Consider a fuel cell operated with acetylene and pure oxygen. In the fuel cell,

Question

Consider a fuel cell operated with acetylene and pure oxygen. In the fuel cell, this equation is split up into two parts, at the Cathode, C_2 H_2 + 8. H_2 O right arrow 4. CO_2 + 20. e^- + 20. H^+ The electrons produced by the reaction flow through the external load, and the positive ions migrate through the electrolyte to the cathode, where the reaction is 20. e^- + 20. H^+ + 5. O_2 right arrow 10. H_2O Calculate the reversible work and the reversible EMF (electromotive force or voltage) for the fuel cell operating at 25 degree C, 100 kPa.

Explanation / Answer

reaction 1. 2C2H2 + 8H2O --> 4CO2 + 20e- + 20H+

reaction 2. 20e- + 20H+ + 5O2 -> 10H2O

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Overall:   2C2H2 + 5O2  --> 4CO2 + 2H2O

Now DeltaHo and DeltaSo can be calculated Hess's Law as follows

DeltaHo at 25oC = 4(393.51) + 2(285.830) - 2(+227.48) + 5(0) = -2600.66 KJ

DeltaSo at 25oC =  4(213.6) + 2(69.9) - 2(200.9) + 5(205.0) =-432.6 KJ/k

DeltaGo = DeltaHo - T *DeltaSo

DeltaGo = 2600.66 KJ - 298.15 (-432.6 KJ/k)

DeltaGo = -131580.35 KJ

Wrev = -DeltaGo - 0 = -(-131580.35) = 131580.35 KJ

DeltaGo = nFEocell

Eocell =DeltaGo/ nF = 131580.35 KJ/ (20 * 96485) = 0.0681 V

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