The balanced reaction equation between sodium thiosulfate (Na₂S₂O₃) and molecular iodine (I₂) is:

2 Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2 NaI

Here's the breakdown:

* Reactants:

* Sodium thiosulfate (Na₂S₂O₃) - a white crystalline solid

* Molecular iodine (I₂) - a purple-black solid

* Products:

* Sodium tetrathionate (Na₂S₄O₆) - a colorless solution

* Sodium iodide (NaI) - a colorless solution

Explanation:

The reaction involves a redox process where iodine (I₂) is reduced and thiosulfate (S₂O₃²⁻) is oxidized.

* Oxidation: Two thiosulfate ions (S₂O₃²⁻) are oxidized to tetrathionate (S₄O₆²⁻) by losing electrons.

* Reduction: One iodine molecule (I₂) is reduced to two iodide ions (I⁻) by gaining electrons.

This reaction is often used in analytical chemistry to determine the concentration of iodine solutions (iodometry). The disappearance of the brown iodine color is used to indicate the endpoint of the reaction.