Justify that the following reactions are redox reactions:
$(a) \ CuO_{(s)} + H_{2(g)} \to Cu_{(s)} + H_2O_{(g)}$
$(b) \ Fe_2O_{3(s)} + 3CO_{(g)} \to 2Fe_{(s)} + 3CO_{2(g)}$
$(c) \ 4BCl_{3(g)} + 3LiAlH_{4(s)} \to 2B_2H_{6(g)} + 3LiCl_{(s)} + 3AlCl_{3(s)}$
$(d) \ 2K_{(s)} + F_{2(g)} \to 2K^+F^-_{(s)}$
$(e) \ 4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_2O_{(g)}$

(A) $(a) \ CuO_{(s)} + H_{2(g)} \to Cu_{(s)} + H_2O_{(g)}$
Oxidation number of $Cu$ decreases from $+2$ to $0$ (reduction) and $H$ increases from $0$ to $+1$ (oxidation). Thus,it is a redox reaction.
$(b) \ Fe_2O_{3(s)} + 3CO_{(g)} \to 2Fe_{(s)} + 3CO_{2(g)}$
Oxidation number of $Fe$ decreases from $+3$ to $0$ (reduction) and $C$ increases from $+2$ to $+4$ (oxidation). Thus,it is a redox reaction.
$(c) \ 4BCl_{3(g)} + 3LiAlH_{4(s)} \to 2B_2H_{6(g)} + 3LiCl_{(s)} + 3AlCl_{3(s)}$
Oxidation number of $B$ decreases from $+3$ to $-3$ (reduction) and $H$ increases from $-1$ to $+1$ (oxidation). Thus,it is a redox reaction.
$(d) \ 2K_{(s)} + F_{2(g)} \to 2K^+F^-_{(s)}$
Oxidation number of $K$ increases from $0$ to $+1$ (oxidation) and $F$ decreases from $0$ to $-1$ (reduction). Thus,it is a redox reaction.
$(e) \ 4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_2O_{(g)}$
Oxidation number of $N$ increases from $-3$ to $+2$ (oxidation) and $O$ decreases from $0$ to $-2$ (reduction). Thus,it is a redox reaction.