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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)

Respuesta :

Edufirst

Answer:

  • I₂ (aq)    +   2e⁻    →    2I⁻(aq)  

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)

       ===============================

        (the electrons are canceled)

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