Why does oxygen in OH have an oxidation state of 2-? I am watching videos on Khan Academy in order to go over redox reactions (a really weak area for me.) In Sal's video, "Oxidation state trends in periodic table" at 7:10, he says:


  If you had to pretend this was an ionic bond, then maybe this hydrogen
  would fully lose an electron, so it would get an oxidation state of
  plus 1. It would be oxidized by the oxygen.
  
  And the oxygen actually has fully gained one electron. And if we're
  forced to think about this is an ionic bond, we'd say it fully gains
  two electrons. So we'd have an oxidation state of negative 2.


I'm having a hard time understanding how he is able to deduce that oxygen has an oxidation state of 2- given the above reasoning. He does add that the states add up to the overall charge which is helpful, but I think I need a little more clarity than that. My reasoning for oxygen being 2- is that "it usually is because my textbook says it usually is," which is probably an even worse explanation.

Could anyone clarify this for me?
First step: draw the Lewis structure of $\ce{OH-}$ with all lone pairs. You should be getting three lone pairs on oxygen and one $\ce{O-H}$ bond.

Second step: ‘[cleave] [all bonds] heterolytically in such a way that the electronegative partner gets both electrons.’[1] You should now have eight electrons around the oxygen and zero on hydrogen.

Third step: count the electrons on both (or all) atoms and check what the difference is to their elemental valence electron count. Oxygen formally has eight electrons on it which is two more than the six it would in its elemental state. The two extraneous electrons give an oxidation state of $\mathrm{-II}$. Likewise, hydrogen has lost its electron for an oxidation state of $\mathrm{+I}$.

Control step: add up the oxidation states of all atoms. If you did everything correctly, they will add up to the entire charge of the ion. Here, this works out:

$$(-2) + (+1) = -1$$



[1]: https://chemistry.stackexchange.com/a/66242