# How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide?

In the following reaction: $$\ce{2O2- (aq) + 2H+ (aq) <=> H2O2 (aq) + O2(g)}$$

I see where some oxygen atoms have lost electrons, but I do not see what has gained electrons.
On the reactant side, the superoxide has a $$-1$$ charge per oxygen atom. On the product side, half the oxygen atoms now have a $$0$$ charge, so they lost an electron each (right?). For it to be a true redox reaction, does not a species/atom have to gain electron(s)?

On the reactant side, the superoxide has a $$-1$$ charge per oxygen atom.

No, the superoxide has a $$-\frac{1}{2}$$ charge per oxygen atom.

The reaction is better thought of as two separate reactions:

\begin{align} \ce{O2- &<=> O2 + e-}\tag1\\ \ce{O2- + 2H+ + e- &<=> H2O2}\tag2 \end{align}

This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $$\ce{H2O2}$$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both.

Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.