PEM Fuel Cell Calculations

In summary, the conversation discusses the theoretical voltage of a commercially available PEM fuel cell, which is calculated using the Nernst equation. The partial pressures of the reactants and products are important in determining the voltage, and the standard potential for this reaction is 1.229V. However, due to operating conditions and assumptions, the theoretical voltage is lower than the standard potential. The conversation also mentions that the measured voltage is different from the theoretical value, raising questions about the efficiency of the fuel cell.
  • #1
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.
 
Chemistry news on Phys.org
  • #2
clementlee87 said:
(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?
 
  • #3
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?
 
  • #4
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.
 
Last edited:
  • #5



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,
 

1. What is a PEM fuel cell and how does it work?

A PEM (Proton Exchange Membrane) fuel cell is a type of electrochemical device that converts chemical energy from a fuel, typically hydrogen, into electrical energy. It consists of an anode and a cathode separated by a proton exchange membrane. The hydrogen fuel is fed to the anode where it is split into protons and electrons. The protons pass through the membrane to the cathode while the electrons travel through an external circuit, creating an electrical current. At the cathode, the protons and electrons combine with oxygen to form water, the only byproduct of the reaction.

2. What are the key factors to consider when calculating the performance of a PEM fuel cell?

The key factors to consider when calculating the performance of a PEM fuel cell include the operating conditions (temperature and pressure), the type and concentration of the fuel, the type and thickness of the membrane, the catalyst material and loading, and the design of the cell components (flow channels, gas diffusion layers, etc.). These factors can affect the efficiency, power output, and durability of the fuel cell.

3. How do you calculate the efficiency of a PEM fuel cell?

The efficiency of a PEM fuel cell can be calculated by dividing the electrical power output by the fuel cell by the energy content of the fuel consumed. The energy content of hydrogen is typically measured in kilojoules (kJ) and the power output in watts (W). The higher the efficiency, the more efficient the fuel cell is at converting the chemical energy in the fuel into electrical energy.

4. How do you determine the power output of a PEM fuel cell?

The power output of a PEM fuel cell can be determined by multiplying the voltage and current generated by the fuel cell. This can be measured using a voltmeter and an ammeter. The power output can also be affected by the load resistance, which can be adjusted to maximize the power output of the fuel cell. Additionally, the power output can be influenced by the operating conditions and design of the fuel cell.

5. What are the most common challenges in PEM fuel cell calculations?

Some of the most common challenges in PEM fuel cell calculations include accurately modeling and predicting the behavior of the cell under different operating conditions, accounting for losses and inefficiencies in the system, and selecting appropriate materials and components to optimize the performance of the fuel cell. Other challenges may include accounting for the effects of impurities in the fuel, managing the water balance in the cell, and accounting for degradation over time.

Suggested for: PEM Fuel Cell Calculations

Replies
10
Views
1K
Replies
2
Views
1K
Replies
1
Views
553
Replies
17
Views
2K
Replies
1
Views
940
Replies
8
Views
2K
Replies
4
Views
2K
Back
Top