Calculate the solubility product:

Problem:

Calculate the solubility product of AgCl from the data

Ag⁺+ le → Ag(s)                                            E⁰_Ag+,Ag = +0.799V

AgCl(s) + le → Ag (s) + Cl^-                           E⁰_AgCl,Ag = +0.222V

Answer:

cell reaction can be written as:

AgCl(s) +le→ Ag(s) +Cl

  Ag(s) → Ag (s) +le

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AgCl (s) =→ Ag⁺ +C^l-    

Log K_sp = 1[0.222-(+0.799)]/ 0.0591

K_sp = 1.82× 10^-10