How do you balance disproportionation reactions?

1 Answer
Aug 10, 2016

Answer:

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:

#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.