There is more redox chemistry in the workup. Excess iodine reacts with thiosulfate to form iodide and dithionate: I2 (aq) + 2 S2O3 2– (aq) → 2 I– (aq) + S4O6 2– (aq) What is the practical advantage of reducing excess iodine to iodide (i. e. how does this make it easier to collect pure product)?
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There is more redox chemistry in the workup. Excess iodine reacts with thiosulfate to form iodide an...
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