Question

Fluorine reacts with ice and results in the change:
$H_2O(s) + F_2{(g)} \rightarrow HF{(g)} + HOF{(g)}$ Justify that this reaction is a redox reaction.

Answer 1

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:
$\overset{+1}{H_2}\overset{-2}{O}+\overset{0}{F_2}\rightarrow \overset{+1}{H}\overset{-1}{F}+\overset{+1}{H}\overset{-2}{O}\overset{+1}{F}$
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.