Question

Justify that the following reaction is redox reaction: $CuO(s)+H_{2}(g) \longrightarrow Cu(s)+H_{2}O(g)$

Answer 1

Step 1: Write the reaction along with oxidation number of each element that is present in the reactants as well as the products of the reaction.

$\begin{matrix} \quad\quad+2\quad\quad -2\quad \quad \quad 0\quad \quad 0\quad \quad +1\quad\quad -2 \\ CuO(s)+{ H }_{ 2 }(g)\longrightarrow Cu(s)+{ H }_{ 2 }O(g) \end{matrix}$

Step 2: Notice that the oxidation number of copper (Cu) is +2 in the reactants but it is 0 in the products. So, there is a decrease in the oxidation number of copper during the chemical reaction. Hence it is evident that CuO is reduced to Cu.

Step 3: Notice that the oxidation number of hydrogen (H) is 0 in the reactants but it is +1 in the products. So, there is an increase in the oxidation number of hydrogen during the chemical reaction. Hence it is evident that ${ H }_{ 2 }$ is oxidized to ${ H }_{ 2 }O$.

Thus, it is proved that the given reaction is a redox reaction.