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
Correct Answer:
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