Find the emf of the cell in which the following reaction takes place at 298   K Ni ( s ) + 2 Ag + ( 0 . 001   M ) → Ni 2 + ( 0 . 001 M ) + 2 Ag ( s ) (Given that E ° cell = 1 . 05   V ,   2 . 303   RT F = 0 . 059 at 298   K )

Step-by-step solution

Given reaction: Ni ( s ) + 2 Ag + ( 0 . 001   M ) → Ni 2 + ( 0 . 001 M ) + 2 Ag ( s ) Applying Nernst equation we have: E cell = E cell 0 - 2 . 303   RT nF log Ni 2 + [ Ag ] Ag + 2 [ Ni ] Active mass of solid is taken to be unity so Ni s = Ag s = 1 E c e l l = E c e l l 0 - 0.059 n log N i 2 + A g + 2 = 1.05 - 0.0591 2 log 0.001 0.001 2 = 1.05 - 0.0295   log ( 1 × 10 3 ) = 1.05 - 0.0295   × 3 = 0.9615   V Therefore, the emf of the cell is 0 . 9615   V . Note: Question i