Start by determining the oxidation numbers of each element on both sides of the equation CuO(s) + H2 (g) → Cu(s) + H2O (l)
Let's start with the reactants.
In the compound CuO(s), the oxidation number of O is -2 and the oxidation number of Cu is +2. (Remember, oxygen's oxidation number is -2 unless it's in a peroxide)
In H2(g) the oxidation number of H is 0 because it is in its elemental form.
Now onto the products
The oxidation number of Cu(s) is 0 because it is in its elemental form
In H2O(l) the oxidation number of oxygen is -2, and the oxidation number of H is +1.
Now let's summarize the change in the oxidation numbers of each element.
Cu: started as +2 and became 0
O: started as -2 and became -2 (didn't change)
H: started as 0 and became +1
To determine which element was oxidized, we look at which element lost electrons, and thus became more positive. So, we can determine that H was oxidized in this reaction, because it started with the oxidation number 0 and became +1.