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

Oct 6, 2016

A redox reaction occurs. Sulfite should be oxidized to sulfate. Permanganate ion should be reduced to $M {n}^{2 +}$.

Explanation:

$\text{Reduction half equation: } M n \left(V I I +\right) \rightarrow M n \left(I I +\right)$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i\right)$

$\text{Oxidation half equation: } S \left(I V +\right) \rightarrow S \left(V I +\right)$

$S {O}_{3}^{2 -} + {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 2 {H}^{+} + 2 {e}^{-}$ $\left(i i\right)$

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 $2 \times \left(i\right) + 5 \times \left(i i\right)$:

$2 M n {O}_{4}^{-} + 5 S {O}_{3}^{2 -} + 6 {H}^{+} \rightarrow 2 M {n}^{2 +} + 5 S {O}_{4}^{2 -} + 3 {H}_{2} O$

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

So what would you observe in this reaction? $M n {O}_{4}^{-}$ is strongly coloured, and gives a beautiful deep purple solution. On the other hand, $M {n}^{2 +}$ is almost colourless (very concentrated solutions are a pale rose). And thus this redox reaction is self-indicating, and proposes that $2$ $\text{equiv}$ permanganate reacts with $5$ $\text{equiv}$ 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.