Respuesta :

This is an oxidation-reduction (redox) reaction:

2 Cu² + 2 e- → 2 Cu¹ (reduction)

2 I⁻¹ - 2 e- → 2 I⁰ (oxidation)

CuSO₄ is an oxidizing agent, KI is a reducing agent. Therefore Copper is being reduced and Iodine oxidized.