Question

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

Answer 1

The reaction between $C{u}^{2+}$ ion and $KI$ will take place as:

$\begin{array}{cc}2C{u}^{2+}\left(aq\right)+4I^{-}\left(aq\right)\to & C{u}_2{I}_2\left(s\right)+I_2\\ & white\\ & precipitate\end{array}$


In the aqueous solution, $KI$ dissociates to give $K^{+}$ and $I^{-}$.

Since iodide ion $\left(I^{-}\right)$ is a strong reducing
agent, it reduces $C{u}^{2+}$ to $C{u}^{+}$and a white precipitate of $C{u}_2{I}_2$ is formed.