Fluorine reacts with ice and results in the change:
$H_{2} O (s) + F_{2} (g) ⟶ H F (g) + H O F (g)$
i) Justify that this reaction is a redox reaction.
ii) Write the substances oxidized and reduced.
iii) Write the name of the substances which acts as reducing agent and oxidizing agent.

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Solution

$H_{2} O + F_{2} \to H F + H O F$
oxidation number of $F$ in $F_{2} = 0$
oxidation number of $F$ in $H F = - 1$
oxidation number of $F$ in $H O F = + 1$
Here,we see oxidation number of $F e$ decreases from zero to $- 1 (H F)$ and increase from $+ 1 (H O F)$ . This shows that $F_{2}$ is both reduced as well as oxidised. hence, it is a redox reaction and more specifically , it is a disproportionation redox reaction.

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Fluorine reacts with ice and results in the change: H2O(s)+F2(g)⟶HF(g)+HOF(g)i) Justify that this reaction is a redox reaction.ii) Write the substances oxidized and reduced.iii) Write the name of the substances which acts as reducing agent and oxidizing agent.

Fluorine reacts with ice and results in the change:
$H_{2} O (s) + F_{2} (g) ⟶ H F (g) + H O F (g)$
i) Justify that this reaction is a redox reaction.
ii) Write the substances oxidized and reduced.
iii) Write the name of the substances which acts as reducing agent and oxidizing agent.