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.

Writing the O.N. of each atom above its symbol, we have,          

here, the O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Therefore, F₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.

<p>Writing the O.N. of each atom above its symbol, we have, &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1741860595phpSvrLGc_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1741860595phpSvrLGc.jpeg" alt="" width="209" height="30"></picture></div></div><p>here, the O.N. of F decreases from 0 in F₂&nbsp;to -1 in HF and increases from 0 in F₂&nbsp;to +1 in HOF. Therefore, F₂&nbsp;is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.</p>

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