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Calculate the potential of an indicator ...

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`

Text Solution

Verified by Experts

The correct Answer is:
1.39 V
`{:(,MnO_(4)^(-),+,8H^(+),+,5e^(-),to,Mn^(2+),+,H_(2)O),("Initially",0.1,,0.8,,,,0,,0),("After reaction",0.1-0.1xx0.9,,0.8-0.09xx8,,,,0.09,,0.09),(,0.10-0.09,,0.8-0.09xx8,,,,0.09,,0.09),(,0.01,,0.08,,,,0.09,,0.09):}`
`E=E^(@)-(0.059)/(5)log(([Mn^(2+)])/([MnO_(4)^(-)][H^(+)]^(8)))=1.51-(0.059)/(g)log_(10)(((0.09))/((0.01)(0.08)^(8)))=1.39V`
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