Question

Find the emf of the cell
Ni(s) + 2 Ag+ (0.001 M) → Ni2+ (0.001 M) + 2Ag(s)
(Given that E°cell = 1.05 V, $\frac{2.303RT}{F} = 0.059$ at 298 K)

• A. 1.0385 V
• B. 1.385 V
• C. 0.9615 V
• D. 1.05 V

Answer 1

Answer: (C)

0.9615 V

Answer 2

The correct option is (C): 0.9615 V