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## Feb 25 Disproportionation Reactions

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.

David Witten