Find the emf of the cell in which the following reaction takes place at 298 K Ni(s) + 2Ag+ (0.001 M) Ni2+ (0.001 M) + 2Ag(s) (Given that E cell = 1.05 V, 2.303 RTF = 0.059 at 298 K)
- 1.0385 V
- 1.385 V
- 0.9615 V
- 1.05 V
Answer
(3) 0.9615 V
Solution
Here n = 2. Nernst equation:
Ecell = E cell - 0.059n [Ni2+][Ag+]2
= 1.05 - 0.0592 10-3(10-3)2
= 1.05 - 0.0295 103
= 1.05 - 0.0885
= 0.9615 V
Answer: (3)