The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}$$ 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 $\ce{O2 +2e^-->2O-}$ doesn't work because the equations won't balance out then.
By the way, this chemical equaltion is used in the Winkler titration.