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?

  • $\begingroup$ 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. $\endgroup$ Commented Dec 25, 2015 at 7:16

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In principle you could find a lot of equations, all leading to the same result. But these are only (half-)equations. Most of the times the reaction mechanism is much more complicated than just these reaction equations.

The decomposition of hydrogenperoxide is still an active field of research in various media with various catalysts.

Unfortunately with the massive overflow of the above, I was only able to find a quite old PhD thesis by Trice Walter Haas (Iowa State University, 1960)

Here is a small line-up of reaction equations that will play a role in the decomposition, at least to that part that I can imagine it. And I am just assuming that there are radicals involved, because there usually are. \begin{align} \ce{H2O2 &<=> H+ + HO2- }\\ \ce{HO2- + H2O2 &<=> HO2 + HO + OH-}\\ \ce{HO2 + OH &<=> H2O + O2}\\ \end{align}


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