I need some help with partial pressures of reactions for a fuel cell project. The overall reaction for a fuel cell is $$\ce{H_{2} +\frac{1}{2}O_{2}->H_{2}O}$$

Which means that the mole fraction of hydrogen, oxygen and water are 0.66, 0.33 and 1, respectively. If the operating pressure of the fuel cell is 1 bar (100 kPa) then the partial pressures of the species are 66.66 kPa, 33.33 kPa and 100 kPa.

Now the above is the case for when the fuel cell is being fed pure oxygen and pure hydrogen. I am doing a project and we are designing to feed the fuel cell ambient air which is composed of 21% oxygen and then rest is gases that don't participate in the reaction.

Does this mean that the mole fraction of the oxygen stays the same at 0.33 since that's what is reacting, or will it decrease? If it decreases, what does it decrease to?

  • $\begingroup$ Could you describe the conditions in which the gasses are contained. I assume there's a seal contained of some sort. Additionally are you trying to calculate the partial pressures before or after the reaction? $\endgroup$ – Julia Gonzalez Jan 28 '13 at 5:51

A fuel cell designed to be fed pure hydrogen and pure oxygen works by separating the combustion of hydrogen into a pair of electrochemical processes that occur at separate electrodes. By design, the hydrogen and the oxygen are separate from each other. They enter the fuel cell in difference places and never come in contact with each other. The protons flow through an ion-conductive membrane.

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$$\text{Reduction:} \ \ \ce{O2 + 4H+ + 4e- -> 2H2O}\ \ E^\circ_{\text{red}} = +1.229 \text{ V}$$ $$\text{Oxidaton:} \ \ \ce{2H2 -> 4H+ + 4e-} \ \ E^\circ_{\text{ox}} = 0 \text{ V, defined}$$ $$\text{Overall:} \ \ \ce{2H2 + O2 -> 2H2O} \ \ E^\circ_{\text{cell}} = +1.229 \text{ V}$$

Since the gases are separated, there is no mixture. The mole fraction of hydrogen in the hydrogen stream is 1. The mole fraction of oxygen in the oxygen stream is 1. The partial pressures of the gases are equal to whatever pressure they are set at from the tanks.

In your case, the oxygen is replaced by air, which has a mole fraction of 0.21 (=21%). The partial pressure of oxygen then depends on this mole faction and the operating pressure of 100 kPa. I am assuming you would still use hydrogen as the fuel, so the hydrogen line would still have a mole fraction of 1 and a partial pressure of 100 kPa.

Just because the stoichiometric coefficients are 2:1 does not mean that the gasses will actually be mixed in that ratio - they just react in that ratio. They could be deliberately mixed in other ration to influence the rate of the reaction or the position of equilibrium. In your case, you are using air (21% oxygen) because it is easier than pure oxygen, so your mole fraction is set to the composition of air.

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