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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- Pb(s) E = -0.126 V
PbCl2(s) + 2 e- Pb(s) + 2 Cl-(aq) E = -0.267 V
Fe3+(aq) + e- Fe2+(aq) E = +0.771 V
Fe2+(aq) + e- Fe(s) E = -0.44 V
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