Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s) + 2 Ag+ (0.001 M) → Ni2+ (0.001 M) + 2 Ag (s)
(Given that \(\mathrm{E}_{\text {cell }}^{\circ}\) = 1.05 V, \(\frac{2.303 \mathrm{RT}}{\mathrm{F}}\) = 0.059 at 298 K)
1
1.05 V
2
1.0385 V
3
1.385 V
4
0.9615 V