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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.
संबंधित प्रश्न
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\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]
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\[\ce{Br2 + 2Cl- -> Cl2 + 2Br-}\]
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\[\ce{2KClO3 → 2KCl +3O2}\]
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