The following reaction is from https://ncert.nic.in/textbook/pdf/kech202.pdf, p. 11:

$$\ce{H2O2 + HOCl -> H3O+ + Cl- + O2}$$

I have to find if $\ce{H2O2}$ is acting like a reducing agent or an oxidising agent in this reaction.

If we see, the $\ce{Cl}$ atom in $\ce{HOCl}$ got reduced from $+1$ to $-1$ oxidation state. That tells me that $\ce{H2O2}$ is acting as a reducing agent. But if I see the $\ce{O}$ atom, it is oxidised from $-2$ to $0$, now this tells me that $\ce{H2O2}$ is acting as a oxidising agent.

But how is this possible? I know I'm doing something fundamentally wrong.

  • $\begingroup$ Evaluate formal oxidation numbers of chlorine and oxygen, the answer to the question should be clear from the result. $\endgroup$
    – Poutnik
    Mar 16, 2022 at 11:57

1 Answer 1


You are not doing much wrong except for assigning improper oxidation number to oxygen of the peroxo group. If the redox reaction “feels” wrong, you can check yourself using a table of standard reduction potentials [1, p. 5-79]:

$$ \begin{align} \ce{H2\overset{-1}{O}_2 &<=> \overset{0}{O}_2 + 2 H+ + 2 e-} &\quad E^\circ_1 &= \pu{-0.695 V} \tag{1}\\ \ce{H\overset{+1}{Cl}O + H+ + 2 e- &<=> \overset{-1}{Cl}^- + H2O} &\quad E^\circ_2 &= \pu{+1.482 V} \tag{2}\\ \hline \ce{H2\overset{-1}{O}_2 + H\overset{+1}{Cl}O &-> H3O+ + \overset{-1}{Cl}^- + \overset{0}{O}_2} &\quad E^\circ &= +\pu{0.787 V} \tag{3} \end{align} $$

Since resulting $E^\circ = \pu{0.787 V} > 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ < 0,$ and the redox reaction where hydrogen peroxide is oxidized by chlorate(I) can be considered a thermodynamically favorable process.

Alternatively, you can use Latimer diagram, which comes in handy for disproportionation and synproportionation reactions.


  1. Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

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