Iodine Titration with Free SO2 in wine To determine the free $\ce{SO2}$ in wine, a common method used is called the Ripper Titration Method. 

The equation is as follows: $$\ce{SO2 (aq) + I2 (aq) + 2H2O (l) -&gt; 4H+ (aq) + SO4^2- (aq) + 2I- (aq)}$$

This method involves titrating an Iodine solution into a sample of wine. I have found that triiodide is made from reacting $\ce{I-}$ with $\ce{I2}$. However, if it is the triiodide we are titrating, how is the iodine reacting with the wine?

Is it triiodide ($\ce{I3-}$) that is being titrated with the wine sample or iodine ($\ce{I2}$)?
Reaction goes along the lines $\ce{3 I2 -&gt; 2 I3- -&gt; 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.