Calcualte the emf of the following cell at `25^(@)` C
`Ag(s) | Ag^(+) (10^(-3) M) || Cu^(2+) (10^(-1) M) | Cu(s)`
Given: `E_("cell")^(@) = +0.46 V` and `log 10^(n) =n`.

Explanation

The cell reaction is :
`Cu + 2Ag^(+) to Cu^(2+) + 2Ag`
EMF of the cell can be obtained by using the Nernst equation:
`E_("cell") = E_("cell")^(@) -0.059/2 log ([Cu^(2+)])/([Ag^(+)]^(2))`
Substituting the values, we get
`E_("cell") = 0.46 V - 0.059/2 log ([10^(-1)])/([10^(-3)])^(2)`
`=0.46 - 0.0295 log 10^(5)`
`=0.46 - 0.1475`
or `E_("cell") = 0.3125` V