Question

What would occur in the reaction between `"potassium permanganate"`, `KMnO_4` and `"sodium sulfite"`, `Na_2SO_3`?

Answer 1

A redox reaction occurs. Sulfite should be oxidized to sulfate. Permanganate ion should be reduced to `Mn^(2+)`.

Explanation:

`"Reduction half equation: "Mn(VII+) rarr Mn(II+)`

`MnO_4^(-) +8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O` `(i)`

`"Oxidation half equation: "S(IV+) rarr S(VI+)`

`SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^(-)` `(ii)`

Both equations are (I think) balanced with respect to mass and charge, as they must be if they reflect reality. The overall redox reaction excludes the electrons; so we take `2xx(i)+5xx(ii)`:

`2MnO_4^(-) +5SO_3^(2-)+6H^(+)rarr 2Mn^(2+) + 5SO_4^(2-)+3H_2O`

Which is (I think) balanced with respect to mass and charge.

So what would you observe in this reaction? `MnO_4^-` is strongly coloured, and gives a beautiful deep purple solution. On the other hand, `Mn^(2+)` is almost colourless (very concentrated solutions are a pale rose). And thus this redox reaction is self-indicating, and proposes that `2` `"equiv"` permanganate reacts with `5` `"equiv"` of sulfite.

How did I know the oxidation/reduction products? Experience and practice, and actually doing the titrations and observing and explaining the colour change.