# Disproportionation of hydrogen peroxide

Hydrogen peroxide is decomposed as follows: $$\ce{2H2O2 -> 2H2O +O2}$$ This is a disproportionation redox reaction of $\ce{H2O2}$ involving the 2 half reactions $$\ce{H2O2 -> O2 + 2H^+ + 2e-}$$ $$\ce{H2O2 + 2e- + 2H^+ -> 2H2O}$$ But I noticed semantically that it can also be the sum of the two following half reactions: $$\ce{2H2O2 -> 2O2 + 4H^+ + 4e-}$$ $$\ce{O2 + 4H^+ + 4e- -> 2H2O}$$ Both reactions are feasible according to the following:

$$\begin{array}{ccc} \text{Oxidised species} & \text{Reduced species} & E^\circ (\mathrm{V}) \\ \hline \ce{H2O2} & \ce{H2O} & 1.763 \\ \ce{O2} & \ce{H2O} & 1.23 \\ \ce{O2} & \ce{H2O2} & 0.695 \\ \end{array}$$

Can it happen in the second way? If yes, can you explain the mechanism in simple terms?

• If it is what I think it is, and the fourth equation is just a typo (H2O should be H2O2), then yes, it is formally correct. However, if you just keep H2O2 by itself, without any oxygen, it will still decompose. This isn't captured by your proposed pair of half-equations, which essentially says that you need oxygen gas for the total reaction (i.e. sum of 2 half-reactions) to proceed. – orthocresol Dec 25 '15 at 7:16