Answer:
Cu is oxidized and H₂ is reduced.
Explanation:
Given chemical equation,
[tex]$Cu + H_{2}O \to CuO + H_{2}$[/tex]
On observing the equation, if we write the oxidation states of all elements in the compounds, we will get
[tex]$Cu^{0} + H_{2}^{+1}O^{-2} \to Cu^{+2}O^{-2} + H_{2}^{0}$[/tex]
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.
