Justify that the following reactions are redox reactions: $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g) $ $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) $ $4BCl_3(g) + 3LiAlH_4(s) \rightarrow 2B_2H_6(g) + 3LiCl(s) + 3 AlCl_3 (s) $ $2K(s) + F_2(g) \rightarrow 2K+F^- (s) $ $4 NH_3(g) + 5 O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$

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(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. (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. (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. (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. (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.

Solution and Explanation

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Concepts Used:

Redox Reactions

Redox Reaction:

Types of Redox Reactions:

Combination Reaction:

Decomposition Reaction:

Displacement Reaction:

Disproportionation Reaction: