Question

Justify that the following reactions are redox reactions: 

(a) CuO_(s) + H2_(g) → Cu_(s) + H₂O_(g) 

(b) Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g) 

(c) 4BCl_3(g) + 3LiAlH_4(s) → 2B2H_6(g) + 3LiCl_(s) + 3 AlCl_{3 (s}) 

(d) 2K_(s) + F_2(g) → 2K+F– (s) 

(e) 4 NH_3(g) + 5 O_2(g) → 4NO_(g) + 6H₂O(g) 

Answer 1

(a) 

Let us write the oxidation number of each element involved in the given reaction as: 

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 H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O. Hence, this reaction is a redox reaction. 

(b) 

Let us write the oxidation number of each element in the given reaction as: 

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂. Hence, the given reaction is a redox reaction. 

(c) 

The oxidation number of each element in the given reaction can be represented as: 

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to –3 in B₂H₆. i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from –1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction. 

(d) 

The oxidation number of each element in the given reaction can be represented as: 

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 F₂ to – 1 in KF i.e., F₂ is reduced to KF. 

Hence, the above reaction is a redox reaction. 

(e) 

The oxidation number of each element in the given reaction can be represented as: 

Here, the oxidation number of N increases from –3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to –2 in NO and H₂O i.e., O₂ is reduced. Hence, the given reaction is a redox reaction.