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E(MnO4^(-)// Mn^(+2), H^(+)) can be 0.48...

`E_(MnO_4^(-)// Mn^(+2), H^(+))` can be 0.48 V greater than from `E^@` value if the `[MnO_4^(-)]` is equal to `[Mn^(+2)]` at a `P^(H)` of ___________

Text Solution

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`MnO_4 + 8 H^(oplus) + 5e^(-) to Mn^(+2) + 4H_2O , E= E^(0) - (0.0591)/(5) log""([Mn^(+2)])/([MnO_4^(-)][H^(oplus)]^(8))`
` (E-E^(0)) = (-0.0591)/(2) log [H^(oplus)]^(-8) , 0.48 = -(0.0591)/(5) xx (-8) log[H^(oplus)]`
` P^(H) - log[H^(oplus)] = [(0.048 xx 5)/(0.0591xx 8)] = 5.076`
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