Show all work for full credit. Using literature values for the half-cell reactio

Question

Show all work for full credit. Using literature values for the half-cell reactions of the following metals, calculate the total cell potentials for each of the 10 possible cells Assume complete reduction of the metals. Show all your work, please. Copy the resulting potentials into your lab notebook. The standard potential of a galvanic cell (E degree_cell) can be determined from thermochemical data such as the free energy of formation (Delta G degree_t) of each of the species in the reaction. For example, given the reaction: Fe^3+(aq) + Al(s) doubleheadarrow Al^3+(aq) + Fe(s) First use the data to the right to calculate the delta G degree_rxn. Then calculate E degree_cell from Delta G degree_rxn

Explanation / Answer

1) Total cell potential for,

Fe3+/Fe and Pb2+/Pb

Eo = Ecathode - Eanode = -0.037 - (-0.123) = 0.086 V

Fe3+/Fe and Zn2+/Zn

Eo = Ecathode - Eanode = -0.037 - (-0.763) = 0.726 V

Fe3+/Fe and Al3+/Al

Eo = Ecathode - Eanode = -0.037 - (-1.66) = 1.623 V

Fe3+/Fe and Mg2+/Mg

Eo = Ecathode - Eanode = -0.037 - (-2.37) = 2.333 V

Pb2+/Pb and Zn2+/Zn

Eo = Ecathode - Eanode = -0.123 - (-0.763) = 0.640 V

Pb2+/Pb and Al3+/Al

Eo = Ecathode - Eanode = -0.123 - (-1.66) = 1.537 V

Pb2+/Pb and Mg2+/Mg

Eo = Ecathode - Eanode = -0.123 - (-2.37) = 2.247 V

Zn2+/Zn and Al3+/Al

Eo = Ecathode - Eanode = -0.763 - (-1.66) = 0.897 V

Zn2+/Zn and Mg2+/Mg

Eo = Ecathode - Eanode = -0.763 - (-2.37) = 1.607 V

Al3+/Al and Mg2+/Mg

Eo = Ecathode - Eanode = -1.66 - (-2.37) = 0.71 V

2) For the given cell,

dGo = dGo(products) - dGo(reactants) = -481.2 - (-10.5) = -470.7 kJ/mol

dGo = -nFEo

Eo = -470700/-3 x 96485 = 1.626

dGorxn = dGo + RTlnK

RTlnK = nFEo

dGorxn = -470700 + 3 x 96485 x 1.626 = -46.17 J/mol

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