In acid medium, `MnO_(4)^(c-)` is an oxidizing agent.
`MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O`
If `H^(o+)` ion concentration is doubled, electrode potential of the half cell `MnO_(4)^(c-), Mn^(2+)|Pt` will

Options

A. Increase by `28.36mV`
B. Decrease by `28.36mV`
C. Increase by `14.23mV`
D. Decrease by `142.30mV`

Answer

a

Explanation

`MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O`
`E_(MnO_(4)^(c-)|Mn^(2+))`
`=E^(c-)._(MnO_(4)|Mn^(2+))-(0.059)/(5)log(([Mn^(2+)])/([MnO_(4)][H^(o+)]^(8)))`
`=E^(c-)._(MnO_(4)^(c-)|Mn^(2+))-(8)/(5)xx0.059pH`
`-(0.059)/(5)log(([Mn^(2+)])/([MnO_(4)^(c-)]))`
`implies[H^(o+)]` is doubled, `i.e., pH` is reduced by `log 2 -= 0.3,` then `E_(MnO_(4)^(c-)|Mn^(2+))` will be changed `(` increase `)` by
`(8)/(5)xx0.059xx0.3V=28.36mV`