Question

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

Answer 1

Reduction of $$\mathrm{Cu}{\phantom{A}}^{2+}$$ to $$\mathrm{Cu}{\phantom{A}}^{+}$$ takes place due to reaction with F ions.

This is given by the equation.

$$2\mathrm{Cu}{\phantom{A}}^{2+}+4\mathrm{I}{\phantom{A}}^{-}⟶\underset{(\mathrm{White}\mathrm{precipitate})}{\mathrm{Cu}{\phantom{A}}_2\mathrm{I}{\phantom{A}}_2+\mathrm{I}{\phantom{A}}_2}$$