Question

Iodine, I2, has many uses, including the production of dyes, antiseptics, photographic film, pharmaceuticals, and medicinal soaps. It forms when chlorine, Cl2, reacts with iodide ions in a sodium iodide solution. Which of the following half-reactions for this oxidation‑reduction reaction describes the oxidation, and which one describes the reduction? Cl2 + 2e‒ → 2Cl‒ 2I‒ → I2 + 2e‒

Answer 1

• 2I ⁻ → I₂ + 2e⁻ describes the oxidation.

• Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.

Step-by-step explanation:

Oxidation-reduction reaction is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.

The oxidation half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.

In this case, the process that shows the loss of electrons is:

• 2I⁻ → I₂ + 2e⁻

That reaction shows:

• Two I⁻ ions lose two electrons (one each) to be oxidized to I₂.

• The change in the oxidation number is from -1 to 0.

• Hence this half-reaction is the oxidation reaction.

On the other hand, the reduction half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.

In this case, the process that shows the gain of electrons is:

• Cl₂ + 2e⁻ → 2Cl⁻

That reaction shows:

• Two Cl atoms gain two electrons (one each) to be reduced to Cl⁻.

• The change in the oxidation number is from 0 to - 1.

• Hence, this half-reaction is the reduction reaction.

Summary:

• 2I ⁻ → I₂ + 2e⁻ describes the oxidation.

• Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.