Question- The concentrations of NaCl and KI are at 0.150 M and 0.100 M, respectively, in the following electrochemical cell:
Cu(s) | CuI(s) | I-(aq) || Cl-(aq) | AgCl(s) | Ag(s)
I got 0.396V for cell voltage
B) Now, calculate the cell voltage using the following half-reactions. The solubility products (Ksp) for AgCl and CuI are 1.8×10-10 and 1.0×10-12, respectively.
Cu+ + e- <--> Cu(s) Eo = 0.518V
Ag+ + e- <--> Ag(s) Eo = 0.7993V
Illustrate structures to prove your interpretation.