Consider the redox reaction as represented by the following equation: 12(aq) + 2S2O32-(aq) → 21-(aq) +…

Analyze the question: The equation (with the typos corrected) is I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq). "12" is I₂ and "1-" is I⁻. We need to find the species that is reduced.

Concept: Reduction is gain of electrons, which means the oxidation number decreases. The species that is reduced is the one whose oxidation number goes down.

Solution:

1. Iodine: • In I₂, the oxidation number is 0. • In I⁻, the oxidation number is −1. • It goes from 0 to −1, so I₂ gains electrons and is reduced.

2. Sulfur: • In S₂O₃²⁻: 2(S) + 3(−2) = −2, so S = +2. • In S₄O₆²⁻: 4(S) + 6(−2) = −2, so S = +2.5. • The oxidation number increases, so thiosulfate is oxidized. It is the reducing agent.

Half-equations:

• Reduction: I₂ + 2e⁻ → 2I⁻
• Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻

Checking the options:

• A. S₄O₆²⁻ is a product of oxidation, not reduced.
• B. This is a typo for thiosulfate, S₂O₃²⁻, which is oxidized.
• C. I₂ is reduced. Correct.
• D. I⁻ is the product of the reduction, not the species being reduced.

Tip: Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The reactant whose oxidation number drops is reduced. In iodine/thiosulfate titrations, iodine is always the oxidizing agent, and thiosulfate is the reducing agent.