2H₂O(l) + 2F₂(g) → 4HF(aq) + O₂(g). In the reaction above, the substance that is being reduced is

Correct answer: F₂(g)

Explanation: Track oxidation states:
Reactants: H in H₂O = +1, O in H₂O = −2, F in F₂ = 0.
Products: H in HF = +1, F in HF = −1, O in O₂ = 0.

Fluorine goes from 0 → −1 (gain of electrons) so F₂ is reduced. Oxygen in water goes from −2 → 0 (loss of electrons), so H₂O is oxidized to O₂. Therefore, the species that is reduced is F₂(g) (and it acts as the oxidizing agent).

• O₂(g) — Incorrect; it is formed by oxidation of water.
• H₂O(l) — Incorrect; water is oxidized (O: −2 → 0).
• F₂(g) — Correct; F: 0 → −1 (reduction).
• HF(aq) — Product containing reduced F, but the substance being reduced is F₂.