# Stability of cuprous iodide in aqueous solution

An aqueous solution of $$\ce{Cu+}$$ is usually unstable, but $$\ce{CuI}$$ is stable. Explain. $$E^\circ(\ce{Cu^2+/Cu+}) = \pu{0.153 V}$$ and $$E^\circ(\ce{Cu+/Cu}) = \pu{0.521 V}.$$

I found this question while solving a paper and I am confused as to how I should answer. For the first part at least the disproportionation of $$\ce{Cu+}$$ into $$\ce{Cu^2+}$$ and $$\ce{Cu}$$ explains why it is unstable in aqueous solution. (since it is neither oxidised nor reduced by water) $$\ce{2Cu+ <=> Cu^2+ + Cu}$$ However, for the stability of $$\ce{CuI}$$ which remains as a precipitate in aqueous solution, I do not understand what reasoning I should provide for its stability in reference to redox reactions and electrode/formal potentials.