Question

96. A galvanic cell consists of a iron electrode in 0.0521 M Fe(NO₃)₂ and a copper electrode in 0.349 M Cu(NO₃)₂. What is the cell potential (emf, in V) of this cell at 25^oC?

Answer 1

96)

Fe / Fe+2 . Eo = - 0.44 V

Cu/Cu+2 ,   Eo = 0.34 V

Oxidation : Fe -------------> Fe+2 + 2e-

Reduction : Cu+2 + 2e-   ----------> Cu

_______________________________-

overall : Fe + Cu+2   -----------> Fe+2 + Cu

Eocell = Eored - Eooxidation

           = 0.34 - (0.44)

Eocell = 0.78 V

Ecell = Eo - 0.05916 / 2 log [Fe+2 / Cu+2]

         = 0.78 - 0.05916 / 2 log [0.0521 / 0.349]

Ecell = 0.804 V

cell potential = 0.804 V