Coulometric vitamin C Titration, pretty basic

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SUMMARY

The discussion focuses on the secondary coulometric titration of vitamin C, where an applied voltage generates a current that oxidizes iodide to iodine, which subsequently oxidizes vitamin C. The current flows from the voltage source to the anode, then to the cathode, and back to the voltage source. At the cathode, a reduction reaction occurs, accepting electrons from the solution. Reversing the leads transforms the setup into an electrolytic cell, where the current direction opposes electron flow.

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These questions relate to the secondary coulometric titration of vitamin C. In this titration an applied voltage is used to generate a current that oxidizes Iodide to iodine. Iodine then oxidizes vitamin C. A starch indicator detects excess iodine via the formation of I3-

My questions are:

1) What direction does the current flow in this type of experiment? I would think from the voltage source, to the anode, then to the cathode then to the amperometer (sp?) then back to the voltage source?

2) What is happening at the cathode here? I know at the anode Iodide is oxidized. But I can't figure out what's going on at the cathode.

3) If the leads are reversed, what happens at the anode? No idea about this one.
 
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Electrons are pulled from the iodide into the circuit while the same number of electrons are delivered back into the solution. The process is not spontaneous so the cell is not a galvanic cell but is instead an electrolytic cell. Remember, the current flows in the opposite direction as the electrons.

As to what happens at the electrode opposite the one where iodide is oxidized to iodine, you need to determine what, in solution, is available to accept an electron?

Hint: remember that this experiment is used to determine ascorbic acid.
 

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