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`

Explanation

Let us consider Galvanic cell
`Pt, H_(2)|H^(+) ("1 molar")||MnO_(4)^(-) (H^(+)) |Mn^(2+), Pt`
Anode half cell: `H_(2) (g) to 2H^(+) 2e^(-)`
Cathode half cell" `MnO_(4)^(-)+8H^(+) +5e^(-) to Mn^(+2) +4H_(2)O`
Initial Conc: `0.1" "0.8" "0" "0`
Alter Complete reaction with `Fe^(+2) (0.1 -(0.1 xx 90)/(100)) (0.8 -(0.1 xx 90)/(100) xx 8)" "(0.1 xx 90)/(100)`
`" (0.01) (0.08) (0.09)"`
So, electrode potential of indicator electrode
`E_(MnO_(4)^(-)//Mn^(+2)) =E_(MnO_(4)^(-)//Mn^(+2))^(@)-(0.0591)/(5) log ([Mn^(2+)])/([MnO_(4)^(-)] [H^(+)]^(8))`
`=1.51 -(0.0591)/(5) log ((0.09))/((0.01) (0.08)^(8))`
`=1.51 -(0.0591)/(5) log (9)/(1.67 xx 10^(-9))`
`=1.51 -(0.0591)/(5) log (5.36 xx 10^(9))`
=1.51 -0.1149
=1.395 V
Thus, potential of as indicator electrode versus the SHE is 1.395 V because `E_(SHE)=0`