$CuI _{2}$ is not stable and thus does not exist iodide ion will act as reducing agent that converts $Cu(II)$ to $Cu(I)$ and part of iodide ion will be oxidised to iodine. The first reaction involved is as follows:
$2 Cu ^{2+}(a q)+4 I ^{-}(a q) \longrightarrow 2 \operatorname{CuI}(s)+ I _{2}(a q)$