Question

Justify that the following reactions are redox reactions:
$CuO\left(s\right)+H_2\left(g\right)\to Cu\left(s\right)+H_{2}O\left(g\right)$

$F{e}_2{O}_3\left(s\right)+3CO\left(g\right)\to 2Fe\left(s\right)+3C{O}_2\left(g\right)$

$4BC{l}_3\left(g\right)+3LiAl{H}_4\left(s\right)\to 2B_2{H}_6\left(g\right)+3LiCl\left(s\right)+3AlC{l}_3\left(s\right)$

$2K\left(s\right)+F_2\left(g\right)\to 2K+F–\left(s\right)$

$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$

Answer 1

$CuO\left(s\right)+H_2\left(g\right)\to Cu\left(s\right)+H_{2}O\left(g\right)$
Let us write the oxidation number of each element involved in the given reaction as:
$\stackrel{+2}{Cu}{\stackrel{-2}{O}}_{\left(s\right)}+\stackrel{0}{H_{2\left(g\right)}}⟶{\stackrel{0}{Cu}}_{\left(s\right)}+{\stackrel{+1-2}{H_{2}O}}_{\left(g\right)}$
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 $H_2$ to +1 in $H_{2}O$ i.e., $H_2$ is oxidized to $H_{2}O.$ Hence, this reaction is a redox reaction.

$F{e}_2{O}_3\left(s\right)+3CO\left(g\right)\to 2Fe\left(s\right)+3C{O}_2\left(g\right)$
Let us write the oxidation number of each element in the given reaction as:
${\stackrel{+3}{Fe}}_2{\stackrel{-2}{O}}_{3\left(s\right)}⟶2{\stackrel{0}{Fe}}_{\left(x\right)}+3{\stackrel{+4-2}{CO}}_2\left(g\right)$
Here, the oxidation number of Fe decreases from +3 in $F{e}_2{O}_3$ to 0 in Fe i.e., $F{e}_2{O}_3$ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in $C{O}_2$ i.e., CO is oxidized to $C{O}_2.$ Hence, the given reaction is a redox reaction.

$4BC{l}_3\left(g\right)+3LiAl{H}_4\left(s\right)\to 2B_2{H}_6\left(g\right)+3LiCl\left(s\right)+3AlC{l}_3\left(s\right)$
The oxidation number of each element in the given reaction can be represented as:
$4{\stackrel{+3-1}{BCl}}_{3\left(g\right)}+3\stackrel{+1}{L}i\stackrel{+3}{A}l{\stackrel{-1}{H}}_{4\left(s\right)}⟶2\stackrel{-3}{B_2}{\stackrel{+1}{H}}_{6\left(g\right)}+3\stackrel{+1}{L}i{\stackrel{-1}{Cl}}_{\left(s\right)}+3\stackrel{+3}{A}l{\stackrel{-1}{Cl}}_{3\left(s\right)}$
In this reaction, the oxidation number of B decreases from $+3inBC{l}_3$ to –3 in $B_2{H}_6$. i.e.,
$BC{l}_3$ is reduced to $B_2{H}_6$. Also, the oxidation number of H increases from –1 in $LiAl{H}_4$ to +1
in $B_2{H}_6$ i.e., $LiAl{H}_4$ is oxidized to $B_2{H}_6$. Hence, the given reaction is a redox reaction.

$2K_{\left(s\right)}+F_{2_{\left(s\right)}}⟶2K^{+}{F}_{\left(s\right)}^{-}$
The oxidation number of each element in the given reaction can be represented as:
$2{\stackrel{0}{K}}_{\left(s\right)}+{\stackrel{0}{F_2}}_{\left(s\right)}⟶2\stackrel{+1}{K^{+}}{\stackrel{-1}{F^{-}}}_{\left(s\right)}$
In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in $F_2$ to – 1 in KF i.e., $F_2$ is reduced to KF. Hence, the above reaction is a redox reaction.

$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$
The oxidation number of each element in the given reaction can be represented as:
$4\stackrel{-3}{N}{\stackrel{+1}{H}}_{3\left(g\right)}+5{\stackrel{0}{O}}_2\left(g\right)⟶4\stackrel{+2}{N}{\stackrel{-2}{O}}_{\left(g\right)}+6\stackrel{+1}{H_2}{\stackrel{-2}{O}}_{\left(g\right)}$
Here, the oxidation number of N increases from –3 in $N{H}_3$ to + 2 in NO. On the other hand,
the oxidation number of $O_2$ decreases from 0 in $O_2$ to –2 in NO and $H_{2}O$ i.e., $O_2$ is reduced.
Hence, the given reaction is a redox reaction.