Calculate the cell potential for the following galvanic cell:
The first electrode consists of `Fe^(3+)|Fe^(2+)` complete in which `[Fe^(3+)]=1M` and `[Fe^(2+)]=[0.1M]`
The second electrode consists of `MnO_4(-)|Mn^(2+)` couple in acidic solution in which
`[MnO_4^-]=1times 10^-2 M, [Mn^(2+)]=1 times 10^-4 M and [H^+]=1 times 10^-3 M`
`[E_(Fe^(3+),Fe^(2+))^@=0.771V, E_(MnO_4^-, Mn^(2+))^@=1.51V)`

Explanation

0.42V