Let us calculate the emf of the following cell at `25^@C` using Nernst equation.
`Cu(s)|Cu^(2+)(0.25 aq. M)||Fe^(3+)(0.05 aq M)|Fe^(2+)(0.1 aq M)|pt(s)`

Explanation

Given : `(E^@)_(Fe^(3+)|Fe^(2+)) = 0.77 V and (E^@)_(Cu^(2+)|cu) = 0.34V`
Half reaction are `Cu(s) to Cu^(2+) (aq) + 2e^(-) " " ………(1)`
`2Fe^(3+) (aq) + 2e^(-) to 2Fe^(2+)(aq)" " ……..(2)`
the overall reaction is `Cu(s) + 2Fe^(3+)((aq)) to Cu^(2+)(aq) + 2Fe^(2+)(aq)`, and n = 2 apply Nernst equation at `25^@C`
`E_("cell") = E_("cell")^(@) - (0.0591)/(2) log ([Cu^2][Fe^(2+)]^2)/([Fe^(3+)]^(2)) " " [ :. Cu(S) = 1]`
`E_("cell")^(@) = (E_("ox")^(@))_(Cu|Cu^(2+)) + (E_("red")^(@))_(Fe^(3+)|Fe^(2+))`
Given standared reduction potential of `Cu^(2+)|Cu` is 0.34 V
`:. (E_("ox")^@)_(Cu|Cu^(2+)) = -0.34 V`
`(E_("cell")^(@))_(Fe^(3+)|Fe^(2+)) = 0.77 V`
`:. E_("cell")^(@) = -0.34 + 0.77 = 0.43 V`
`:. E_("cell") = 0.43- (0.0591)/2 xx log ((0.25)(0.1)^2)/((0.005)^2)`
`= 0.43 - (0.0591)/2 xx 2`
`= 0.43 - 0.0591`
`= 0.3709V`
`= log((0.25)(0.1)^2)/((0.005)^2)`
`= log(25 xx 10^(-2)xx 1 xx 10^(-2))/(25 xx 10^(-6))`
`= log 10^2`
`= 2 log_(10)10 = 2`.