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Question
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
A.-0.504 V
B.-0.062 V
C.+0.504 V
D.+1.038 V
E.+1.604 V
Answer
This answer is hidden. It contains 2 characters.
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