How do you balance disproportionation redox reactions?

1 Answer
anor277
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 #(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)#.

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