Final answer:
The stoichiometry of the reaction between iodine and thiosulfate involves each mole of triiodide (I3-) reacting with two moles of thiosulfate (S2O32-), based on the principle of electron conservation in redox reactions.
Step-by-step explanation:
The stoichiometry for the ionic reaction between iodine and thiosulfate involves the reduction of triiodide (I3−) to iodide ions (I−) and the oxidation of thiosulfate ions (S2O32−) to tetrathionate ions (S4O62−). Reduction of triiodide to iodide requires two electrons per iodine molecule as the oxidation state changes from -½ to −1. Simultaneously, each sulfur in thiosulfate is oxidized, changing the oxidation state from +2 to +2.5 and releasing one electron per molecule of S2O32−. Based on the conservation of electrons, the stoichiometry demands that each mole of triiodide (I3−) react with two moles of thiosulfate (S2O32−).
When iodide ions are oxidized by peroxydisulfate ion (S2O82−), a slow reaction that is catalyzed by iron(II) or iron(III) ions, iodine is formed. The iodine then reacts with iodide to produce triiodide, which can be titrated with thiosulfate. The precise stoichiometry required for titrations can be deduced from balanced reaction equations and the principle of electron conservation in redox reactions.