When $Cu^{2+}$ ion is treated with $KI$,a white precipitate is formed. Explain the reaction with the help of a chemical equation.

(N/A) $I^{-}$ is a strong reducing agent. It reduces $Cu^{2+}$ ions to $Cu^{+}$ ions.
The reaction is as follows:
$2Cu^{2+} (aq) + 4I^{-} (aq) \rightarrow Cu_{2}I_{2} (s) + I_{2} (s)$
The white precipitate formed is $Cu_{2}I_{2}$ (Copper$(I)$ iodide).