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Exercises · 7.4

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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Oxygen is oxidised (from −2-2 in H₂O to 00 in HOF) while fluorine is reduced (from 00 in F₂ to −1-1, 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 −2-2 state in all products.

Step-by-step oxidation state analysis

  1. Assign oxidation states in the reactants

    In H2O(s)\text{H}_2\text{O}(s):

    • Hydrogen: +1+1 (as always in compounds with non-metals)
    • Oxygen: −2-2 (its standard state in most compounds)

    In F2(g)\text{F}_2(g):

    • Fluorine: 00 (elemental form)
  2. Assign oxidation states in the products

    In HF(g)\text{HF}(g):

    • Fluorine: −1-1 (fluorine always takes −1-1 in compounds)
    • Hydrogen: +1+1

    In HOF(g)\text{HOF}(g) (hypofluorous acid):

    This is the critical molecule. Fluorine being more electronegative than oxygen forces an unusual assignment:

    • Fluorine: −1-1 (most electronegative, always −1-1)
    • Hydrogen: +1+1
    • Oxygen: must be 00 to balance (since +1+0+(−1)=0+1 + 0 + (-1) = 0)
  3. Identify the changes

    ElementInitial stateFinal state(s)Change
    Oxygen−2-2 in H2O\text{H}_2\text{O}00 in HOF\text{HOF}Oxidation (loses 2e⁻)
    Fluorine00 in F2\text{F}_2−1-1 in both productsReduction (gains 1e⁻ per atom)
  4. Verify electron balance …

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