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## Main Question or Discussion Point

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.

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.

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

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.