A disproportionate reaction is a redox reaction, where the same substance is both oxidized and reduced. For example, in the decomposition of hydrogen peroxide (H2O2), it looks like this:
2H2O2 -> 2H2O + O2. In this reaction, the O.S. of H2O2 is 0, so hydrogen (+) balances the oxygen, so the original O.S. of oxygen is -1. In 2(H 2 O), the O.S, is -2, as usual, and in O2, it's zero, because it's neutral. So, it oxidizes and reduces at the same time.
Here is an example problem (Chap.5 , Problem 41A in General Chemistry, by Petrucci)
Cl2-> Cl- + ClO 3
- (Basic Solution)
First, you split the reaction up:
Cl 2 -> Cl - Cl 2 -> ClO 3
- Obviously, those aren't balanced, and their oxidiation states are equal, so you must balance them.
Cl2+ e- (electron) -> Cl-
Cl2+ 2e- -> 2Cl - So, now we're done for this half equation, but we still have to do the other half equation.
Cl2 -> ClO3-
Since O = -2, the charge of Cl goes from 0 to 5, oxidizing.
Cl2 -> ClO3- + 5e-
Cl2 -> 2ClO3- + 10e-
Now, we have to add water to balance out the oxygens.
Cl2 + 6H2O -> 2 ClO3 + 10e-
Now, we add 12 H+ atoms on the right side
Cl2 + 6H2O -> 2 ClO3 + 12H+ 10e-
Since it is reacting in a basic solution, we can't have 12 H+, we need to add 12 OH- to both sides.
Cl2 + 6H2O + 12OH- -> 2 ClO3 + 12H2O + 10e- (12 H+ + 12 OH- = 12HOH = 12H2O)
Subtract by 6H2O on both sides
Cl2 + 12OH- -> 2ClO3 + 6H2O + 10e-
Now, we're done with both, so now we just scale up the original to make the electrons equal.
5Cl2 + 10e- -> 10Cl-
We add them together, and cross out like terms
6Cl2 + 12OH- -> 10Cl- + 2ClO3 + 6 H2O
Now, we divide by two:
3Cl2 + 6OH- -> 5Cl- + ClO3 + 3H2O
That's our final answer, and we have no H atoms. With an acidic solution, that would be acceptable, however in a base, we just have to add the extra step of adding the hydroxide.