How do you balance disproportionation reactions?

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Answer 1 · 2016-08-10T05:16:23

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

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

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

But also chlorate ion is produced, an oxidation from $Cl_2^(0)$ to $Cl^(V+)$

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

We add $6xxHO^-$ to get:

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

And $5xx(i)+(ii)$ 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.