Question

What happens when Copper sulphate reacts with potassium iodide?

Answer 1

When potassium iodide, is added to a copper sulphate solution, a redox reaction occurs in which the iodide ion converts cupric ion, to cuprous ion, (blue to white precipitate) before being oxidised to iodine gas.

Complete answer:
When copper sulphate combines with $KI$ solution, complete iodine separates, forming a white cuprous iodide precipitate.
$2CuS{O_4} + KI \to 2{K_2}S{O_4} + C{u_2}{I_2} + {I_2}$
When copper sulphate in water reacts with potassium iodide in solid form, cuprous iodide develops as a precipitate, releasing iodine gas and forming potassium sulphate in water.
$Cu\left( {II} \right)$ is reduced to $Cu\left( I \right)$ by the iodide ions, and the iodide is oxidised to iodine. Potassium triiodide is formed when the iodine formed reacts with potassium ions in the reaction mixture. As a result, a white copper $\left( I \right)$ iodide precipitate will form in a brown potassium triiodide solution.

Additional Information:
A redox reaction is a chemical reaction in which electrons are exchanged between two reactants that are involved. Changes in the oxidation states of the reacting species can be used to identify this electron transfer.

Note:
Oxidizing agents are electron-accepting entities that tend to undergo a reduction in redox processes. A reducing agent is an electron-donating species that tends to hand over electrons. Oxidation is a common occurrence in several species. Any redox process can be decomposed into two half-reactions, namely the oxidation half-reaction and the reduction half-reaction.