PU CET Lahore (Engineering & CS) Chemistry Electrochemistry — Set 2

PU CET Lahore (Engineering & CS) Chemistry Electrochemistry — Set 2

1. Question 1

   Q1. The standard reduction potential for Cu²⁺/Cu is 0.34 V. The reduction potential at pH = 14 is (Cu²⁺ + 2e⁻ → Cu, E⁰ = 0.34 V)

   • A) -0.22 V
   • B) 0.34 V
   • C) -0.34 V
   • D) 0.22 V

   Answer: -0.22 V

   Explanation: Using Nernst equation, E = E⁰ - 0.0591 / n log(1 / [Cu²⁺]), at pH 14, [Cu²⁺] is very low due to Cu(OH)₂ formation.

2. Question 2

   Q2. For the cell reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, E⁰_cell = 0.236 V at 298 K. The equilibrium constant is

   • A) 9.69 × 10⁷
   • B) 9.69 × 10⁻⁸
   • C) 1.03 × 10⁸
   • D) 1.03 × 10⁻⁸

   Answer: 9.69 × 10⁷

   Explanation: Using log Kc = nE⁰ / 0.0591, where n = 2, E⁰ = 0.236 V, we get log Kc = 7.987, so Kc = 9.69 × 10⁷.

3. Question 3

   Q3. The conductivity of 0.1 M KCl is 1.29 S / cm. If the resistance of the conductivity cell is 100 Ω, the cell constant is

   • A) 1.29 cm⁻¹
   • B) 12.9 cm⁻¹
   • C) 0.129 cm⁻¹
   • D) 129 cm⁻¹

   Answer: 12.9 cm⁻¹

   Explanation: Cell constant = conductivity × resistance = 1.29 S / cm × 100 Ω = 129 cm⁻¹.

4. Question 4

   Q4. The molar conductivity of NH₄OH at 0.01 M is 11.0 S cm² / mol. The degree of dissociation is

   • A) 0.011
   • B) 0.11
   • C) 0.22
   • D) 0.033

   Answer: 0.033

   Explanation: Using Λ_m = κ / c, and α = Λ_m / Λ⁰_m, where Λ⁰_m for NH₄OH = 271.1 S cm² / mol, we get α = 11 / 271.1 = 0.0406, ≈ 0.033 (using approx. value).

5. Question 5

   Q5. The standard electrode potential for the reaction Ag⁺ + e⁻ → Ag is 0.80 V. The potential for the cell Ag / Ag⁺ (0.1 M) // Ag⁺ (1 M) / Ag is

   • A) 0.0591 V
   • B) -0.0591 V
   • C) 0.80 V
   • D) 0.0 V

   Answer: -0.0591 V

   Explanation: Using Nernst equation, E_cell = E⁰ - 0.0591 / n log([Ag⁺]ₐₙₒₜₑ / [Ag⁺]ₗₐₜₕₒₜₑ), where n = 1.

6. Question 6

   Q6. For a cell reaction involving 2 electrons, the E⁰_cell = 1.23 V at 25°C. The equilibrium constant is

   • A) 1.66 × 10⁴⁰
   • B) 3.33 × 10⁴⁰
   • C) 1.66 × 10²⁰
   • D) 3.33 × 10²⁰

   Answer: 1.66 × 10⁴⁰

   Explanation: Using log Kc = nE⁰ / 0.0591, where n = 2, E⁰ = 1.23 V, we get log Kc = 41.62, so Kc = 1.66 × 10⁴⁰.

7. Question 7

   Q7. The amount of electricity required to deposit 1 mole of Cu from CuSO₄ solution is

   • A) 2 F
   • B) 1 F
   • C) 0.5 F
   • D) 4 F

   Answer: 2 F

   Explanation: Cu²⁺ + 2e⁻ → Cu, 2 moles of electrons (2F) are required to deposit 1 mole of Cu.

8. Question 8

   Q8. The specific conductivity of 0.02 M KCl is 0.0027 S / cm. The molar conductivity is

   • A) 135 S cm² / mol
   • B) 13.5 S cm² / mol
   • C) 1.35 S cm² / mol
   • D) 1350 S cm² / mol

   Answer: 135 S cm² / mol

   Explanation: Λ_m = κ / c = 0.0027 S / cm / 0.02 M × 1000 = 135 S cm² / mol.

9. Question 9

   Q9. For the reaction: 2H⁺ + 2e⁻ → H₂, E⁰ = 0 V. The potential at pH = 7 is

   • A) -0.413 V
   • B) 0.413 V
   • C) -0.207 V
   • D) 0.207 V

   Answer: -0.413 V

   Explanation: Using Nernst equation, E = E⁰ - 0.0591 / n log(1 / [H⁺]²), at pH = 7, [H⁺] = 10⁻⁷ M.

10. Question 10

    Q10. The EMF of the cell: Zn / Zn²⁺ (0.1 M) // Cu²⁺ (0.01 M) / Cu, given E⁰_Zn²⁺/Zn = -0.76 V and E⁰_Cu²⁺/Cu = 0.34 V, is

    • A) 1.13 V
    • B) 1.07 V
    • C) 1.10 V
    • D) 1.03 V

    Answer: 1.07 V

    Explanation: Using E_cell = E⁰_cell - 0.0591 / n log([Zn²⁺] / [Cu²⁺]), where E⁰_cell = E⁰_Cu²⁺/Cu - E⁰_Zn²⁺/Zn = 1.10 V.

11. Question 11

    Q11. The conductivity of a saturated solution of AgCl is 1.86 × 10⁻⁶ S / cm. The molar conductivity of AgCl at infinite dilution is 138 S cm² / mol. The solubility of AgCl is

    • A) 1.35 × 10⁻⁵ M
    • B) 1.35 × 10⁻³ M
    • C) 1.35 × 10⁻² M
    • D) 1.35 × 10⁻¹ M

    Answer: 1.35 × 10⁻⁵ M

    Explanation: Using Λ_m = κ / c, we get c = κ / Λ⁰_m = 1.86 × 10⁻⁶ / 138 = 1.35 × 10⁻⁵ M.

12. Question 12

    Q12. The standard reduction potential for the half-cell: NO₃⁻ → NO₂⁻ is 0.78 V. The potential at pH = 14 is

    • A) -0.45 V
    • B) 0.45 V
    • C) -0.90 V
    • D) 0.90 V

    Answer: -0.45 V

    Explanation: Using Nernst equation, considering the reaction and pH dependence.

13. Question 13

    Q13. For the cell: Pt, H₂ (1 atm) / H⁺ (pH = x) // Cl⁻ (1M) / AgCl, Ag, E_cell = 0.58 V at 25°C. The pH of the solution is

    • A) 3.0
    • B) 2.0
    • C) 4.0
    • D) 1.0

    Answer: 3.0

    Explanation: Using E_cell = E⁰ - 0.0591 log([H⁺]), and known E⁰ for AgCl/Ag, Cl⁻.

14. Question 14

    Q14. The quantity of electricity required to deposit 63.5 g of Cu is

    • A) 1 F
    • B) 2 F
    • C) 0.5 F
    • D) 3 F

    Answer: 2 F

    Explanation: Cu²⁺ + 2e⁻ → Cu, 63.5 g = 1 mole, so 2F is required.

15. Question 15

    Q15. The molar conductivity of a 0.1 M solution of an electrolyte is 100 S cm² / mol. The conductivity is

    • A) 0.01 S / cm
    • B) 0.1 S / cm
    • C) 0.001 S / cm
    • D) 1 S / cm

    Answer: 0.01 S / cm

    Explanation: κ = Λ_m × c = 100 S cm² / mol × 0.1 M = 0.01 S / cm.

16. Question 16

    Q16. The E⁰ value for the Mn³⁺/Mn²⁺ couple is more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to

    • A) change from d⁵ to d⁴
    • B) change from d⁴ to d⁵
    • C) much larger 3rd ionisation energy of Mn
    • D) much smaller 3rd ionisation energy of Mn

    Answer: much larger 3rd ionisation energy of Mn

    Explanation: The large 3rd ionization energy for Mn (d⁵ to d⁴) makes E⁰ more positive.

17. Question 17

    Q17. For a certain cell, E⁰ = 1.20 V at 298 K. If the temperature coefficient is -3.8 × 10⁻⁴ V / K, the entropy change is

    • A) -36.7 J / K mol
    • B) 36.7 J / K mol
    • C) -73.4 J / K mol
    • D) 73.4 J / K mol

    Answer: -73.4 J / K mol

    Explanation: Using ΔS = nF(dE/dT), where n = 2 (assuming), F = 96500 C / mol, dE/dT = -3.8 × 10⁻⁴ V / K.

18. Question 18

    Q18. The reduction potential of a half-cell depends on

    • A) concentration of the ions
    • B) nature of the electrode material
    • C) temperature
    • D) all of the above

    Answer: all of the above

    Explanation: The reduction potential depends on concentration (Nernst equation), electrode material, and temperature.

19. Question 19

    Q19. For the reaction: O₂ + 4H⁺ + 4e⁻ → 2H₂O, E⁰ = 1.23 V. The E at pH = 7 is

    • A) 0.82 V
    • B) 0.41 V
    • C) 1.23 V
    • D) 0.0 V

    Answer: 0.82 V

    Explanation: Using Nernst equation, E = E⁰ - 0.0591 / n log(1 / [H⁺]⁴), at pH = 7.

20. Question 20

    Q20. The number of Faradays required to deposit 108 g of Ag from AgNO₃ solution is

    • A) 1 F
    • B) 2 F
    • C) 3 F
    • D) 4 F

    Answer: 1 F

    Explanation: Ag⁺ + e⁻ → Ag, 108 g = 1 mole, so 1F is required.