What are the half-equations of manganese and oxygen in the following redox reaction? The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -&gt; 2MnO2(s) + 2H2O(l)}$$
I know that the Manganese (II) gets oxidized to Manganese (IV), but I'm not sure about the other half. I know oxygen gets reduced, but using simply $\ce{O2 +2e^--&gt;2O-}$ doesn't work because the equations won't balance out then.

By the way, this chemical equaltion is used in the Winkler titration.
The half-equations should look like this:

$$\ce{2Mn^2+(aq) \to 2Mn^4+(aq) + 4e^-}$$

and

$$\ce{4OH^-(aq) + O2(aq) + 4e^- -&gt; 4O- + 2H2O(l)}$$