Cu$$^{2+}$$ salt reacts with potassium iodide to give:

When $$Cu^{2+}$$ salt reacts with potassium iodide (KI), a redox reaction takes place. $$Cu^{2+}$$ is a sufficiently strong oxidising agent to oxidise $$I^-$$ to $$I_2$$, while being itself reduced to $$Cu^+$$.

The reaction is:

$$2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$$

The $$Cu^+$$ ions combine with $$I^-$$ to form cuprous iodide ($$Cu_2I_2$$), which is a white precipitate. This can also be written as $$2CuI$$, but the formula of cuprous iodide in its crystal form is $$Cu_2I_2$$.

Therefore, the product is $$Cu_2I_2$$.