InterviewSolution
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InterviewSolution
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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` willA. Increase by `28.36mV`B. Decrease by `28.36mV`C. Increase by `14.23mV`D. Decrease by `142.30mV` |
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Answer» Correct Answer - a `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` |
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