ°_NaBr = Λ°_NaCl + Λ°_KBr − Λ°_KCl
2. = 126 + 152 − 150 = 128 S cm² mol⁻¹

Teacher's Note:
1. Pick the combination in which K⁺ and Cl⁻ cancel, leaving Na⁺ and Br⁻.

Answer: (b) 0.02

Explanation:
1. Λ°_CH3COOH = Λ°_CH3COONa + Λ°_HCl − Λ°_NaCl
2. = 90 + 425 − 125 = 390 mho cm² mol⁻¹
3. α = \( \frac{\Lambda_m^c}{\Lambda_m^\circ} \) = \( \frac{7.8}{390} \) = 0.02

Teacher's Note:
1. Degree of dissociation of a weak electrolyte = molar conductivity at the given concentration / limiting molar conductivity.

Answer: (b) stops diffusion of ions from one electrode to another

Explanation:
1. The salt bridge does not take part in the cell reaction. It keeps the two electrolytic solutions apart (stops direct diffusion of ions from one electrode solution to the other) while completing the inner circuit and maintaining electrical neutrality.

Teacher's Note:
1. Without a salt bridge the charge builds up in each half-cell and the current stops almost at once.

Answer: (c) F₂ and I⁻

Explanation:
1. Higher the reduction potential, more easily it is reduced and hence stronger is the oxidizing agent. F₂ has highest reduction potential. Hence, F₂ is strongest oxidizing agent.
2. Considering the reverse reactions (oxidation reactions), higher the oxidation potential, more easily it is oxidized and hence stronger is the reducing agent.
3. Oxidation potential of I⁻ (−0.53 V) is highest out of −2.85, −1.36, −1.06 and −0.53. Hence, I⁻ is strongest reducing agent.

Teacher's Note:
1. The reducing agent must be on the right-hand side (I⁻, not I₂) – a common slip in such options.

Answer: (d) NO₃⁻ is reduced in preference to hydronium ion

Explanation:
1. NO₃⁻ ion has a higher reduction potential than H₃O⁺ ion, so it is reduced (to oxides of nitrogen or NH₄⁺) instead of H⁺ being reduced to H₂.

Teacher's Note:
1. HNO₃ is a strong acid; it fails to give H₂ because it is an oxidizing acid, not because it is weak.

Answer: (d) A is false, but R is true.

Explanation:
1. Conductivity decreases with decrease in concentration for both weak and strong electrolytes, because the number of ions per unit volume decreases.
2. So the Assertion is incorrect and the Reason is correct.

Teacher's Note:
1. The weak/strong difference shows up in molar conductivity, not in conductivity.

Answer: (c) A is true, but R is false.

Explanation:
1. At the cathode, the reaction with the higher E° is preferred: reduction of water (E° = −0.83 V) is preferred over reduction of Na⁺ (E° = −2.71 V).
2. So the Assertion is correct and the Reason is incorrect.

Teacher's Note:
1. Cathode: higher E° is reduced first. This is why sodium metal is not obtained from aqueous NaCl.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Zinc (E° = −0.76 V) is more easily oxidised than iron (E° = −0.44 V), so zinc corrodes in preference and protects iron (sacrificial protection).
2. Thus the Reason correctly explains the Assertion.

Teacher's Note:
1. Zinc protects iron even after the coating is scratched; tin, being less reactive than iron, does not.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Conductivity is the conductance of unit volume of solution; on dilution, the number of ions per unit volume falls, so conductivity decreases.

Teacher's Note:
1. Do not confuse with molar conductivity, which increases on dilution.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. Chlorine has higher reduction potential than water but lower discharge potential.
2. Cl₂ is liberated at the anode because of the overpotential of oxygen, not because Cl⁻ has a higher oxidation potential. So A is true but R is false.

Teacher's Note:
1. E°_ox of Cl⁻ (−1.36 V) is actually lower than that of water (−1.23 V); overpotential is the real reason Cl₂ forms.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A.
2. With a copper anode, Cu → Cu²⁺ + 2e⁻ occurs in preference to oxidation of water or SO₄²⁻, so no O₂ is liberated.

Teacher's Note:
1. The nature of the electrode can change the anode product: inert Pt gives O₂, active Cu dissolves.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Molar conductance increases because with dilution in a weak electrolyte, dissociation increases while in a strong electrolyte, interionic attractions decrease.
2. Both statements are true, but R alone does not explain A for all electrolytes.

Teacher's Note:
1. For weak electrolytes the main cause is the increase in the number of ions (degree of dissociation), not ionic speed.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Cell constant is best found by measuring the conductance (resistance) of a solution whose conductivity is known (because cell constant = conductivity/conductance).
2. So A is false while R is true.

Teacher's Note:
1. In practice a standard KCl solution of known κ is used: cell constant = κ × R.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Λ°_m (BaCl₂) = λ°_Ba²⁺ + 2λ°_Cl⁻
2. So A is false; R (for equivalent conductivity) is true.

Teacher's Note:
1. For molar conductivity, multiply each ionic conductivity by the number of ions in the formula unit.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A.
2. As [Zn²⁺] increases and [Cu²⁺] decreases, the two electrode potentials approach each other; when they become equal, E_cell = 0.

Teacher's Note:
1. A dead cell is a cell whose reaction has reached equilibrium (E_cell = 0).