Question

Refer to the tabulated values of ΔG∘f in Appendix IVB in the textbook to calculate E∘cell for a fuel cell that employs the reaction between ethene gas (C2H4) and oxygen to form carbon dioxide and water (gaseous).

Answer 1

The reaction: C2H4 + O2 = CO2 + H2O

balanced reaction

C2H4 + 3O2 = 2CO2 + 2H2O

Now, dG value

dG = Gproducts - Greactant

Values: C2H4 = 68.4 kJ/mol, O2 = 0kJ/mol, CO2= -394.4 kJ/mol, H2O= -228.6kJ/mol

ΔG∘ = 2ΔG f (CO 2(g) ) + 2ΔG f (H 2 O (g) ) - 3ΔGF((O2(g)) - ΔG F( (C2H4) (g))

ΔG∘ = 2 x (-394.4) + 2 x (-228.6) - 0 - (68.4)

= - 1314.4 kJ/mol

Use ΔG = -nFE∘cell
E∘cell = ΔG/(-nF)

You could balance the following half-reactions:
2H2O + C2H4 ====> CO2 + 8H+ + 8e-
8H+ + 8e- 2O2 ======> 4H2O : two of the 4O in H2O are in the CO2

E°cell = -(- 1314.4) / (8 x 96500) = 0.0017 V