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PEM Fuel Cell Calculations

  1. Mar 22, 2010 #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.
  2. jcsd
  3. Mar 22, 2010 #2


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  4. Mar 23, 2010 #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?
  5. Mar 23, 2010 #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: Mar 23, 2010
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