Consider the galvanic cell shown below with a potential of 1.369 V at 25°C

Pb(s) | PbCl₂(s) | Cl^–(0.20 M) || Cr₂O₇^2–(0.20 M), Cr³⁺(0.20 M), H⁺(0.20 M) | Pt(s)

Determine the K_sp value for PbCl₂(s). 

Cr₂O₇^2–(aq) + 14 H⁺(aq) + 6 e^– → 2 Cr³⁺(aq) + 7 H₂O(l)          E°_red = 1.232 V
Pb²⁺(aq) + 2 e^– → Pb(s)                                                            E°_red = –0.126 V