Question
Justify that the following reactions are redox reactions:
CuO(s) + H2(g) → Cu(s) + H2O(g)
CuO(s) + H2(g) → Cu(s) + H2O(g)
Let us write the oxidation number of each element involved in the given reaction as:
$\stackrel{{+2}\ -2}{\hbox{Cu O}}_{(\text{s})}\ +\stackrel{{0}}{\ \ \ \ \ \ \hbox{H}_{2(\text{g})}}\ \rightarrow\stackrel{0}{\ \ \ \ \hbox{ Cu}_{(\text{s})}}+\stackrel{+1\ \ \ \ -2}{\ \ \ \hbox{H}_2\ \ \text{O}_{(\text{g})}}$ Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.