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 electrode reaction is
`Fe^(3+) + e^(-) to Fe^(2+)`
`E = E^(@) - (0.059)/(1) "log" ([Fe^(2+)])/([Fe^(3+)])`
`E^(@) = 0.771 V`
`E^(@) = 0.771 V`
`[Fe^(3+)] = 0.02 M`
`[Fe^(2+)] = 2.0 M`
`E = 0.771 - (0.059)/(1) "log" ((2.0))/((0.02))`
`=0.771 - 0.059 log 100`
`=0.771 - 0.118 = 0.653` V