CBSE Class 12 Chemistry Electrochemistry MCQs Set 13

Answer: (d) −1.66 V

Explanation:
1. Al is more reactive than Ag, i.e., cell reaction is Al + 3Ag⁺ → Al³⁺ + 3Ag.
2. E°_cell = E°_cathode − E°_anode = E°_Ag⁺/Ag − E°_Al³⁺/Al
3. 2.46 = 0.80 − E°_Al³⁺/Al or E°_Al³⁺/Al = −1.66 V

Teacher's Note:
1. A metal that displaces H₂ from dilute acid must have a negative reduction potential – use this to reject +1.66 V at once.

Answer: (a) increase in Cu²⁺

Explanation:
1. Ti + Cu²⁺ → Ti²⁺ + Cu
2. E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Ti^{2+}]}{[Cu^{2+}]} \)
3. Increasing [Cu²⁺] decreases the log term and hence increases E_cell.

Teacher's Note:
1. Increasing the concentration of a reactant (here Cu²⁺) or decreasing that of a product raises E_cell.

Answer: (a) 1.0 M HCl

Explanation:
1. For 2H⁺ + 2e⁻ → H₂,
2. E = E° − \( \frac{0.0591}{2} \) log \( \frac{1}{[H^+]^2} \) = 0 + \( \frac{0.0591}{2} \) log [H⁺]² = 0.0591 log [H⁺] = −0.0591 pH
3. (a) [H⁺] = 1 M, E = 0.0591 log 1 = 0
4. (b) [OH⁻] = 10⁰, [H⁺] = 10⁻¹⁴ M, pH = 14, E = −0.0591 × 14 = −0.827 V
5. (c) [H⁺] = 10⁻⁷, pH = 7, E = −0.0591 × 7 = −0.413 V
6. (d) pH = 4, E = −0.0591 × 4 = −0.236 V
7. Thus, potential E is maximum for 1 M HCl.

Teacher's Note:
1. For the hydrogen electrode (1 atm H₂), E = −0.0591 pH: the lower the pH, the higher the reduction potential.

Answer: (b) p(H₂) = 2 atm and [H⁺] = 1.0 M

Explanation:
1. For hydrogen electrode, H⁺ (aq) + e⁻ → \( \frac{1}{2} \)H₂ (g)
2. E = E° − \( \frac{0.0591}{1} \) log \( \frac{(p_{H_2})^{1/2}}{[H^+]} \) = −0.0591 log \( \frac{(p_{H_2})^{1/2}}{[H^+]} \) (∵ E° = 0)
3. E will be −ve when \( \frac{(p_{H_2})^{1/2}}{[H^+]} \) is +ve, i.e., (p_H2)^1/2 > [H⁺], which is so in case of (b).

Teacher's Note:
1. More precisely, E is negative when the ratio inside the log is greater than 1; in (b) it is √2/1 > 1, in (c) √2/2 < 1.

Answer: (c) increase the E and shift equilibrium to the right

Explanation:
1. E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Zn^{2+}] \times p_{H_2}}{[H^+]^2} \)
2. On adding H₂SO₄, [H⁺] will increase, therefore E_cell will increase. The equilibrium Zn + 2H⁺ ⇄ Zn²⁺ + H₂ will shift towards right on increasing the concentration of H⁺ ions.

Teacher's Note:
1. Nernst equation and Le Chatelier’s principle always agree: adding a reactant raises E_cell and pushes the reaction forward.

Answer: (a) H₂, O₂

Explanation:
1. At cathode, water is reduced in preference to Na⁺: 2H₂O + 2e⁻ → H₂ + 2OH⁻.
2. At anode, water is oxidised in preference to SO₄²⁻: 2H₂O → O₂ + 4H⁺ + 4e⁻.

Teacher's Note:
1. Electrolysis of aqueous Na₂SO₄ is really electrolysis of water; the salt only makes the solution conducting.

Answer: (a) Cu will deposit at cathode

Explanation:
1. Cu²⁺ has a higher reduction potential (+0.34 V) than water, so Cu²⁺ + 2e⁻ → Cu occurs at the cathode.
2. At the platinum anode, water is oxidised to O₂.

Teacher's Note:
1. Cathode means reduction – a metal can never deposit at the anode, and O₂ can never be released at the cathode.

Answer: (c) Dilute NaCl

Explanation:
1. In a very dilute NaCl solution, the concentration of Cl⁻ is very low, so OH⁻ ions (from water) are discharged and O₂ is evolved at the anode.
2. Molten NaCl has no OH⁻ ions and solid NaCl does not conduct at all.

Teacher's Note:
1. Concentration matters in electrolysis: concentrated brine gives Cl₂, very dilute NaCl gives mainly O₂ at the anode.

Answer: (d) 1 M KCl solution, 1 M NaCl solution

Explanation:
1. (a) CuSO₄ → Cu + O₂, CuCl₂ → Cu + Cl₂
2. (b) KCl → H₂ + Cl₂, KI → H₂ + I₂
3. (c) AgNO₃ → Ag + O₂, Cu(NO₃)₂ → Cu + O₂
4. (d) KCl → H₂ + Cl₂, NaCl → H₂ + Cl₂
5. Thus, KCl and NaCl solutions give the same products at anode and cathode.

Teacher's Note:
1. K⁺ and Na⁺ are both far harder to reduce than water, so both solutions behave alike.

Answer: (b) Mg

Explanation:
1. Mg has a very low reduction potential (E° = −2.37 V). In aqueous solution water is reduced to H₂ in preference to Mg²⁺, so Mg is obtained only by electrolysis of its fused salt.

Teacher's Note:
1. Highly electropositive metals (Na, K, Mg, Ca, Al) are always extracted by electrolysis of molten compounds, never aqueous solutions.

Answer: (a) 38%

Explanation:
1. The electrolyte in a lead storage battery is about 38% (by mass) H₂SO₄ solution (density about 1.30 g mL⁻¹).

Teacher's Note:
1. The H₂SO₄ concentration falls during discharge, so measuring its density tells how far the battery is charged.

Answer: (c) Zn

Explanation:
1. Galvanisation is applying a coating of zinc metal to prevent corrosion.

Teacher's Note:
1. Zinc protects iron even after a scratch because zinc, being more reactive, is oxidised in preference (sacrificial protection).

Answer: (c) H₂ is anode and Cu is cathode

Explanation:
1. In Cu–H₂ cell, hydrogen loses electrons (oxidation occurs). Hence, H₂ acts as anode whereas Cu is cathode.

Teacher's Note:
1. E°(Cu²⁺/Cu) = +0.34 V is higher than 0.00 V for SHE, so copper is the cathode.

Answer: (d) Cu + Hg²⁺ → Cu²⁺ + Hg

Explanation:
1. Cu → Cu²⁺ + 2e⁻ (Oxidation)
2. Hg²⁺ + 2e⁻ → Hg (Reduction)
3. Cu + Hg²⁺ → Cu²⁺ + Hg

Teacher's Note:
1. In cell notation the left electrode is the anode (oxidation) and the right electrode is the cathode (reduction).

Answer: (c) KCl

Explanation:
1. This is because Cl⁻ ions react with Ag⁺ ions to form a precipitate of AgCl.

Teacher's Note:
1. The salt bridge electrolyte must not react with the ions of either half-cell.

Answer: (b) electrode potentials of both the electrodes become equal

Explanation:
1. As the cell works, the concentrations change: the anode potential rises and the cathode potential falls. When the two electrode potentials become equal, E_cell = 0 and the cell stops working (the cell reaction has reached equilibrium).

Teacher's Note:
1. A “dead” cell has E_cell = 0 because the reaction is at equilibrium, not because the electrode potentials are zero.

Answer: (c) 0.771 − 0.536 = 0.235 volt

Explanation:
1. E°_cell = E°_Ox (I₂) + E°_Red (Fe)
2. = −0.536 + 0.771 = 0.235 V

Teacher's Note:
1. Electrode potential is an intensive property – it is not multiplied when the half-reaction is multiplied by 2.

Answer: (b) +0.76 V

Explanation:
1. E°_Cell = E°_{Red (Ag⁺/Ag)} + E°_{Ox (Zn/Zn²⁺)}
2. 1.56 = 0.80 + E°_{Ox (Zn/Zn²⁺)}
3. ∴ E°_{Ox (Zn/Zn²⁺)} = 1.56 − 0.80 = +0.76 V

Teacher's Note:
1. Read the question carefully: oxidation potential of Zn is +0.76 V, its reduction potential is −0.76 V.

Answer: (d) CuSO₄ solution in zinc vessel

Explanation:
1. Zn is more reactive than Cu. Hence, Zn can displace Cu from CuSO₄ solution, i.e., reaction occurs. Hence, CuSO₄ solution can not be stored in Zn vessel.
2. This is not so in the first three cases.

Teacher's Note:
1. A solution cannot be stored in a vessel made of a metal that lies below it in the series of reduction potentials (more reactive metal).

Answer: (b) 0.89 V

Explanation:
1. The reactions taking place are:
2. At cathode: Cu²⁺ (aq) + 2e⁻ → Cu (s), E° = 0.34 V
3. At anode: H₂O (l) → 2H⁺ (aq) + \( \frac{1}{2} \)O₂ (g) + 2e⁻, E° = −1.23 V
4. Net reaction: Cu²⁺ (aq) + H₂O (l) → Cu (s) + \( \frac{1}{2} \)O₂ (g) + 2H⁺ (aq), E°_Cell = −0.89 V
5. Hence, minimum voltage required to bring about electrolysis is 0.89 V.

Teacher's Note:
1. A negative E°_cell means the reaction is non-spontaneous; at least that much external voltage (ignoring overpotential) must be applied.

Answer: (b) −0.059 V

Explanation:
1. H⁺ + e⁻ → \( \frac{1}{2} \)H₂
2. E = E° − \( \frac{0.059}{n} \) log \( \frac{p_{H_2}^{1/2}}{[H^+]} \)
3. = 0 − \( \frac{0.059}{1} \) log \( \frac{(100)^{1/2}}{1} \) = −0.059 V

Teacher's Note:
1. Raising the pressure of H₂ (a product of the reduction) lowers the reduction potential; take the square root because only ½ H₂ appears.

Answer: (b) −0.591 V

Explanation:
1. For hydrogen electrode, H⁺ + e⁻ → \( \frac{1}{2} \)H₂. Applying Nernst equation,
2. E = E° − \( \frac{0.0591}{n} \) log \( \frac{1}{[H^+]} \) = 0 − \( \frac{0.0591}{1} \) log \( \frac{1}{10^{-10}} \)
3. (pH = 10 means [H⁺] = 10⁻¹⁰ M)
4. = −0.0591 × 10 = −0.591 V

Teacher's Note:
1. Shortcut for a hydrogen electrode at 1 atm: E = −0.0591 × pH.

Answer: (b) decreasing the temperature

Explanation:
1. Mⁿ⁺ + ne⁻ → M
2. E_Mⁿ⁺/M = E°_Mⁿ⁺/M − \( \frac{2.303\,RT}{nF} \) log \( \frac{1}{[M^{n+}]} \)
3. If T is decreased, magnitude of second factor on RHS decreases so that E_Mⁿ⁺/M increases.

Teacher's Note:
1. Electrode potential does not depend on electrode size; it depends on concentration and temperature only.

Answer: (a) 2.5 V

Explanation:
1. E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Zn^{2+}]}{[Ag^+]^2} \)
2. = 2.57 − \( \frac{0.0591}{2} \) log \( \frac{0.28}{(0.2)^2} \) = 2.57 − \( \frac{0.0591}{2} \) log 7
3. = 2.57 − \( \frac{0.0591}{2} \) × 0.8451 = 2.57 − 0.025 ≈ 2.5 V

Teacher's Note:
1. Square the Ag⁺ concentration because its coefficient is 2, and take n = 2 for the overall reaction.

Answer: (b) 3.92 × 10¹⁵

Explanation:
1. E°_cell = \( \frac{0.0591}{n} \) log K. For the given reaction, n = 2, E°_cell = 0.46 V
2. ∴ 0.46 = \( \frac{0.0591}{2} \) log K_c or log K_c = \( \frac{0.46 \times 2}{0.0591} \) = 15.5668
3. K_c = Antilog 15.5668 = 3.7 × 10¹⁵ ≈ 3.92 × 10¹⁵ (with 0.059 in place of 0.0591)

Teacher's Note:
1. Only the power of 10 really decides the answer here: log K ≈ 15.6 means K is of the order of 10¹⁵.

Answer: (b) −212.3 kJ mol⁻¹

Explanation:
1. ΔG° = −nFE°_Cell
2. = −2 × 96500 C mol⁻¹ × 1.10 V = −212300 CV mol⁻¹
3. = −212300 J mol⁻¹ (1 CV = 1 J) = −212.3 kJ mol⁻¹

Teacher's Note:
1. Remember n = 2 for the Daniell cell and convert J to kJ at the end.

Answer: (a) 4.6

Explanation:
1. Specific conductivity = Obs. conductivity × Cell const
2. = \( \frac{1}{2.5 \times 10^3} \) × 1.15 = 4.6 × 10⁻⁴ ohm⁻¹ cm⁻¹
3. Λ_eq = \( \frac{\kappa \times 1000}{\text{Normality}} \) = \( \frac{4.6 \times 10^{-4} \times 1000}{0.1} \)
4. = 4.6 Ω⁻¹ cm² eq⁻¹

Teacher's Note:
1. Conductivity = cell constant / resistance; for equivalent conductance divide by normality, for molar conductance by molarity.

Answer: (c) Ag at cathode, dissolution of Ag from anode

Explanation:
1. Probable reactions at cathode (Reduction)
2. Ag⁺ + e⁻ → Ag, E°_red = 0.80 V; 2H₂O + 2e⁻ → H₂ + 2OH⁻, E°_red = −0.83 V
3. Ag⁺ ion has higher reduction potential than H₂O. Hence, Ag⁺ ions are reduced more easily and deposited as Ag.
4. Probable reactions at anode (Oxidation)
5. Ag (from anode) → Ag⁺ (aq) + e⁻, E°_ox = −0.80 V; H₂O (l) → \( \frac{1}{2} \)O₂ (g) + 2H⁺ (aq), E°_ox = −1.23 V
6. First reaction has higher oxidation potential. Hence, Ag from anode dissolves to form Ag⁺ ions in the solution.

Teacher's Note:
1. With an active (non-inert) anode, the anode metal itself is often oxidised; with inert Pt electrodes the same solution gives O₂.

Answer: (c) 0.66 g

Explanation:
1. Cl⁻ → \( \frac{1}{2} \)Cl₂ + e⁻
2. Quantity of electricity passed = 1 × 30 × 60 C = 1800 C
3. 96500 C liberate Cl₂ = \( \frac{1}{2} \) mole
4. 1800 C will liberate Cl₂ = \( \frac{1}{2} \) × \( \frac{1}{96500} \) × 1800 = 0.0093 mol = 0.0093 × 71 g = 0.66 g

Teacher's Note:
1. Watch the units in the options: 0.66 appears both as mol and as g – only the mass is correct.

Answer: (c) 2H₂ + 4OH⁻ → 4H₂O + 4e⁻

Explanation:
1. At anode, H₂ is oxidised in the alkaline (KOH) electrolyte: 2H₂ (g) + 4OH⁻ (aq) → 4H₂O (l) + 4e⁻
2. At cathode: O₂ (g) + 2H₂O (l) + 4e⁻ → 4OH⁻ (aq); option (a) is the overall reaction.

Teacher's Note:
1. An electrode reaction must show electrons; the overall cell reaction never does.

Answer: (a) ΔG = −ve, E° = +ve

Explanation:
1. ΔG = −nFE_cell. For a spontaneous cell reaction ΔG is negative, which requires E_cell to be positive.

Teacher's Note:
1. E = 0 means equilibrium (ΔG = 0); a positive ΔG with negative E means a non-spontaneous reaction.

Answer: (b) 115 S cm² mol⁻¹

Explanation:
1. From the graph (intercepts on the y-axis), Λ°_m (HCl) = 425 S cm² mol⁻¹, Λ°_m (KCl) = 150 S cm² mol⁻¹
2. Λ°_m (CH₃COOK) = Λ°_m (CH₃COOH) + Λ° (KCl) − Λ° (HCl)
3. = 390 + 150 − 425 = 115 S cm² mol⁻¹

Teacher's Note:
1. Read limiting molar conductivity at the point where the line meets the y-axis (√c = 0), not at any plotted point.

Answer: (d) All the above are true

Explanation:
1. During discharge: Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻; Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
2. Both electrodes dip in the same 38% H₂SO₄ solution, so all three statements are true.

Teacher's Note:
1. PbSO₄ forms at both electrodes during discharge; that is why the battery can be recharged.

Answer: (b) Zn – Cu Cell : Daniell cell :: Zn – Zn cell : Concentration cell

Explanation:
1. (a) is incorrect because units of molar conductivity are S m² mol⁻¹.
2. (c) is incorrect because conductivity decreases with dilution.
3. (d) is incorrect because EMF of a cell increases with increase in concentration of reactants.

Teacher's Note:
1. A concentration cell uses the same electrodes in solutions of different concentration, so its E°_cell is zero.

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

Explanation:
1. Correct A. Electrolytic cells convert electrical energy into chemical energy (Remember: Galvanic cells and electrolytic cells are collectively called electrochemical cells).
2. So A is false and R is true.

Teacher's Note:
1. Only galvanic (voltaic) cells convert chemical energy into electrical energy.

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

Explanation:
1. Correct A. Copper sulphate solution cannot be kept in zinc vessel.
2. Zinc, being more reactive, displaces copper: Zn + CuSO₄ → ZnSO₄ + Cu. So A is false and R is true.

Teacher's Note:
1. In the electrochemical series (arranged by increasing reduction potential), a metal higher up displaces metals below it from their salt solutions.

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

Explanation:
1. Correct A. Copper does not liberate hydrogen from dilute hydrochloric acid solution.
2. E°(Cu²⁺/Cu) = +0.34 V is higher than that of hydrogen, so A is false and R is true.

Teacher's Note:
1. Only metals with negative reduction potentials (Zn, Fe, Mg...) liberate H₂ from dilute acids.

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

Explanation:
1. Correct R. Reduction potential of F₂ is greater than that of Cl₂.
2. So A is true but R is false.

Teacher's Note:
1. A strong oxidizing agent is itself easily reduced, so compare reduction potentials.

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. OH⁻ ions used up at the anode (2H₂ + 4OH⁻ → 4H₂O + 4e⁻) are regenerated at the cathode (O₂ + 2H₂O + 4e⁻ → 4OH⁻); the net reaction 2H₂ + O₂ → 2H₂O does not involve OH⁻.

Teacher's Note:
1. The voltage stays constant because no ion in the cell expression changes concentration; the reactants are fed continuously.

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

Explanation:
1. Both A and R are correct and R is the correct explanation of A.
2. At the anode Cu → Cu²⁺ + 2e⁻, so copper dissolves instead of O₂ being evolved.

Teacher's Note:
1. This is the basis of electrorefining of copper: the anode dissolves while pure copper deposits on the cathode.

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

Explanation:
1. Correct explanation. Electrochemical equivalent of a substance is the amount deposited by one coulomb of charge.
2. Since 96500 C (1 F) deposit one gram equivalent, E = 96500 × Z. Both statements are true, but R does not explain A.

Teacher's Note:
1. Remember Z = E/F: electrochemical equivalent is the mass deposited by 1 coulomb, equivalent weight is the mass deposited by 1 faraday.

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

Explanation:
1. Correct explanation. Specific conductivity = Conductance × Cell constant. As conductance of an electrolytic solution increases with temperature therefore, specific conductivity increases with temperature.
2. R is only the definition and does not explain A.

Teacher's Note:
1. Electrolytic conduction increases with temperature (ions move faster), whereas metallic conduction decreases.

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

Explanation:
1. Correct A. Specific conductivity of a solution decreases with dilution (because number of ions/cc in the solution decreases).
2. So A is false and R is true.

Teacher's Note:
1. Total number of ions may increase on dilution, but the number per cm³ decreases – and κ depends on ions per cm³.

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

Explanation:
1. Correct A. During discharge of lead storage battery, Pb is oxidized to PbSO₄ while PbO₂ is reduced to PbSO₄.
2. During recharge, PbSO₄ is reduced to Pb at one electrode and oxidized to PbO₂ at the other, so R is true.

Teacher's Note:
1. Oxidation numbers settle it: Pb (0) → Pb (+2) is oxidation, PbO₂ (+4) → PbSO₄ (+2) is reduction.