Practical investigations

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I am doing an investigation into the Vitamin C concentrations of different types of orange juices. I carried out a direct titration using iodine and starch. I understand the simple equation and ion exchange. However, I don't understand how the iodine reacts with the ascorbic acid molecule, i.e. what bonds it breaks, abd where the electrons are taken from.
 
I also carried out a colorimetry study. I followed a method that used sulphuric acid to convert the iodide ions back to iodine. Why is it necessary for this reaction to happen? Thanks.

Igloo writes ..........
 
I don’t think “ion exchange” is relevant here. Did you mean electron transfer? The reaction between ascorbic acid and iodine takes the form:
 
Ascorbic Acid + I₂ ----> Dehydro-ascorbic acid + 2HI
 
This is a semi “word equation”, of course, but you can obtain the relevant chemical formulae from text books or by searching for the two names at Google, and then you will find that it balances properly. A simple way of seeing that redox has occurred is to realise that the oxidation number of iodine changes from 0 to -1 during this process, which means that iodine molecules have been reduced. Logically, therefore, the ascorbic acid (in losing two hydrogen atoms) must have been oxidised.
 
The half equation for the conversion of iodide ions to iodine molecules is:
 
2I^-(aq) ----> I₂ (aq) + 2e^-
 
This oxidation process can only occur if an oxidising agent is used (which you failed to mention). Perhaps this was a solution containing iodate(V) ions, IO₃-(aq). The appropriate half-equation for the relevant reduction process is:
 
2IO₃^-(aq) + 12H⁺(aq) + 10e^- ----> I₂ (aq) + 6H₂O(l)
 
You can see that the hydrogen ions are being used to dislodge the oxygen atoms from the iodate(V) ions, and that they are an essential part of the reduction process. The hydrogen ions are usually supplied using dilute sulphuric acid – hence its use in your case.
 
The two half equations above can be combined together to give the overall redox reaction:
 
5I^- (aq) + IO₃^-(aq) + 6H⁺(aq) ----> 3I₂ (aq) + 3H₂O(l)
 
If you used another oxidising agent, since you needed sulphuric acid as well, you will find that the inclusion of hydrogen ions is essential for redox to occur. In other words the redox only occurs under acidic conditions.
 
Always carry out a risk assessment and check with your teacher before starting any practical work.

Risk assessment
Before attempting any practical work based on the advice and suggestions on this website, you must do the following. Identify any hazards, assess the risks from these hazards, and then decide appropriate control measures to reduce the risks. You must have these approved by those in authority in your school or college laboratory. Do not rely on what is said on this website. For further guidance see our tutorial on Risk Assessment.

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updated: 17 April 2007