Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: dgt20 on 26/04/2018 12:03:22
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What is the redox reaction occuring in the ripper titration method (wine SO2 determination), between SO2 and I2 and the use of starch indicator. How does this method determine the amount of SO2 both free and bound? What would the half reactions be and what is oxidised and reduced.
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Both half reactions that the combined reaction look good to me (though I would show the negative charge on the electrons)
The first half reaction shown is a reduction because the electrons show up on the reactants side or the arrow, and the second half reaction shown is an oxidation because the electrons are on the products side of the arrow.
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For the titration to reveal the amount of SO2, you would have to know how much iodine you are adding. Because the SO2 and I2 react in a 1:1 molar ratio, then you just have to keep track of how much I2 is going into the solution. As the color disappears, you know that the I2 is reacting, removing equivalent amounts of SO2. As soon as the color remains, you know you have run out of SO2. The amount of I2 reacted must be equal to the amount of SO2 that was in there when you started the titration.
(there are more complicated ways of doing multiple titrations, even electrochemical titrations, but I just used the most simple setup for my explanation--don't let me confuse you if you're doing a more complex titration.)