(a) \(\stackrel{+2}{\mathrm{Cu}}\stackrel{-2}{\mathrm{O(s})}+\mathrm{H}_2\) (g) \(\longrightarrow \mathrm{Cu}(\mathrm{s})+\stackrel{+1}{\mathrm{H}_2}\stackrel{-2}{\mathrm{O(g})}\)
Here, \(\mathrm{O}\) is removed from \(\mathrm{CuO}\), therefore, it is reduced to \(\mathrm{Cu}\) while \(\mathrm{O}\) is added to \(\mathrm{H}_2\) to form \(\mathrm{H}_2 \mathrm{O}\), therefore, it is oxidized. Further, O. N. of \(\mathrm{Cu}\) decreases from + 2 in \(\mathrm{CuO}\) to 0 in \(\mathrm{Cu}\) but that of \(\mathrm{H}\) increases from 0 in \(\mathrm{H}_2\) to \(+1\) in \(\mathrm{H}_2 \mathrm{O}\). Therefore, \(\mathrm{CuO}\) is reduced to \(\mathrm{Cu}\) but \(\mathrm{H}_2\) is oxidized to \(\mathrm{H}_2 \mathrm{O}\). Thus, this is a redox reaction.
(b) \(\stackrel{+3}{\mathrm{Fe}_2}\stackrel{-2}{ \mathrm{O}_3(\mathrm{s})}+3 \stackrel{+2}{\mathrm{C}} \stackrel{-2}{\mathrm{O}(\mathrm{g})} \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \stackrel{+4}{\mathrm{C}}\stackrel{-2}{\mathrm{O}_2(\mathrm{g})}\)
Here O.N. of Fe decreases from \(+3\) is \(\mathrm{Fe}_2 \mathrm{O}_3\) to 0 in Fe while that of \(\mathrm{C}\) increases from \(+2\) in \(\mathrm{CO}\) to \(+4\) in \(\mathrm{CO}_2\). Further, oxygen is removed from \(\mathrm{Fe}_2 \mathrm{O}_3\) and added to \(\mathrm{CO}\), therefore. \(\mathrm{Fe}_2 \mathrm{O}_3\) is reduced while \(\mathrm{CO}\) is oxidized. Thus, this is a redox reaction.
(c) \(4 \stackrel{+3}{\mathrm{~B}}\stackrel{-1}{\mathrm{Cl}_3}(\mathrm{~g})+2 \stackrel{+1}{\mathrm{Li}} \mathrm{\stackrel{+3}{Al}}\stackrel{-1}{\mathrm{H}_4}(\mathrm{s}) \longrightarrow 2 \stackrel{-3}{\mathrm{B}_2} \stackrel{+1}{\mathrm{H}_6}(\mathrm{~g})+3 \mathrm{\stackrel{+1}{Li}\stackrel{-1}{Cl}}(\mathrm{s})+3 \stackrel{+3}{\mathrm{Al}}\stackrel{-1}{\mathrm{Cl}_3}(\mathrm{s})\)
Here, \(\mathrm{O}, \mathrm{N}\). of \(\mathrm{B}\) decreases from \(+3\) in \(\mathrm{BCl}_3\) to \(-3\) in \(\mathrm{B}_2 \mathrm{H}_6\) while that of \(\mathrm{H}\) increases from \(-1\) in \(\mathrm{LiAlH}_4\) to \(+1\) in \(\mathrm{B}_2 \mathrm{H}_6\). Therefore, \(\mathrm{BCl}_3\) is reduced while \(\mathrm{LiAlH}_4\) is oxidized. Further, \(\mathrm{H}\) is added to \(\mathrm{BCl}_3\) but is removed from \(\mathrm{LiAlH}_4\), therefore, \(\mathrm{BCl}_3\) is reduced while \(\mathrm{LiAlH}_4\) is oxidized. Thus, it is a redox reaction.
(d) \(2 \mathrm{~K}\) (s) \(+\mathrm{F}_2\) (g) \(\longrightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}\)(s)
Here, each \(\mathrm{K}\) atom has lost one electron to form \(\mathrm{K}^{+}\) while \(\mathrm{F}_2\) has gained two electrons to form two \(\mathrm{F}^{-1}\) ions. Therefore, \(\mathrm{K}\) is oxidized while \(\mathrm{F}_2\) is reduced. Thus, it is a redox reaction.
(e) \(4 \mathrm{\stackrel{-3}{N}\stackrel{+1}{H}}_3(\mathrm{~g})+5 \stackrel{0}{\mathrm{O}}_2(\mathrm{~g}) \longrightarrow \mathrm{\stackrel{+2}{4N}}\stackrel{-2}{\mathrm{O}(\mathrm{g})}+6 \stackrel{+1}{\mathrm{H}_2} \stackrel{-2}{\mathrm{O}(\mathrm{g})}\)
Here, O. N. of \(\mathrm{N}\) increases from \(-3\) in \(\mathrm{NH}_3\) to \(+2\) in \(\mathrm{NO}\) while that \(\mathrm{O}\) decreases from 0 in \(\mathrm{O}_2\) to \(-2\) in \(\mathrm{NO}\) or \(\mathrm{H}_2 \mathrm{O}\). Therefore, \(\mathrm{NH}_3\) is oxidized while \(\mathrm{O}_2\) is reduced. Further \(\mathrm{H}\) has been removed from \(\mathrm{NH}_3\) but added to \(\mathrm{O}_2\). Therefore, \(\mathrm{NH}_3\) has been oxidized while \(\mathrm{O}_2\) is reduced. Thus, this is a redox reaction.