Given the data at 250C , Ag+I−→AgI+e−;E0=+0.152 VAg→Ag++e−;E0=−0.8 VWhat is the value of log Ksp  for AgI  ?(2.303RTF=0.059V)

Given   Ag→Ag++e−;E0=−0.8⇒(1)                         Ag+I−→AgI+e−;E0=+0.152⇒(2)_     (1)−(2)⇒AgI →Ag++I−;E0=−0.952 V                                    E0=2.303RTnF.logKsp                              −0.952=0.0591.logKsp                              log.Ksp=−0.9520.59≈−16.108