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) When $Cu^{2+}$ ions are treated with $KI$,the $Cu^{2+}$ ions are reduced to $Cu^{+}$ ions by iodide ions $(I^-)$.
The resulting $Cu^{+}$ ions react with the remaining iodide ions to form a white precipitate of copper$(I)$ iodide $(Cu_2I_2)$.
The balanced chemical equation for this reaction is:
$2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \rightarrow Cu_2I_{2(s)} + I_{2(s)}$