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 28.36 mV
B. Decrease by 28.36mV
C. Increase by 14.23 mV
D. Decrease by 142 .23mV

Answer

A, B

Explanation

`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4h_(2)O`
`E_(Mno_(4)^(1)//Mn^(2+)) = E_(MnO_(4)^(-)//Mn^(2+))(0.059)/(5) log_(10) ([Mn^(2+)])/([MnO_(4)^(-) ][H^(+)]^(8))`
`= E_(MnO_(4)^(-)//Mn^(2+)) - (8)/(5) xx0.059 PH - (0.059)/(5) log _(10) ([Mn^(2+)])/([MnO_(4)//(-)])`
If `[H^(+)]` is double i.e PH is reduced by `log_(10) 2equiv0.3` Then `E_(MnO_(4)^(-)//Mn^(2+))`will be changed (increased) by `(8)/(5)0.059 xx0.3V = 28.36mV`