Answer:
Cu is oxidized and H₂ is reduced.
Step-by-step explanation:
Given chemical equation,
![$Cu + H_(2)O \to CuO + H_(2)$](https://img.qammunity.org/2020/formulas/chemistry/middle-school/cca1024p7rpsiev8dlmtjzgrtamu1j6qhd.png)
On observing the equation, if we write the oxidation states of all elements in the compounds, we will get
![$Cu^(0) + H_(2)^(+1)O^(-2) \to Cu^(+2)O^(-2) + H_(2)^(0)$](https://img.qammunity.org/2020/formulas/chemistry/middle-school/z1wrp0pipz3ztuxgghunc7xjyxkhwev63m.png)
In this equation,
Oxidation number of 'Cu' is increased from 0 to 2. Hence it is oxidized. While the oxidation number of 'H₂' is decreased from +1 to 0. Hence it is reduced.