I am studying biochemistry at university, as part of the course we do several titration experiments.
In our handouts, we were given the following half equations:
IO3- + 6H+ + 4e- I+ + 3H2O
C6H8O6 C6H6O6 + 2H+ + 2e-
Which can be balanced to give:
2C6H8O6 2C6H6O6 + 4H+ + 4e-
My issue is, in this experiment we observe a pink colour in chloroform. Now I don't see how any of the above species can be soluble in chloroform AND give that colour. But other universities state that I2 is the cause of this colour in the chloroform layer.
So I devise equations to explain this colour:
Combine the next two equations to give formation of I2 reaction:
Iodate ions are reduced to form iodine:
2 * (IO3- + 6H+ -> 1/2I2 + 3H2O)
While iodide ions are oxidised to form Iodine:
5 * (2I- -> I2 + 2e-)
Combining these two and cancelling like species (including electrons) gives:
2IO3- + 10I- + 12H+ -> 6I2 + 6H2O
Dividing this by a factor of 2 gives the following:
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
Then I combined the half equations for oxidation of ascorbic acid by iodine:
Iodine molecules are reduced to form Iodide ions:
I2 + 2e- -> 2I-
While ascorbic acid is oxidised to form dehydroascorbic acid:
C6H8O6 -> C6H6O6 + 2H+ + 2e-
Combining those two gives:
C6H8O6 + I2 -> C6H6O6 + 2H+ + 2I-
Now combine Iodine formation reaction with reduction of ascorbic acid:
3* (C6H8O6 + I2 -> C6H6O6 + 2H+ + 2I-)
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
This gives:
3C6H8O6 + IO3- 3C6H6O6 + I- + 3H2O
The difference between the two sets of equations is that the stoichiometric ratio between IO3- and C6H8O6 is different depending on which one you use. 2:1 for the first (which my university says is correct but doesn't explain how we get the I2 colour) and 3:1 for the second (which my university says is NOT correct but DOES explain how we get the I2 colour) .
What is the actual chemistry going on between iodate and ascorbic acid?
Dear Alex Ryder many thanks for the interesting question. As you now from your experiment, the reaction (oxidation) of ascorbic acid (Vitamin C) with iodate is used for the redox titration of Vitamin C. Any pink or purple coloration of your chloroform phase can only be due to the presence of free elemental iodine (I2). No other component in this reaction system can produce a pink or purple color. The oxidation product of ascorbic acid is dehydroascorbic acid (see attached picture and Wikipedia entry: https://en.wikipedia.org/wiki/Dehydroascorbic_acid). Pure solid dehydroascorbic acid forms a beige sold, but in water it immediately forms a colorless hydrate. Thus it cannot be responsible for any pink or purple colors.
For more detailed information about the reaction mechanism please have a look at this nice experiment description from La Salle University:
Redox Titration of Vitamin C
http://www1.lasalle.edu/~prushan/Experiment8-redox%20titration.pdf
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