Loulou74

@chenbeier In that case, can the reduction of iodate to iodine by sulfur dioxide be represented with the following equation (including spectator ions for completion): 2KIO3 + 5SO2 + 4H2O ⟶ K2SO4 + I2 + 4H2SO4 ?

@chenbeier Can iodate ions theoretically produce iodine directly as you said according to the following equation: or should there be potassium iodide present in solution to react with KIO3 and form iodine as follows: KIO3 + 5KI + 3H2O ⟶ 3I2 + 6KOH ?

@chenbeier Thank you very much for pointing out that potassium iodate is used in the method, not KI. I have thought of the following reactions that could be theoretically possible with KIO3: KIO3 + 3SO2 + 4H2O ⟶ KOH + 3H2SO4 + HI KIO3 + 3SO2 + 3H2O ⟶ KHSO4 + 2H2SO4 + HI 2KIO3 + 6SO2 + 6H2O ⟶ K2SO4 + 5H2SO4 + 2HI Do you know which of those three products is more likely to form?

I have worked in a laboratory of a winery where I analyzed the free and total sulfur dioxide in wine with a manual titrator by the Ripper method. This method is usually a simple iodine titration where iodine reacts with sulfur dioxide according to the following equation: SO2+ I2+ 2H2O⟶ H2SO4+ 2HI The reagents used in the method I followed are slightly modified. Instead of titrating the wine sample acidified with sulfuric acid against an iodine solution, I titrated it against a solution of potassium iodide (a starch indicator is used to see the color change and distilled water is added to the sample). I have thought of the following equations to represent this reaction: SO2+ 2H2O+ H2SO4+ 4KI⟶ 2K2SO4+ 4HI SO2+ 2KI+ 2H2O⟶ H2SO4+ 2HI+ 2K+ HSO3−+ KI⟶ KSO3− However, in equation 1 the hydrogen is not balanced, and in equation 2 the charges are not balanced. Equation 3 looks fine at first, but the bisulfite ion would imply that this reaction only occurs when analyzing the free fraction of sulfur dioxide, which is not true. How does potassium iodide in solution react differently to iodine, supporting this explanation with a chemical equation preferably? If you are interested, here you can find the user's manual of the apparatus I used. I contacted the company but their response was not very helpful. They only said that, in their method, the common iodine solution is replaced by potassium iodide solution because as a product it is cheaper and more stable.