Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained `0.1M MnO_(4)^(-)` and `0.8M H^(+)` and which was treated with `Fe^(2+)` necessary to reduce `90%` of the `MnO_(4)` to `Mn^(2+)`
`MnO_(4)^(-) +8H^(+) +5e rarr Mn^(2+) +H_(2)O, E^(@) = 1.51V`

Answer

`1.39V`

Explanation

`MnO_(4)^(-) +8H^(+) +2e^(-) rarr Mn^(2+) +4H_(2)O +E^(@) = 1.51V`
`{:(0.1,0.8),(0.01,0.08):}`
`bar(E=1.51 -(0.0591)/(5)log.(0.09)/(0.01xx(0.08)^(8))`
`E = 1.51 -0.15`
`= 1.39V`