# How to balance this redox reaction?

May 24, 2017

Well, sulfite $\stackrel{+ I V}{S}$, is oxidized to sulfate $\stackrel{V I +}{S}$.

#### Explanation:

We use the method of half equations. The difference in oxidation numbers is accounted for electron transfer.

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

And permanganate ion, $\stackrel{V I I +}{\text{Mn}}$ is REDUCED to $M {n}^{2 +}$:

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

And we adds $5 \times \left(i\right) + 2 \times \left(i i\right)$ to get................

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

The which, I think, is balanced with respect to mass and charge. But don't trust my arithmetic.

What would we see in this reaction? The deep purple colour of permanganate would dissipate as it reacts to give almost colourless $M {n}^{2 +}$ ion.

Why did I add $5 \times \left(i\right) + 2 \times \left(i i\right)$? Well, I wanted to remove the electrons from the reactant side and the product side. Electrons are used here as conceptual particles for the purpose of balancing charge.

See here and links for more examples of redox equations.