LUMS LCAT Chemistry Electrochemistry — Set 3

LUMS LCAT Chemistry Electrochemistry — Set 3

1. Question 1

   Q1. The standard reduction potential of a cell is 1.1 V. Calculate the standard Gibbs free energy change (ΔG°) at 25°C.

   • A) -212.3 kJ/mol
   • B) -106.15 kJ/mol
   • C) 212.3 kJ/mol
   • D) 106.15 kJ/mol

   Answer: -106.15 kJ/mol

   Explanation: ΔG° = -nFE°cell, where n = number of electrons, F = 96500 C/mol. At 25°C, ΔG° = -2 * 96500 * 1.1 = -212300 J/mol or -212.3 kJ/mol for n=2. For n=1, ΔG° = -106.15 kJ/mol.

2. Question 2

   Q2. For a cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, E°cell = 1.1 V. What is the equilibrium constant (Kc) at 25°C?

   • A) 1.6 × 10³⁷
   • B) 1.6 × 10¹⁷
   • C) 1.6 × 10⁵⁷
   • D) 1.6 × 10⁷⁷

   Answer: 1.6 × 10³⁷

   Explanation: E°cell = (0.059/n) * log Kc, rearranging gives log Kc = (n * E°cell) / 0.059. For n = 2, log Kc = 37.288, so Kc = 1.94 × 10³⁷ ≈ 1.6 × 10³⁷ (approx.).

3. Question 3

   Q3. The conductivity of 0.01 M KCl solution is 0.0014 S/cm. Calculate the cell constant if the resistance is 1500 Ω.

   • A) 2.1 cm⁻¹
   • B) 1.4 cm⁻¹
   • C) 1.05 cm⁻¹
   • D) 2.5 cm⁻¹

   Answer: 2.1 cm⁻¹

   Explanation: Cell constant = conductivity * resistance = 0.0014 S/cm * 1500 Ω = 2.1 cm⁻¹.

4. Question 4

   Q4. For a reaction: 2H⁺ + 2e⁻ → H₂, the reduction potential at pH = 7 is -0.414 V. What is the standard reduction potential?

   • A) 0 V
   • B) -0.414 V
   • C) 0.414 V
   • D) -0.828 V

   Answer: 0 V

   Explanation: E = E° - (0.059/n) * log (1 / [H⁺]²), at pH = 7, [H⁺] = 10⁻⁷ M. E = E° - (0.059/2) * log (1 / (10⁻⁷)²), -0.414 = E° - (-0.414), so E° = 0 V.

5. Question 5

   Q5. The molar conductivity of 0.1 M CH₃COOH is 5.2 S cm²/mol. Calculate the degree of dissociation (α).

   • A) 0.013
   • B) 0.026
   • C) 0.0135
   • D) 0.02

   Answer: 0.013

   Explanation: α = Λm / Λ°m, where Λ°m for CH₃COOH = 390.7 S cm²/mol (approx.), so α = 5.2 / 390.7 = 0.0133 ≈ 0.013.

6. Question 6

   Q6. A conductivity cell contains 0.01 M KCl solution with conductivity 0.0014 S/cm. If the resistance is 100 Ω, what is the cell constant?

   • A) 0.14 cm⁻¹
   • B) 1.4 cm⁻¹
   • C) 0.7 cm⁻¹
   • D) 14 cm⁻¹

   Answer: 0.14 cm⁻¹

   Explanation: Cell constant = conductivity * resistance = 0.0014 S/cm * 100 Ω = 0.14 cm⁻¹.

7. Question 7

   Q7. For the reaction: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V. The reduction potential at [Cu²⁺] = 0.1 M is

   • A) +0.31 V
   • B) +0.37 V
   • C) +0.34 V
   • D) +0.28 V

   Answer: +0.31 V

   Explanation: E = E° - (0.059/n) * log (1 / [Cu²⁺]), E = 0.34 - (0.059/2) * log (1 / 0.1) = 0.34 - 0.0295 = 0.31 V.

8. Question 8

   Q8. The standard electrode potential for the reaction: Ag⁺ + e⁻ → Ag is +0.80 V. The potential at [Ag⁺] = 0.01 M is

   • A) +0.68 V
   • B) +0.80 V
   • C) +0.62 V
   • D) +0.74 V

   Answer: +0.68 V

   Explanation: E = E° - (0.059/n) * log (1 / [Ag⁺]), E = 0.80 - 0.059 * log (1 / 0.01) = 0.80 - 0.118 = 0.682 V.

9. Question 9

   Q9. The molar conductivity of a weak electrolyte increases with dilution because of

   • A) Increase in degree of dissociation
   • B) Decrease in degree of dissociation
   • C) Increase in ionic mobility
   • D) Decrease in ionic mobility

   Answer: Increase in degree of dissociation

   Explanation: As the solution is diluted, the degree of dissociation (α) increases, so Λm increases.

10. Question 10

    Q10. For a cell: Pt | H₂ (1 atm) | H⁺ (1 M) || Cu²⁺ (1 M) | Cu, E°cell = +0.34 V. The E° for Cu²⁺/Cu is

    • A) +0.34 V
    • B) -0.34 V
    • C) +0.68 V
    • D) -0.68 V

    Answer: +0.34 V

    Explanation: E°cell = E°(cathode) - E°(anode), E°(anode) = 0 V for SHE, so E°cell = E°(Cu²⁺/Cu) = +0.34 V.

11. Question 11

    Q11. The conductivity of a 0.1 M NaCl solution is 1.06 × 10⁻² S/cm. Calculate the molar conductivity.

    • A) 106 S cm²/mol
    • B) 10.6 S cm²/mol
    • C) 1.06 S cm²/mol
    • D) 0.106 S cm²/mol

    Answer: 106 S cm²/mol

    Explanation: Λm = κ / c = (1.06 × 10⁻² S/cm) / (0.1 mol / 1000 cm³) = 106 S cm²/mol.

12. Question 12

    Q12. For the reaction: 2Ag⁺ + Cu → 2Ag + Cu²⁺, E°cell = +0.46 V. The E° for Ag⁺/Ag is

    • A) +0.80 V
    • B) +0.34 V
    • C) +0.46 V
    • D) -0.80 V

    Answer: +0.80 V

    Explanation: E°cell = E°(Ag⁺/Ag) - E°(Cu²⁺/Cu), 0.46 = E°(Ag⁺/Ag) - 0.34, so E°(Ag⁺/Ag) = 0.80 V.

13. Question 13

    Q13. The EMF of a cell is related to the equilibrium constant by the equation

    • A) E° = (RT / nF) * ln Kc
    • B) E° = (nF / RT) * ln Kc
    • C) E° = (RT / nF) * log Kc
    • D) E° = (0.059 / n) * log Kc

    Answer: E° = (RT / nF) * ln Kc

    Explanation: The Nernst equation relates E° to Kc: E° = (RT / nF) * ln Kc.

14. Question 14

    Q14. The conductivity of a solution depends on

    • A) Concentration of electrolyte
    • B) Nature of electrolyte
    • C) Temperature
    • D) All of the above

    Answer: All of the above

    Explanation: Conductivity depends on concentration, nature of electrolyte, and temperature.

15. Question 15

    Q15. For a strong electrolyte, the molar conductivity

    • A) Decreases with increase in concentration
    • B) Increases with increase in concentration
    • C) Remains constant with concentration
    • D) Is independent of concentration

    Answer: Decreases with increase in concentration

    Explanation: For strong electrolytes, Λm decreases with increase in concentration due to ion-ion interactions.

16. Question 16

    Q16. The Nernst equation is used to calculate

    • A) E°cell
    • B) Ecell
    • C) ΔG°
    • D) Kc

    Answer: Ecell

    Explanation: The Nernst equation is used to calculate Ecell under non-standard conditions: E = E° - (RT / nF) * ln Q.

17. Question 17

    Q17. The standard hydrogen electrode has a potential of

    • A) +1 V
    • B) 0 V
    • C) -1 V
    • D) +0.5 V

    Answer: 0 V

    Explanation: By definition, the standard hydrogen electrode has a potential of 0 V.

18. Question 18

    Q18. The cell constant is given by

    • A) l / a
    • B) a / l
    • C) l × a
    • D) 1 / (l × a)

    Answer: l / a

    Explanation: The cell constant is the ratio of the distance between electrodes (l) to the area of electrodes (a), i.e., l / a.

19. Question 19

    Q19. The limiting molar conductivity is the sum of

    • A) Molar conductivity of cation and anion
    • B) Limiting molar conductivity of cation and anion
    • C) Conductivity of cation and anion
    • D) None of the above

    Answer: Limiting molar conductivity of cation and anion

    Explanation: According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivity of its ions.

20. Question 20

    Q20. The conductivity of a 0.1 M solution of KCl is 1.29 S / m. Calculate its molar conductivity.

    • A) 129 S cm² / mol
    • B) 12.9 S cm² / mol
    • C) 1.29 S cm² / mol
    • D) 1290 S cm² / mol

    Answer: 129 S cm² / mol

    Explanation: Molar conductivity = conductivity * 1000 / M = 1.29 * 1000 / 0.1 = 12900 S cm² / mol, then convert to S cm² / mol.