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Copper ions are reduced by the iodide ions to copper(I). The best way to look at this is via the half equations:
Cu2+ + 1e --> Cu+ 2I- --> I2 + 2e
to balance these you double the first equation (to equalise the electrons) and add them together.
2Cu2+ + 2e --> 2Cu+ 2I- --> I2 + 2e------------------------------------------------------------- add2Cu2+ + 2I- --> 2Cu+ + I2
Now add in the spectator ions:
2CuSO4 + 2KI --> Cu2SO4 + I2 + K2SO4
So the white precipitate is copper(I) sulphate and the brownish coloration is iodine solution...
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