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If we compare the standard reduction potentials of $\ce{H_2O_2}$ with that of $\ce{Cl_2}$:

$\ce{H_2O_2 + 2H^+ + 2e^- <=> 2H_2O : E^o = 1.78V}$

$\ce{Cl_2 + 2e^- <=> 2Cl^- : E^o = 1.36V}$

(Source : Wikipedia)

From the above data, as value of $\ce{E^o}$ of $\ce{H_2O_2}$ is greater, its oxidizing power is greater than that of $\ce{Cl_2}$.

But if $\ce{H_2O_2}$ is a better oxidizing agent, why does it act as a reducing agent when reacted with $\ce{Cl_2}$ (in the following reaction)?

$\ce{H_2O_2 + Cl_2 -> 2HCl + O_2}$

The above reaction is feasible, but my doubt is why does it preferentially occur even though $\ce{H_2O_2}$ is the better oxidizing agent.

Could anyone please explain this?

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  • $\begingroup$ The first half reaction is in acidic medium. I don't think it will take place and after all these are standard state values. $\endgroup$ Jun 19 at 7:10
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    $\begingroup$ Because H2O2 is a reducing agent as well, so what? $\endgroup$ Jun 19 at 7:21
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    $\begingroup$ @IvanNeretin So then, how can we determine that $\ce{H_2O_2}$ is going to act as a reducing agent here, instead of an oxidizing agent? $\endgroup$
    – Pal
    Jun 19 at 7:48
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    $\begingroup$ @Pal A point to note that( though I doubt whether it is useful here) just because a reaction is thermodynamically feasible does not necessarily mean, it will be observed necessarily. $\endgroup$
    – Rishi
    Jun 19 at 7:52
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    $\begingroup$ It is a problem of activation energy, not of thermodynamics.. $\endgroup$
    – Maurice
    Jun 19 at 8:21