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How is the following reaction a disproportionation reaction?

$$\ce{2 K\overset{-1/2}{O}_2 + 2 H2\overset{-2}{O} ->2 KOH + H2\overset{-1}{O}_2 + \overset{0}{O}_2}$$

In this, the OS are $-1/2$, $-2$, $-1$, $0$, respectively, but in disproportionation the element$ O $both oxidizes and reduces. How do we know that this happens here? $\ce{KOH}$ may get oxidized, and $\ce{H2O}$ get reduced. How do we know the same compound has undergone both?

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    $\begingroup$ Compounds don't get oxidized or reduced; elements do. $\endgroup$
    – Ivan Neretin
    May 21, 2019 at 5:10
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    $\begingroup$ Note that it is adviced against MathJax or other intensive formatting in the title. Rather, plain text titles are preferred, if applicable. E.g."Disproportionation at potassium superoxide hydrolysis" $\endgroup$
    – Poutnik
    May 21, 2019 at 7:08
  • $\begingroup$ @Poutnik I will take a note of that $\endgroup$
    – Mr.HiggsBoson
    May 21, 2019 at 9:45

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The disproportionation happens to elements, not to compounds.

In this particular case, to the part of oxygen,

$$ \overset{\ce{2 KO2}}{ 4 \times -0.5}= \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$

while the other part keeps the oxidation state constant:

$$ \overset{\ce{2 H2O}} {2 \times -2}= \overset{\ce{2 KOH}} {2\times -2}$$

$\ce{H2O}$ does not get directly involved in disproportionation.

$\ce{KO2}$ oxidizes and at the same time reduces itself.

This is the nature of isproportionation, which happens , if the particular compound of the element in the intermediate oxidation state has higher Gibbs energy then particular compounds of this element in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.

Typical cases are $\ce{2 Cu^{I}->Cu^{II} + Cu^{0}}$ and $\ce{2 Hg^{I}->Hg^{II} + Hg^{0}}$

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  • $\begingroup$ Thanks , but I got confuse with if $\ce{K\overset{-1/2}{O}_2}$ is getting oxidised to $\ce{O2}$ and $\ce{H2\overset{-2}{O}}$ to $\ce{H2O2}.$ $\endgroup$
    – Mr.HiggsBoson
    May 21, 2019 at 9:51
  • $\begingroup$ No, it does not. H2O does not get directly involved in disproportionation. KO2 oxidizes and at the same time reduce itself. This is the nature of isproportionation, which happens , if the compounds with the element in intermediate oxidation state have higher Gibbs energy then compounds in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred. $\endgroup$
    – Poutnik
    May 21, 2019 at 10:40

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