I have run some tests with a PEM fuel cell and the voltage we have found was $1.44\ \mathrm{V}$. Given that the standard potential for a fuel cell is $1.229\ \mathrm{V}$ what is causing the difference in our measurement?

Initially we provide power to the fuel cell so it can generate oxygen and hydrogen into separate containers. These containers are submerged in water so there is water pressure keeping them contained.

All measurements were done in standard conditions, except that the pressure of each container is slightly different (hydrogen displaced twice the amount of water since there are twice the moles of atoms present).

I am not sure how to calculate this using the Nernst equation since the conditions of both the oxygen and hydrogen are slightly different (each gas is of a different concentration in its respective container since there is different water pressure and water vapour pressure present).

My equation is:

$$E_\text{cell} = 1.229\ \mathrm{V} - \frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{0.5}}\right)$$

For $[\ce{H2}]$ I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

I also do the same thing for $[\ce{O2}]$

I'm not sure what to use for $[\ce{H2O}]$

The answer I am getting is $1.24\ \mathrm{V}$. What am I missing?

  • 1
    $\begingroup$ At first glance your approach looks decent. Is your solution concentrated? The Nernst equation only works well with dilute solutions if using concentrations, otherwise you have to use activities. $\endgroup$ May 14, 2015 at 15:59
  • 1
    $\begingroup$ When you are applying power to the cell, you are most definitely applying a higher voltage. How long do you wait after you turn it of before you measure the potential? It might be that the cell didn't come to equilibrium after the electrolysis and the voltage you are measuring is just leftover from the previous process $\endgroup$ Jun 10, 2015 at 8:39
  • $\begingroup$ Check your equipment and gas flow measurement set up. Storage under water will deviate from atmospheric pressure. If the gas deliveries are not properly regulated, then the partial pressures and concentrations will vary. This is the variable in the Nernst equation that will change your E value. It is highly likely to be this. $\endgroup$
    – MikePb
    Mar 7, 2023 at 10:51

2 Answers 2


Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement?

Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load.

Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "stacked", or placed in series, to create sufficient voltage to meet an application's requirements.

So it is surprising that the voltage is higher than the standard potential. Perhaps two cells were stacked to provide about 1.4 V, similar to the voltage of a standard alkaline battery, and about twice what you would expect from a single cell under load.

For [H2] I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

For the standard hydrogen electrode, the standard state is defined as a partial pressure of 1 atm. This is much lower than a concentration of 1 M. In any case, the Nernst equation does not predict a voltage that is significantly higher than the standard potential.


Seems your cell is operating as,
Anode: $\ce{H2 = 2H + 2e'}$
Cathode: $\ce{2H(+) + 2e' = 2H}$, and then $\ce{2H + 1/2O2(g) = H2O(l)}$ -Heat.

E = 1.4 V gives, 1.4 = [2.303RT/F] log R, where R = ratio H activity between cathode and anode R = 10^-23.65 if your anode assumes as activity H = 1.0, at the cathode H activity is 10^-23.65 the removal H by the oxygen as, $\ce{2H + 1/2 O2(g) = H2O(l)}$. Atomic H is highly reactive as $\ce{H + H = H2}$ or $\ce{H + O = OH}$, or $\ce{2H + 1/2O2(g) = H2O(i,g)}$, etc.

$\ce{H2O(l)}$ blocks Oxygen contacting cathode (electrode reaction), i.e., $\ce{2H+ + 2e' + 1/2O2(g) = H2O(l)}$. Actually if current is OK, it is better do Fuel Cell operation in concentration cell that the oxygen is removal hydrogen by the H2O(g) vapor which is high T fuel Cell. Therefore we need high T Fuel Cell that is more appropriate in science and technology with Fuel Efficiency, CHP module.

Currently, I am developing such as:
[H-eChemEng = H-fuel Cell + H-Thermal to Electric converter]

you have good measurement data!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.