What is the reaction mechanism of the reaction:

$$\ce{H2O2 + O3 -> H2O + 2O2}$$

I have tried searching this on Google but am not getting good answers.


1 Answer 1


This is basically an excerpt of the answer I have given to another quite similar question.

Ben is quite right, neither the reaction nor the mechanism is simple. It involves several different radical chains all resulting in the decomposition of ozone.[a]

It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}$$

In presence of Hydrogen donors like $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[a] Those are forming the main chain carriers will be $\ce{HO, HO2}$. Finally stopping at some point by recombination of these species. \begin{aligned}\ce{ O + H2O2 &-> OH + HO2\\ HO + O3 &-> HO2 + O2\\ HO2 + O3 &-> HO + 2O2\\ 2HO2 &->}\text{stable products} \end{aligned}

In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}$$

Especially important might also be protolysis of hydrogen peroxide. $$\ce{H2O2 + H2O <=> HO2- + H3+O}$$

Steahelin and Hoigné propose several hypotheses in which way ozone will be decomposed, but I will not go into absurd detail here.[b] The core assumptions involve $\ce{{}^{-}OH}$ as the initiator of a radical chain. Followed by radical (ionic) species like $\ce{.O2-, .O3-, HO2., .OH}$ resulting again in a chain reaction. \begin{aligned}\ce{ O3 + {}^{-}OH &-> .O2- + HO2.\\ O3 + {}^{-}OH &-> HO2- + O2\\ O3 + {}^{-}OH &-> .O3- + .OH\\ HO2. &<=> H+ + .O2-\\ O3 + HO2- &-> .OH + .O2- + O2\\ O3 + .O2- &-> .O3- + O2\\ .O3- + H2O &-> .OH + {}^{-}OH + O2 }\end{aligned}

  • $\begingroup$ Hi Martin ! In this reaction, hydrogen peroxide is the reducing agent,right ? $\endgroup$
    – biogirl
    May 14, 2014 at 10:20
  • $\begingroup$ @biogirl Yes and no. The oxidation state in ozone is $0$, but that is the same as in bimolecular oxygen. The oxidation state of oxygen in hydrogen peroxide is $-1$. So one oxygen is reduced when forming water OS$=-2$ and one is oxidised to $\ce{O2}$ OS$=0$. This kind of reaction is called disproportionation. So referring to itself it is both, referring to ozone it is none of it. $\endgroup$ May 14, 2014 at 10:37
  • $\begingroup$ Ok. Makes sense. $\endgroup$
    – biogirl
    May 14, 2014 at 15:44
  • $\begingroup$ What if you use Barium peroxide instead of Hydrogen peroxide? $\endgroup$ May 4, 2017 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.