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Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

(d) 2K(s) + F2(g) → 2K+F– (s)

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

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(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...
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(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 LiAlH4to +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.

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