Question

Calculate the e.m.f of the following cell reaction at 298 K:

`Mg_(s) + Cu^(2+) (0.0001 M) -> Mg^(2+) (0.0001M) + Cu_(s)`

The standard potential `(E^@)`of the cell is  2.71 V

Answer 1

`Mg  + Cu^(2+) -> Mg^(2+) + Cu`

`Q = [Mg^(2+)]/[Cu^(2+)] = (10^(-3))/(10^(-4)) = 10`

n = 2

`E_"cell" = E_"cell"^0 - 0.0592/n log_10 Q`

`= 2.71 - 0.0592/2 log_10 10`

= 2.71 - 0.0296

`E_"cell" = 2.68 V`