Question

3. Determine E0 Cell for the reaction: Pb2+ + Zn ---> Pb + Zn2+, the half...

3. Determine E0 Cell for the reaction: Pb2+ + Zn ---> Pb + Zn2+, the half reactions are:

Pb2+ + 2e- ---> Pb E0= -0.125

Zn2+ + 2e- ---> Zn E0= -0.763 V

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Answer #1

oxidation:   anode

Zn -----------------------> + Zn2+   , E0= -0.763 V

reduction : cathode

Pb2+ + 2e- -------------------> Pb E0= -0.125

Eo (cell )   =   Eo cathode -Eo anode

                = -0.125 - (-0.763)

               = 0.638 V

Eo (cell )    = 0.638 V

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