What are the half-equations of manganese and oxygen in the following redox reaction?

The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}$$
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.




The half-equations should look like this:


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


and


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


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Let us consider these half equations: 
$$\ce{Mn^2+(aq) + 4OH^- \to MnO\_2(aq) + 2e^- + 2H2O}$$


and


$$\ce{ 2H2O(l) + O2(aq) + 4e^- -> 4OH- }$$
By multiplying the first equation by two and adding it to the second one, we get the overall reaction.


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