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.