Cu2+ salt reacts with potassium iodide to give

Step 1 — Reaction of $\mathrm{Cu^{2+}}$ with $\mathrm{KI}$:

$\mathrm{2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2}$

Or equivalently: $\mathrm{2CuSO_4 + 4KI \rightarrow Cu_2I_2 + I_2 + 2K_2SO_4}$

Step 2 — Analyse the reaction:

• $\mathrm{Cu^{2+}}$ is reduced to $\mathrm{Cu^+}$ (gain of 1e⁻ per Cu)
• $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$ (loss of 1e⁻ per I)
• $\mathrm{Cu^+}$ combines with $\mathrm{I^-}$ to form $\mathrm{Cu_2I_2}$ (cuprous iodide)

Step 3 — Product identification: $\mathrm{Cu_2I_2}$ = cuprous iodide (white precipitate) $\mathrm{I_2}$ = iodine (brown, detected by starch indicator turning blue)

This reaction is the basis of the iodometric determination of $\mathrm{Cu^{2+}}$.

Answer: Option (1) — $\mathrm{Cu_2I_2}$