Question

Identify the oxidized and reduced compound in the given reaction $\underset{\mathrm{Copper}\mathrm{oxide}}{\mathrm{CuO}\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Answer 1

Oxidized and reduced substances:

1. Oxidized compounds are those whose oxidation number increases, or they gain oxygen in a process, or they lose electrons or electronegative species combine with them in a reaction.
2. Reduced substances are those whose oxidation number lowers, meaning they lose oxygen or gain electrons, or electropositive species join with them in a reaction.
3. The oxidation number of copper changes from $+2$ to$0$. Also, the oxygen atom is removed from copper oxide. Therefore copper oxide is reduced here. $\stackrel{+2}{\underset{\mathrm{Copper}\mathrm{oxide}}{\mathrm{CuO}\left(\mathrm{s}\right)}}\to \underset{\mathrm{Copper}}{\stackrel{0}{\mathrm{Cu}\left(\mathrm{s}\right)}}$
4. Hydrogen gas is oxidized here as an oxygen atom is added to it. Also, the oxidation number of hydrogen changes from $0$ to$+1$.

Therefore, in the given reaction oxidized substance is hydrogen and the reduced substance is copper oxide.