Calculate for the Electrochemical Cell Below,
Pb(s)|PbCl₂(s)| Cl^-(aq,1 $\to$ Pb(s) $~~~ ~~~~$

Calculate for the electrochemical cell below,
Pb(s) |PbCl₂(s) | Cl^-(aq,1.0 M) || Fe³⁺(aq,1.0 M) ,Fe²⁺(aq,1.0 M) | Pt(s)
Given the following reduction half-reactions.

Pb²⁺(aq) + 2 e^- $\to$ Pb(s) $~~~ ~~~~$$~~~ ~~~~$$~~~~~~~~$ E ^$\circ$ = -0.126 V
PbCl₂(s) + 2 e^- $\to$ Pb(s) + 2 Cl^-(aq) $~~~~~~~~$ E ^$\circ$ = -0.267 V
Fe³⁺(aq) + e^- $\to$ Fe²⁺(aq) $~~~~ ~$$~~~~~~~~$$~~~~~~~~$ E ^$\circ$ = +0.771 V
Fe²⁺(aq) + e^- $\to$ Fe(s) $~~~~~~~~$$~~ ~~~~~$$~ ~ ~$$~~~~~~~~$ E ^$\circ$ = -0.44 V

A) -0.504 V
B) -0.062 V
C) +0.504 V
D) +1.038 V
E) +1.604 V