Explanation:

This is not a simple reaction but a mixture of processes taking place in the same reaction vessel. If this is a test tube style reaction or it is happening in a beaker where chemicals are simply mixed together, then the quantities will not be exact. LEFT over reagents can account for residual colours.

Yellow / brown - Iodine

Blue - hydrated Cu^(2+)

Green - Both of the above (blue + yellow = green)

First: The reduction of Cu^(2+) to Cu^(+)

Cu^(2+) + 2 I^(-) → 2Cu^(+) + I2 = REDOX reaction

…Reduction of Cu^(2+) to Cu^(+).

…Oxidation of iodide to iodine.

Secondly: pptn of CuI

Cu^(+) + I^(-) → CuI (white ppt)

NB the ppt actually appears a yucky brown sludge because of the iodine. Iodine is sparingly (if at all) soluble in water but it does dissole in the PRESENCE of I^(-) IONS to form a dark yellow / brown solution due to the formation of the complex ion. There is almost certainly some KI left to achieve this…

Thirdly: the dissolving of iodine:

I2 + I^(-) → I3^(-) in solution. Effectively this behaves as iodine solution and this is how iodine is dissolved for titration purposes.

Fourth: (Extension)

If thiosulfate solution is used, then the Iodine is re-reduced back to Iodide and the yucky brown discolouration disappears leaving the white ppt of CuI

I2 + 2 S2O3^(2-) → 2 I^(-) + S2O6^(2-)

…Reduction of iodine to iodide

…Oxidation of sulfate (VI) / thiosulfate to persulfate

All of the processes 1, 2 and 3 are taking place simultaneously. The thiosulfate is just an added part to help illustrate the chemistry and, if NECESSARY, help to obtain a clean sample of CuI