from Nernst formula of electrochimical potential
#E = E° + (RT)/(nF) ln ([Ox]/[Red])# you can see that #Cu^2+# will always be in equilibrium with #Cu^+# otherwise the voltage of solution will became infinite.
Well, since #Cu_2I# would be soluble, while CuI not, the small amount of #Cu^+# present in solution start to form the precipitate moving continuosly l'equilibrium toward the riducted form that is consumed during the precipitation.
Since happen a reduction #Cu^2+ e=Cu^+ # it 'll happen also that
#2I^(-) = I_2 #
