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Fluorine reacts with ice and results in the change: HX2O(s)+FX2(g)⟶HF(g)+HOF(g) Justify that this reaction is a redox reaction.

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प्रश्न

Fluorine reacts with ice and results in the change: \[\ce{H2O(s) + F2(g) → HF(g) + HOF(g)}\]

Justify that this reaction is a redox reaction.

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उत्तर

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:
+1 -2         0              +1 -1        +1 -2 +1
\[\ce{H2O(s) + F2(g) → HF(g) + HOF(g)}\]

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to –1 in HF. Thus, in the above reaction, F is both oxidized and reduced.
Hence, the given reaction is a redox reaction.

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अध्याय 7: Redox Reactions - EXERCISES [पृष्ठ २८०]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 11
अध्याय 7 Redox Reactions
EXERCISES | Q 8.4 | पृष्ठ २८०

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