How does the oxidation state of O change in the following reaction?

L (S) + NaOH (aq) → LiOH (aq) + Na (s)

Oxygen Doesn't change

However, Li is oxidized (0 to + 1), Na is reduced (+1 to 0)

Explanation:

On reactant side, Oxygen has - 2 oxidation charge because we know common oxidation states such as oxygen - 2, hydrogen + 1 etc.

So NaOH, O is - 2, H is + 1, so Na has to be + 1 to equal total charge of compound

In product side, LiOH, again O has to be - 2, H is + 1, so Li + 1 as well ...

We see that oxygen oxidation state doesn't change. However, for Li it becomes oxidized going from 0 to + 1 whereas, Na is reduced going from + 1 to 0.