Ncert Exercises & Solutions

8.3 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)

(d) 2K(s) + F2(g) → 2K+F– (s)

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

(a) $_{2}$

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 $C u O$ to 0 in $C u$ i.e., $C u O$ is reduced to $C u$ . A/so, the oxidation number of $H$ increases from 0 in $_{2}$ to +1 in $_{2}$ i.e., $_{2}$ is oxidized to $_{2}$ O. 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 $_{2}$ to 0 in $F e$ i.e., $_{2}$ is reduced to $F e$ . On the other hand, the oxidation number of $C$ increases from +2 in $C O$ to +4 in $_{2}$ i.e., $C O$ is oxidized to $_{2}$ . Hence, the given raaction 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 BCl $_{3}$ to –3 in B $_{2}$ H $_{6}$ . i.e., BCl $_{3}$ is reduced to B $_{2}$ H $_{6}$ . Also, the oxidation number of H increases from –1 in LiAlH $_{4}$ to +1 in B $_{2}$ H $_{6}$ i.e., LiAlH4 is oxidized to B $_{2}$ H $_{6}$ . 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 F $_{2}$ to –1 in KF i.e., F $_{2}$ 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 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 $_{2}$ O i.e., O $_{2}$ is reduced. Hence, the given reaction is a redox reaction.

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