Solution:

$\mathrm{CuO}_{(s)}+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(g)}$
Oxidation no. of $\mathrm{Cu}$ and $\mathrm{O}$ in $C u O$ is $+2$ and $-2$ individually.
Oxidation no. of $H_{2}$ is 0 .
Oxidation no. of $\mathrm{Cu}$ is $\mathrm{O}$.
Oxidation no. of $\mathrm{H}$ and $\mathrm{O}$ in $\mathrm{H}_{2} \mathrm{O}$ is $+1$ and $-2$ individually.
The oxidation no. of $\mathrm{Cu}$ diminished from $+2$ in $\mathrm{CuO}$ to 0 in $\mathrm{Cu}$. That is $C u O$ is decreased to $\mathrm{Cu}$.
The oxidation no. of $\mathrm{H}$ expanded from 0 to $+1$ in $H_{2}$. That is $H_{2}$ is oxidized to $\mathrm{H}_{2} \mathrm{O}$.
Thusly, the response is redox response.