# Consider the following reaction: CO_2(g) + H_2(g) -> CO(g) + H_2O(l) What is being oxidized?

Even without going into the valency electrons etc., it is clear that the $C {O}_{2}$ loses one oxygen atom in favour of ${H}_{2} O$.
So the $C {O}_{2}$ is reduced (=oxidator) and the ${H}_{2} O$ is oxidized (=reductor).