# How do you balance disproportionation reactions?

Aug 10, 2016

I usually write 2 separate redox reaction: one an oxidation; the other a reduction.

#### Explanation:

For example, let's take the action of hydroxide on chlorine gas. Chloride ion is produced, a reduction:

$\frac{1}{2} C {l}_{2} + {e}^{-} \rightarrow C {l}^{-}$ $\left(i\right)$

But also chlorate ion is produced, an oxidation from $C {l}_{2}^{0}$ to $C {l}^{V +}$

1/2Cl_2 + 3H_2O rarr ClO_3^− + 6H^+ + 5e^-

We add $6 \times H {O}^{-}$ to get:

1/2Cl_2 + 6HO^(-) rarr ClO_3^− + 3H_2O + 5e^- $\left(i i\right)$
(I add the hydroxides because the reaction was specified to be the action of BASE!)

And $5 \times \left(i\right) + \left(i i\right)$ gives

3Cl_2 + 6HO^(-) rarr ClO_3^− + 5Cl^(-) +3H_2O

This is the same as any redox calculation. But here that stuff that is oxidized is the SAME as that which is reduced.