Page 1 of 1 ELECTROCHEMISTRY 1. Equilibrium constant (K c) for the given cell reaction is 10 Calculate Ecell0: A (s) + B2+ (aq) → A2+ (aq) + B(s) [Ans: 0.0295 V ] 2. Calculate ∆rG0 for the following reaction: Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) [Given: Ecell0 = + 2.71 V, 1 F = 96500 C mol–1] [Ans: – 523.03 KJ mol–1] 3. For the reaction, ∆G0 = –43600 J at 250C. Calculate the emf of the cell. [log 10–n = –n] n = 2, using the formula, ∆r G0 = – n E0 F ⇒ 43600 = – 2 x E0 x 96500 ⇒ E0 = 0.226 V Ecell = Ecell0 – 0.0591
n log [H+]2 [Cl−]2
p[H2] = [0.226 V – 0] – 0.0591
2 log [0.1]2 [0.1]2
1 = 0.3442 V 4. Calculate the emf of the following cell at 298 K: 2Cr (s) + 3 Fe2+ (0.1 M) → 2Cr3+ (0.01M) + 3Fe (s) [Given: ECr3+Cr⁄0 = –0.74 V, EFe2+Fe⁄0 = – 0.44 V] [Ans: 0.31 V ] 5. Calculate Ecell0 for the following reaction at 298 K: 2Al (s) + 3Cu2+ (0.01M) → 2Al3+(0.01M) + 3Cu (s), Given: E cell = 1.98 V [Ans: 2.00 V] 6. Calculate Ecell0 for the following reaction at 250C: A + B2+ (0.001M) → A2+ (0.0001M) + B, [Given: Ecell = 2.6805 V , 1F = 96500 C mol–1] [Ans: 2.651 V] 7. Calculate the emf of the following cell at 250C: Fe (s) | Fe2+ (0.001 M) || H+ (0.01M) | H 2 (g) (1 bar) | Pt(s) [Given: EH+H2⁄0 = 0.00 V, EFe2+Fe⁄0 = – 0.44 V] [Ans: 0.4105 V] 8. Calculate the emf of the following cell at 250C: Zn (s) | Zn2+ (0.001 M) || H+ (0.01M) | H 2 (g) (1 bar) | Pt(s) [Given: EH+H2⁄0 = 0.00 V, E𝑍𝑛2+𝑍𝑛⁄0 = – 0.76 V] [Ans: 0.7305 V] 9. Calculate Ecell0 and ∆rG0 for the following reaction at 250C: A2+ (aq) + B+ (aq) → A3+ (aq) + B (s), [Given: K c = 1010, 1F = 96500 C mol–1] [Ans: 0.591 V, –57.031 kJ mol–1] 10. A voltaic cell is set up at 250C with the following half – cell: Ag+ (0.001M) | Ag and Cu2+ (0.10 M) | Cu (s) What would be the voltage of this cell? (Ecell0 = 0.46 V) [Ans: 0.3123 V]
n log [H+]2 [Cl−]2
p[H2] = [0.226 V – 0] – 0.0591
2 log [0.1]2 [0.1]2
1 = 0.3442 V 4. Calculate the emf of the following cell at 298 K: 2Cr (s) + 3 Fe2+ (0.1 M) → 2Cr3+ (0.01M) + 3Fe (s) [Given: ECr3+Cr⁄0 = –0.74 V, EFe2+Fe⁄0 = – 0.44 V] [Ans: 0.31 V ] 5. Calculate Ecell0 for the following reaction at 298 K: 2Al (s) + 3Cu2+ (0.01M) → 2Al3+(0.01M) + 3Cu (s), Given: E cell = 1.98 V [Ans: 2.00 V] 6. Calculate Ecell0 for the following reaction at 250C: A + B2+ (0.001M) → A2+ (0.0001M) + B, [Given: Ecell = 2.6805 V , 1F = 96500 C mol–1] [Ans: 2.651 V] 7. Calculate the emf of the following cell at 250C: Fe (s) | Fe2+ (0.001 M) || H+ (0.01M) | H 2 (g) (1 bar) | Pt(s) [Given: EH+H2⁄0 = 0.00 V, EFe2+Fe⁄0 = – 0.44 V] [Ans: 0.4105 V] 8. Calculate the emf of the following cell at 250C: Zn (s) | Zn2+ (0.001 M) || H+ (0.01M) | H 2 (g) (1 bar) | Pt(s) [Given: EH+H2⁄0 = 0.00 V, E𝑍𝑛2+𝑍𝑛⁄0 = – 0.76 V] [Ans: 0.7305 V] 9. Calculate Ecell0 and ∆rG0 for the following reaction at 250C: A2+ (aq) + B+ (aq) → A3+ (aq) + B (s), [Given: K c = 1010, 1F = 96500 C mol–1] [Ans: 0.591 V, –57.031 kJ mol–1] 10. A voltaic cell is set up at 250C with the following half – cell: Ag+ (0.001M) | Ag and Cu2+ (0.10 M) | Cu (s) What would be the voltage of this cell? (Ecell0 = 0.46 V) [Ans: 0.3123 V]
