# The hydrogen economy myth-part 2

In the last post we had a look at the background science which is necessary to consider the question of the hydrogen economy.  Before we look at the issue of cell efficiency and some of the other problems we need to introduce one more equation.   In a fuel cell we are dealing with electrical work and the following equation relates free energy (useful work- see last weeks post) to the voltage across the electrodes.

ΔG= -nFE

where E is the voltage difference generated between the anode and the cathode, n is the number of electrons and F is a number known as the Faraday constant which is a measure of the electrical charge per mole of electrons.

If we calculate E for the hydrogen fuel cell reaction given last week we get a theoretical maximum voltage of either 1.229V or 1.18V depending on whether the reactants are in a liquid or gaseous state.  This does not sound like a lot, but in actual fact is not a problem since individual cells can put in parallel or series to build up the current or voltage.

What is more important looking at our hydrogen fuel cell is that of cell efficiency.  Working out the maximum efficiency for the reaction is easy.  ΔG and ΔH are known and are listed in the thermodynamic tables.  Dividing ΔG/ΔH and multiplying 100 gives a value of 83%.  Since this reaction goes both ways its the same efficiency both ways.  So if I was to take 1kW of electricity and use it to convert water to hydrogen its easy to see I would have lost 17% of my energy and be left with the equivalent of “830 Watts” of hydrogen.  Converting it back to electricity using the reverse of the first reaction in my fuel cell would leave me with 83% 0f the 83%, this is 69% of our starting energy (rounded up).

Nearly 70% sounds pretty good.  However there are a whole heap of other losses that have to be taken into account.  These are briefly;

• Internal resistance of the electrical circuitry in the fuel cell and electrolyser.  Unless and until high temperature superconductors are developed there is not really any way of stopping losses of this type.
• Internal (ionic) resistance of the chemical reactants in the electrolyte.  There is no way round this problem.
• Activation energy losses of the chemical reactants at the electrodes.  Energy is needed to get a reaction going (think of striking a match).  The platinum electrode reduces the energy required but cannot eliminate it.  This can be influenced by electrode and   physical conditions in the fuel cell but never completely eliminated.
• Mass transport losses.  As reactants are used up localised concentration gradients are set up in the cathode and anode compartment.  Reaction conditions may be able to reduce these losses but not reduce them.
• Oxygen can cross the semi-permeable membrane into the anode compartment.  Cell design can reduce losses from this source.

All this reduces the theoretical efficiency above massively to less than 50% for a PEM fuel cell.  If the losses in the electrolyser are as high then its an easy calculation to see only 25% of the original electrical starting energy is returned and that the hydrogen economy starts to look like a myth.

Yet another post will be necessary to look at some other practical problems.

Neil This entry was posted in other, Renewables and tagged . Bookmark the permalink.