In the iron oxidation that results in rust, the first step is to steal 2 electrons from $\ce{Fe}$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $\ce{OH-}$ according to the reaction $\ce{O2 + 2H2O + 4e- -> 4OH-}$.
What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $\ce{4 OH-}$ comes out of the collision. I imagine that there must be some steps between the reactants and $\ce{4OH-}$. Is it that, first, $\ce{O2}$ steals the two electrons from the $\ce{Fe}$ lattice becoming $\ce{2O-}$ and then, later, each $\ce{O-}$ takes an $\ce{H}$ from $\ce{H2O}$ to form $\ce{2OH-}$? That does not seem right to me, because this means that what is happening is:
$$\ce{O2 + 4e- -> 2O-}$$
$$\ce{2O- + 2H2O -> 4OH-}$$
But the reaction "$\ce{O2 + 4e- -> 2O-}$" implies that $\ce{2O-}$ is more stable than $\ce{O2}$. Is that right?