Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in `2.0M Fe^(2+)` and `0.02M Fe^(3+)` solution. Given `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771 V`.

Explanation

The half cell reaction is :
`Fe^(3+)+e to Fe^(2+)` (or take `Fe^(2+) to Fe^(3+)+e`)
Thus `E_(Fe^(3+)//Fe^(2+))=+0.771+(0.059)/1 log(0.02)/(2.0)`
`=0.771+(0.059)/(1)log10^(-2)=0.653V`