Question

A direct hydrogen fuel cell uses hydrogen and air (oxygen) to produce electricity, as we showed in class.

If you wanted a voltage of 2.2 V for your fuel cell device (which can be made of multiple cells), what mix of conversion and/or number of cells would you want to use? Remember: the voltage of cells in series adds up. There are a variety of possible answers.

Answer 1

I have no information, what your teacher tought or showed you in the class.

Answer is 2 cells (theoretical)

Answer is 3 cells (Practically)

Please read my explanation carefully.

The theoretical open circuit voltage of a hydrogen-oxygen fuel cell is 1.23 V at 298 K, in practice it is around 1 V at open circuit. Under load conditions, the cell voltage is between 0.5 and 0.8 V.

There are three major loss factors in a fuel cell, including kinetic loss, ohmic loss and mass transfer loss. I assume that you are interested in PEM fuel cell.

Firstly, at the low load region, you will see a large drop of voltage from a theoretical 1.23V to about 0.8V or so. This is the kinetic loss due to the very slow reaction kinetic of oxygen reduction in the cathode side. This in turn requires a large over potential (voltage loss) to drive any practical current density. Pt is currently the best commercial catalyst, but this is still not efficient enough. In this region, the ohmic loss and mass transfer loss are not significant.

Then with increasing load, while the kinetic loss continues taking place, there is an ohmic loss adding in as well. This is due to the internal resistance of the fuel cell, mainly from the PEM membrane.

Lastly, at high current density, you may see another quick drop of voltage. This is due to the mass transport issue that the reactants cannot be delivered to the catalyst sites quickly enough, either due to the low porosity of the electrode, or due to water flooding, or anything that prevents the reactant flows.

Thanks