The reaction of iodine with sodium hydroxide is a disproportionation reaction, meaning the iodine is both oxidized and reduced. Here's the balanced equation:

3 I₂ (s) + 6 NaOH (aq) → 5 NaI (aq) + NaIO₃ (aq) + 3 H₂O (l)

Explanation:

* Reactants: Iodine (I₂) is a solid, and sodium hydroxide (NaOH) is an aqueous solution.

* Products:

* Sodium iodide (NaI) is formed as a result of the reduction of iodine.

* Sodium iodate (NaIO₃) is formed as a result of the oxidation of iodine.

* Water (H₂O) is also produced.

Here's how the reaction works:

1. Disproportionation: Iodine (I₂) is both oxidized and reduced.

* Oxidation: Some iodine atoms lose electrons and become iodate ions (IO₃⁻).

* Reduction: Other iodine atoms gain electrons and become iodide ions (I⁻).

2. Reaction with NaOH: The iodide ions react with sodium hydroxide to form sodium iodide (NaI) and water. The iodate ions also react with sodium hydroxide to form sodium iodate (NaIO₃) and water.

Important Notes:

* The reaction is usually carried out in a hot solution to speed up the process.

* The reaction produces a yellow solution due to the formation of triiodide ions (I₃⁻).

* This reaction is often used to prepare sodium iodate, which is used in various applications, including as a reagent in analytical chemistry.