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:

2H_{2}O_{2} -> 2H_{2}O + O_{2}. In this reaction, the O.S. of H_{2}O_{2}
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)

Cl_{2}-> 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.

Cl_{2}+ e^{-} (electron) -> Cl^{-}

Cl_{2}+ 2e^{-}
-> 2Cl
^{-}
So, now we're done for this half equation, but we still have to do the other half equation.

Cl_{2} -> ClO3^{-}

Since O = -2, the charge of Cl goes from 0 to 5, oxidizing.

Cl2 -> ClO_{3}^{-} + 5e-

Cl2 -> 2ClO_{3}^{-} + 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.