Question

Why KMnO4 in alkaline solution accept 3e per Mn and act as mild oxidizing agent while in acidic solution can gain 5 electron per Mn act as strong oxidizing agen???

Answer 1

Consider the Pourbaix diagram of `"Mn"`:

If it's easy to reduce `"MnO"_4^-`, it's a good oxidizing agent. Two of these are possible:

`stackrel(color(blue)(+7))"Mn""O"_4^(-)(aq) + 2"H"_2"O"(l) + 3e^(-) rightleftharpoons stackrel(color(blue)(+4))"Mn""O"_2(s) + 4"OH"^(-)(aq)`

`stackrel(color(blue)(+7))"Mn""O"_4^(-)(aq) + 8"H"^(+)(aq) + 5e^(-) rightleftharpoons stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4"H"_2"O"(l)`

In acidic solution, specifically below `"pH"` `4`, `stackrel(color(blue)(+7))("Mn")"O"_4^(-)` could get reduced all the way to `"Mn"^(2+)`, i.e. from `+7` to `+2`, a change in oxidation state of `-5` in the reduction half-reaction.

The diagonal `"Mn"^(2+)//"MnO"_2` equilibrium line shows a dependence on both voltage and `"pH"`. As the equilibrium between `stackrel(color(blue)(+4))"Mn""O"_2(s)` and `"Mn"^(2+)(aq)` is `bb"pH"`-dependent...

`"MnO"_2(s) + 4"H"^(+)(aq) + 2e^(-) rightleftharpoons "Mn"^(2+)(aq) + 2"H"_2"O"(l)`

The more acidic the `pH` (the more `H^+` there is), the more the equilibrium shifts to `Mn^(2+)` by Le Chatelier's principle, favoring a `5e^(-)` reduction.

In basic solution (above `"pH"` `7.5`), `"Mn"^(2+)` does not exist; as the `"pH"` increases, the `["OH"^(-)]` increases...

`"MnO"_2(s) + 2"H"_2"O"(l)(aq) + 2e^(-) rightleftharpoons "Mn"^(2+)(aq) + 4"OH"^(-)(aq)`

And that shifts the equilibrium left, towards `stackrel(color(blue)(+4))(Mn)O_2(s)`, favoring a `3e^(-)` reduction.

And so, in basic solution, we only have a change in oxidation state of `-3` in the reduction half-reaction: `+7` to `+4`.