# How do I calculate partial pressure here?

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?

• 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? – Julia Gonzalez Jan 28 '13 at 5:51

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