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Question
Given the following two half-reactions, write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential.
Pb2+(aq) + 2 e- Pb(s) E = -0.126 V
Fe3+(aq) + e- Fe2+(s) E = +0.771 V
A.Pb2+(aq) + 2 Fe2+(s) Pb(s) + 2 Fe3+(aq) 
= +0.897 V
B.Pb2+(aq) + Fe2+(s) Pb(s) + Fe3+(aq) = +0.645 V
C.Pb(s) + 2 Fe3+(aq) Pb2+(aq) + 2 Fe2+(s) = +1.416 V
D.Pb(s) + 2 Fe3+(aq) Pb2+(aq) + 2 Fe2+(s) = +0.897 V
E.Pb(s) + Fe3+(aq) Pb2+(aq) + Fe2+(s) = +0.645 V
Answer
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