Prove that the reaction between fluorine and ice is a disproportionation reaction:
$H_{2}O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$
$H_{2}O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$
(N/A) Assigning oxidation states to the atoms in the reaction:
$H_{2}^{+1}O^{-2} + F_{2}^{0} \to H^{+1}F^{-1} + H^{+1}O^{-2}F^{+1}$
In this reaction,the oxidation state of fluorine $(F)$ changes from $0$ in $F_{2}$ to $-1$ in $HF$ (reduction) and to $+1$ in $HOF$ (oxidation).
Since the same element $(F)$ is simultaneously oxidized and reduced,the reaction is a disproportionation reaction.
$H_{2}^{+1}O^{-2} + F_{2}^{0} \to H^{+1}F^{-1} + H^{+1}O^{-2}F^{+1}$
In this reaction,the oxidation state of fluorine $(F)$ changes from $0$ in $F_{2}$ to $-1$ in $HF$ (reduction) and to $+1$ in $HOF$ (oxidation).
Since the same element $(F)$ is simultaneously oxidized and reduced,the reaction is a disproportionation reaction.
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