Question

Justify that the following reactions are redox reactions:

1. $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g)$
2. $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$
3. $4BCl_3(g) + 3LiAlH_4(s) \rightarrow 2B_2H_6(g) + 3LiCl(s) + 3 AlCl_3 (s)$
4. $2K(s) + F_2(g) \rightarrow 2K+F^- (s)$
5. $4 NH_3(g) + 5 O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$

Answer 1

(a) $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g)$
Let us write the oxidation number of each element involved in the given reaction as:
$\overset{+2}Cu\overset{-2}O(s) + \overset{0}H_2(g) \rightarrow \overset{0}Cu(s) + \overset{+1}H_2\overset{-2}O(g)$

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 $H_2$ to $+1$ in $H_2O$ i.e., $H_2$ is oxidized to $H_2O$.
Hence, this reaction is a redox reaction.

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(b) $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$
Let us write the oxidation number of each element in the given reaction as:
$\overset{+3}Fe_2\overset{-2}O_3(s) + 3\overset{+2}C\overset{-2}O(g) \rightarrow 2\overset{0}Fe(s) + 3\overset{+4}C\overset{-2}O_2(g)$

Here, the oxidation number of $Fe$ decreases from $+3$ in $Fe_2O_3$ to $0$ in $Fe$ i.e., $Fe_2O_3$ is reduced to $Fe$. On the other hand, the oxidation number of $C$ increases from $+2$ in $CO$ to $+4$ in $CO_2$ i.e., $CO$ is oxidized to $CO_2$.
Hence, the given reaction is a redox reaction.

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(c) $4BCl_3(g) + 3LiAlH_4(s) \rightarrow 2B_2H_6(g) + 3LiCl(s) + 3 AlCl_3 (s)$

The oxidation number of each element in the given reaction can be represented as:
$4\overset{+3}B\overset{-1}Cl_3(g) + 3\overset{+1}Li\overset{+3}Al\overset{-1}H_4(s) \rightarrow 2\overset{-3}B_2\overset{+1}H_6(g) + 3\overset{+1}Li\overset{-1}Cl(s) + 3 \overset{+3}Al\overset{-1}Cl_3 (s)$

In this reaction, the oxidation number of $B$ decreases from $+3$ in $BCl_3$ to $-3$ in $B_2H_6$. i.e., $BCl_3$ is reduced to $B_2H_6$. Also, the oxidation number of $H$ increases from $-1$ in $LiAlH_4$ to $+1$ in $B_2H_6$ i.e., $LiAlH_4$ is oxidized to $B_2H_6$.
Hence, the given reaction is a redox reaction.

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(d) $2K(s) + F_2(g) \rightarrow 2K+F^- (s)$
The oxidation number of each element in the given reaction can be represented as:
$2\overset{0}K(s) + \overset{0}F_2(g) \rightarrow 2\overset{+1}K+\overset{-1}F^- (s)$

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.

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(e) $4 NH_3(g) + 5 O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$
The oxidation number of each element in the given reaction can be represented as:
$4 \overset{-3}N\overset{+1}H_3(g) + 5 \overset{0}O_2(g) \rightarrow 4\overset{+2}N\overset{-2}O(g) + 6\overset{+1}H_2\overset{-2}O(g)$

Here, the oxidation number of $N$ increases from $-3$ in $NH_3$ 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 $H_2O$ i.e., $O_2$ is reduced.
Hence, the given reaction is a redox reaction.