Q.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.
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Start your 14-day free trial to unlock the full solution →Oxygen is oxidised (from in H₂O to in HOF) while fluorine is reduced (from in F₂ to , in both HF and HOF). Simultaneous oxidation and reduction confirm this is a redox reaction.
A redox reaction is defined by the transfer of electrons between species, which we track through changes in oxidation states. When oxidation states change, electrons have moved. The challenge here is recognizing that oxygen, typically a stable oxidation-state element in water, is forced into unusual behaviour by fluorine, the most electronegative element in the periodic table.
Fluorine's extreme electronegativity means it can oxidise almost anything — even oxygen itself. This creates a rare situation where oxygen doesn't maintain its usual state in all products.
Step-by-step oxidation state analysis
-
Assign oxidation states in the reactants
In :
- Hydrogen: (as always in compounds with non-metals)
- Oxygen: (its standard state in most compounds)
In :
- Fluorine: (elemental form)
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Assign oxidation states in the products
In :
- Fluorine: (fluorine always takes in compounds)
- Hydrogen:
In (hypofluorous acid):
This is the critical molecule. Fluorine being more electronegative than oxygen forces an unusual assignment:
- Fluorine: (most electronegative, always )
- Hydrogen:
- Oxygen: must be to balance (since )
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Identify the changes
Element Initial state Final state(s) Change Oxygen in in Oxidation (loses 2e⁻) Fluorine in in both products Reduction (gains 1e⁻ per atom) -
Verify electron balance …
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