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In the reaction of tin (IV) iodide in acetone and water, the solution turns a urine yellow.

Is this simply due to the hydrolysis of tin (IV) iodide:

$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$

why is the acetone needed though?

and for the reaction with $\ce{KI}$:

$$\ce{SnI4 + KI -> ?}$$ I understand it forms iodine, but what happens to $\ce{Sn}$ and $\ce{K}$ - do they react with the acetone for instance?

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If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).

$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$

$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$

With $\ce{KI}$, a yellow-brown solution of $\ce{KI3}$ will be formed, which contains the triiodide anion $\ce{I3-}$.

$$\ce{I2 + KI -> KI3}$$

When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (source):

$$\ce{(CH3)2CO + I2 -> CH3C(O)CH2I + HI}$$

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