Consider the following reaction equation: $Br_{2} + 2KI \to 2KBr + I_{2}$. Bromine is acting as

Analyze the question: We need to decide what role Br₂ plays in the reaction Br₂ + 2KI → 2KBr + I₂.

Concept:

• An oxidizing agent gains electrons (it is reduced), and its oxidation number decreases.
• A reducing agent loses electrons (it is oxidized), and its oxidation number increases.

Solution:

1. Assign oxidation numbers: • Br₂ is an element, so its oxidation number is 0. • In KI, iodine is −1. • In KBr, bromine is −1. • I₂ is an element, so its oxidation number is 0.

2. Write the half-reactions: • Bromine: Br₂ + 2e⁻ → 2Br⁻ (0 → −1, reduction, gains electrons) • Iodine: 2I⁻ → I₂ + 2e⁻ (−1 → 0, oxidation, loses electrons)

3. Bromine is reduced, so it takes electrons from the iodide ions. That makes it the oxidizing agent. The iodide ion is the reducing agent.

This also fits the halogen reactivity trend: bromine is more reactive than iodine, so it displaces iodine from its salt.

Options C and D (acid and base) do not apply, because this is a redox reaction, not a proton-transfer reaction.

**Answer: A. an oxidizing agent.

Exam tip: Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The substance that is reduced is the oxidizing agent. In halogen displacement reactions, the more reactive halogen is always the oxidizing agent.