How to Calculate the Theoretical Voltage of a PEM Fuel Cell?

[clementlee87]

Hi, I'm working on a commercially available PEM fuel cell and I'm trying to calculate the theoretical voltage of the cell:

The only electrochemical reaction considered is H2 + 0.5O2 --> H2O

And I understand that I need to use the Nernst equation:

E = EO - (RT / 2F) ln (PH2O / (PH2 * SQRT(PO2))

The anode side in is pure hydrogen from a gas canister at 300 ml/min at 5 psi

The cathode side in is air from ambient at 1 atm

The operating temperature is 50 C, or 323 K.

How can I determine the partial pressure of water produced? If I assume that the water produced is at the vapor pressure of water at my room conditions, it should be at approximately 1584.9 Pa (50% RH at 28 C).

And if I used these pressures,
P H2O = 1584.9 Pa
P H2 = 34483 Pa
P O2 = 21287 Pa

I would get ln (PH2O / (PH2 * SQRT(PO2) = -8.063. This in turn gives me 0.112 V. For this 12 fuel cell stack, that would be equal to 1.344 V. However, the produced voltage when measured with a handheld ammeter reads about 7.2 V.

So wouldn't that mean my efficiency is more than 100%?

Thanks for your help! I need it as soon as I can.

 

[Bystander]

(snip)The anode side in is pure hydrogen from a gas canister at 300 ml/min at 5 psigauge
(snip)P H2 = 34483 Pa+~100kPa
(snip)

questions?

 

[clementlee87]

Ah, thank you, Bystander. A slight on my part.

However, now that I've modified this,

P H2O = 1584.9 Pa
P H2 = 34483 + 101325 = 135808 Pa
P O2 = 21287 Pa

the quotient term ln (PH2O / (PH2 * SQRT(PO2) = -9.434, and my voltage is still only 0.131 V only. Also, I would like to confirm: for this reaction the standard potential Eo is 0 V, right?

 

[Topher925]

Eo is the standard potential of the cell or the potential at standard conditions, i.e. 25C, 1 atm. It is not 0V its 1.229V for a H2-O2 reaction. You can get find this value from any standard reaction table or simple thermodynamic analysis.

The concentrations of the reactants and products are partial pressures over the total pressure. You don't need to determine absolute pressures for each, only the mole faction (Dalton's Law).

So if you are using regular air with H2 at 5 psig you would have,
Eo = 1.229V
PH2 = 1.34 atm
PO2 = 0.21 atm

At a temperature of 323K and 1 atm, water is in liquid form which means you can assume it be incompressible. Because of this you can assume your products, or PH2O to have a pressure of 1 atm.

So using the Nernst Equation (the form I like)

E = Eo + RT/nF ln(PH2 x PO2^0.5 / PH2O)

You get a voltage of

1.222V per cell or 14.667V for your entire 12 cell stack.

Pop quiz, why is your theoretical voltage at your operating conditions less than 1.229V even though you have a higher concentration of H2 on the anode? Also, if your stack voltage has a theoretical value of 14.667V, why are you only reading 7.2V?

BTW, an ammeter measures current. A voltmeter or electrometer measures voltage.

 

[LudusRex]

Hello,

Thank you for sharing your calculations for the theoretical voltage of your PEM fuel cell. It seems like you have a good understanding of the Nernst equation and the factors that affect the voltage of the cell. However, there are a few points that I would like to clarify and some suggestions for further calculations.

First, to determine the partial pressure of water produced, you can use the ideal gas law: PV = nRT. In this case, you know the temperature, volume (300 mL/min), and pressure (1 atm) of the air on the cathode side, so you can calculate the number of moles of air present. Since the only reaction occurring is H2 + 0.5O2 --> H2O, the number of moles of water produced will also be equal to the number of moles of hydrogen consumed. You can then use this information to calculate the partial pressure of water on the anode side.

Second, the vapor pressure of water at 50 C is actually closer to 12,400 Pa (not 1584.9 Pa). This may have been a typo in your calculations. Using this value in the Nernst equation, you should get a voltage of about 0.9 V for a single cell. This would result in a total voltage of 10.8 V for your 12-cell stack.

Third, the voltage measured with a handheld ammeter may not be an accurate representation of the actual voltage produced by the fuel cell. The voltage measured may be affected by factors such as resistance in the circuit and the efficiency of the fuel cell itself. It is possible that your efficiency is greater than 100% due to these factors.

I would suggest double-checking your calculations and taking into account the points mentioned above. It would also be helpful to compare your theoretical voltage with the actual voltage output of similar PEM fuel cells to see if your results are within the expected range. If you continue to have concerns about the efficiency of your fuel cell, you may want to consult with a specialist or conduct further experiments to investigate the issue.

I hope this helps. Good luck with your research!

Best,