Question

How do you balance disproportionation redox reactions?

Answer 1

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 `(i)`:

`1/2Cl_2 + e^(-) rarr Cl^(-)`

Oxidation `(ii)`:

`1/2Cl_2(g) + 3H_2O(l) rarr ClO_3^(-) + 6H^(+) +5e^(-)`

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: `5xx(i)+(ii)`.

`3Cl_2(g) + 3H_2O(l) rarr ClO_3^(-) + 5Cl^(-) +6H^+`

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, `BrF` (`Br(+I) and F(-I)`) to give the interhalogen `BrF_3` and elemental bromine `(Br_2)`? Bromine disproportionates here to `Br^(III+)` and `Br^(0)`.