[KyleH]

Hi, so I have a chem lab where we've created an electrolytic cell using potassium iodide in a U-tube.  Some phenolphthalein was added to see if there was any change in pH.  Inert graphite electrodes were used.  Using a table of standard reduction potentials,  I calculated that at the cathode, hydrogen gas would be produced.  This prediction was proven correct when bubbles were seen to form, and the solution around that electrode turned a deep pink.  (indicating an increase in hydroxide ions because the H2(g) was leaving the solution. 

At the anode, i predicted that solid iodine (I2) would be created.  However, this is where it gets confusing.  The solution around the anode turned a brown-yellow colour, which quickly moved towards the bottom area of the U-tube, indicating a higher density.  However, after looking over the solution, I didn't see that any precipitate had formed.   

So my question is this: when the electrolysis was taking place, did I simply miss the precipitate?  Or is it possible for Iodine to rebond with something else creating the yellow-brown colour that happened at the anode?

any suggestions or places where I could do more research would be greatly appreciated.  I've already done a lot of looking into it myself and seem to be turning up blanks.

[Nobby]

That is easy, you will not find a precipitate, because iodine is soluble in water enhanced to potassioniodide.

KI +I₂ => KI₃ 

This solution has yellow-brown colour.

[Borek]

More precisely what you observe is an equilibrium reaction:

I^- + I₂ <-> I₃^-

with equilibrium shifted to the right.

Iodine as such has low water solubility, but it dissolves pretty nicely in iodides solutions.

[KyleH]

Ohhhh, thank you so much guys.  That helps out a lot