CuSO4 + 2KI -> CuI↓ + 1/2I2 + K2SO4 — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Identify starting materials: Copper(II) sulfate (CuSO4) and potassium iodide (KI) are both soluble salts. Step 2: Observe the redox reaction: Cu2+ oxidizes I- to iodine (I2), and is itself reduced to Cu+, which precipitates as white cuprous iodide (CuI↓). Step 3: Since a solid precipitate (CuI↓) is formed from soluble starting materials, this is classified as a precipitate formation reaction (type A).