[khwcm]

for the reaction:
(1) Cu²⁺ + e^-    Cu⁺    E^o = +0.15v
(2) Cu²⁺ + I^-    CuI_(s) E^o = +0.86v

for the half equation in equation (2), there should be 2 reactions involved:
redox: Cu²⁺ + e^- Cu⁺
ppt:    Cu⁺ + I^-    CuI_(s)

since the precipitation didnt involved change in oxidation number, so it is not a redox reaction => the emf of the cell should not be affected by precipitation in the initial condition.

note that the E^o value have intrinsic property, so why the E^o values in equation (1) is different from that in (2)?

[Tilc]

You should understand that reaction in 1 stage better than in 2. Ligand I^- changes Eº, then it "becomes" E, not Eº standard because there aren't free ions Cu2+ on solution, there are CuI and the environment (envelope?) of Cu its different now.

That's the same reason that allows H₂O₂ oxidize Co²⁺ to Co³⁺ when NH₃ are in solution, but not in water solution:

Co₃₊ +1é   Co²⁺  Eº=1,92v
(Co(NH₃)₆) ³⁺   [Co(NH₃)₆] ²⁺  Eº=0,10v

H₂O₂ 2H⁺ 2é   2H₂O   Eº=1,776v

In water, without NH₃, reaction doesn't occur because E=1,776-1,92<0, but in the presence of NH₃ (ligand or solvent) E=1,776-0,10>0, thats a spontaneous reaction.


*I'm sorry for my translated english, I dont know specific words sometimes

[Dark Vaati]

The potential is different just because the ions involved in the process are also different, as Tilc said. Note that, in addition, the precipitation does not involve any redox process, but it does affect the presence of Cu+ cations in solution, so the equilibrium between Cu2+ and Cu+ is also affected and the reaction towards Cu+ is easier. That's why the potential is higher, meaning the reaction is more favorable.