Ncert Exercises & Solutions

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.

8.4 Fluorine reacts with ice and results in the change:

H2O(s) + F2(g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction.

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

${\overset{+ 1}{H}}_{2} \overset{- 2}{O} + \overset{0}{F_{2}} \to \overset{+ 1}{H} \overset{- 1}{F} + \overset{+ 1}{H} \overset{- 2}{O} \overset{+ 1}{F}$

Here, we have observed that the oxidation number of F increases from 0 in $F_{2}$ to +1 in HOF. Also, the oxidation number decreases from 0 in $F_{2}$ to –1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

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