In the reaction, 2 ~S ₂ O ₃²⁻+ I ₂ S ₄ O ₆²⁻+2 I ⁻

Step-by-step solution

In the given reaction 2 ~S ₂ O ₃²⁻ + I ₂ S ₄ O ₆²⁻ + 2 I ⁻ , we analyze the oxidation states of the elements involved. The oxidation state of iodine in I ₂ is 0 , and in I ⁻ it is -1 . Since the oxidation state decreases from 0 to -1 , I ₂ undergoes reduction and acts as an oxidizing agent. For the thiosulfate ion S ₂ O ₃²⁻ , the average oxidation state of sulfur is calculated as 2x + 3(-2) = -2 , which gives 2x = 4 , so x = +2 . In the tetrathionate ion S ₄ O ₆²⁻ , the average oxidation state of sulfur is 4x + 6(-