Use the following two half-reactions to write an overall balanced redox reaction in the
direction in which it is spontaneous and calculate the standard cell potential.
Fe3+(aq) + e- → Fe2+(s)
Pb2+(aq) + 2e- → Pb(s)
O a. Pb(s) + Fe3+(aq) → Pb2+(aq) + Fe2+(s)
Ob.pb2+(aq) + 2 Fe2+(s) → Pb(s) + 2 Fe3+(aq)
O c. pb2+(aq) + Fe2+(s) → Pb(s) + Fe3+(aq)
E° = 0.771 V
E°=-0.126 V
E=0.645 V
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E=0.897 V
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E=0.645 V
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O d. Pb(s) + 2 Fe3+(aq) → Pb2+(aq) + 2 Fe2+(s)
E=0.897 V
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O e. Pb(s) + 2 Fe3+(aq) → Pb2+(aq) + 2 Fe2+(s)
E = 1.416 V
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