# Nernst equation for a fuel cell when gases are stored separately

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?

• 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. – LordStryker May 14 '15 at 15:59
• 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 – Burak Ulgut Jun 10 '15 at 8:39