Fluorine reacts with ice and results in the change: $H_2O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$. Justify that this reaction is a redox reaction.

(N/A) To justify that the reaction $H_2O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$ is a redox reaction,we assign oxidation numbers to each atom:
$1$. In $H_2O_{(s)}$,the oxidation number of $H$ is $+1$ and $O$ is $-2$.
$2$. In $F_{2(g)}$,the oxidation number of $F$ is $0$.
$3$. In $HF_{(g)}$,the oxidation number of $H$ is $+1$ and $F$ is $-1$.
$4$. In $HOF_{(g)}$,the oxidation number of $H$ is $+1$,$O$ is $-2$,and $F$ is $+1$.
Analysis:
- The oxidation number of $F$ changes from $0$ (in $F_2$) to $-1$ (in $HF$),which is a reduction.
- The oxidation number of $F$ also changes from $0$ (in $F_2$) to $+1$ (in $HOF$),which is an oxidation.
Since the same element $(F)$ is simultaneously oxidized and reduced,this is a disproportionation redox reaction.