# How do you balance disproportionation redox reactions?

Jun 21, 2016

You balance them stoichiometrically, and in fact you follow the same procedure of separate redox processes for oxidation and reduction.

#### Explanation:

Chlorine gas is known to disproportionate to chlorate and chloride ion:

Reduction $\left(i\right)$:

$\frac{1}{2} C {l}_{2} + {e}^{-} \rightarrow C {l}^{-}$

Oxidation $\left(i i\right)$:

$\frac{1}{2} C {l}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right) \rightarrow C l {O}_{3}^{-} + 6 {H}^{+} + 5 {e}^{-}$

So mass and charge are balanced in the half equations. All I have to do is simply combine them in the usual fashion to remove electrons: $5 \times \left(i\right) + \left(i i\right)$.

$3 C {l}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right) \rightarrow C l {O}_{3}^{-} + 5 C {l}^{-} + 6 {H}^{+}$

Now all I have done here is to treat this like a normal redox equation. I considered the oxidation reaction, and the reduction reaction SEPARATELY utilizing electrons as virtual particles of convenience, and I add the oxidation and reductions reactions appropriately to eliminate the electrons.

Can you balance the reaction of the interhalogen compound, $B r F$ ($B r \left(+ I\right) \mathmr{and} F \left(- I\right)$) to give the interhalogen $B r {F}_{3}$ and elemental bromine $\left(B {r}_{2}\right)$? Bromine disproportionates here to $B {r}^{I I I +}$ and $B {r}^{0}$.