Question

For the reaction of thiosulfate anion with iodine, the thiosulfate anion acts as the reducing agent according to the oxidation half-reaction,

2S2O32-(aq) -> S4O62-(aq) + 2e-.

Which of the following reducing half-reactions is correct to give the overall reaction of thiosulfate anion with iodine?

a. I2(aq) + 2e- -> 2I-(aq)

b. I2(aq) + e- -> 2I-(aq)

c. I2(aq) + 2e- -> 2IO3-(aq)

Answer 1

• I₂ (aq) + 2e⁻ → 2I⁻(aq)

Step-by-step explanation:

1. Oxidation half-reaction (given)

• 2S₂O₃²⁻ (aq) → S₄O₆²⁻ + 2e⁻

2. Reduction half-reaction:

Reduction is the gain of electrons with the consequent reduction in the number of oxidation.

The starting reactant is iodine which is diatomic; thus, it is I₂.

Each iodine atom gains one electron; thus, in total 2 electrons are gained and each I atom in solution will become an I⁻ anion.

Half-reaction:

• I₂ (aq) + 2e⁻ → 2I⁻(aq) ↔ answer

The overall reaction of thiosulfate anion with iodine is obtained when you add the two half-reactions:

• 2S₂O₃²⁻ (aq) → S₄O₆²⁻ + 2e⁻

• I₂ (aq) + 2e⁻ → 2I⁻(aq)

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(the electrons are canceled)

• 2S₂O₃²⁻ (aq) + I₂ (aq) → S₄O₆²⁻ + 2I⁻(aq)