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 [**S**uper**o**xide **r**eductase](https://en.wikipedia.org/wiki/Superoxide_reductase) (SOR) that only catalyze the reaction forming $\ce{H2O2}$ and another class of enzymes [**S**uper**o**xide **d**ismutase](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Disproportionation).


See also: [A special case of disproportionation (or 'dismutation') is 'radical disproportionation'](http://goldbook.iupac.org/D01799.html)


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