Water Electrolysis Calculations From Wikipedia (http://en.wikipedia.org/wiki/Water_electrolysis#Efficiency):

The electrolysis of water requires a minimum of 237.13 kJ of electrical energy input to dissociate each mole.

Each mole of water gives you 2 grams of Hydrogen and 16 grams of Oxygen (http://www.lenntech.com/calculators/molecular/molecular-weight-calculator.htm).
The energy density of the Hydrogen is 141.86 MJ/kg (http://en.wikipedia.org/wiki/Energy_density#Energy_densities_ignoring_external_components).
Calculation for 1 kg of water (55.55 moles):

Energy for electrolysis: 237.13 kJ * 55.55 = 13.173 MJ
Energy released by Hydrogen combustion: 0.002 * 55.55 * 141.86 MJ = 15.76 MJ

These calculations are not taking in account efficiency and energy loses, they are purely theoretical.
In various Wikipedia articles there are claims regarding to electrolysis similar to following:

The energy required to generate the oxyhydrogen always exceeds the energy released by combusting it.
Electrolysis-based designs have repeatedly failed efficiency tests and contradict widely accepted laws of thermodynamics (i.e. conservation of energy)

First Question: In theory (in practice we always have less efficiency and must take those in account), is there anything wrong with my calculation?
Second Question: Can someone clarify to me the claims about laws of thermodynamics and conservation of energy - I do not see ANY CONNECTION between energy needed to electrolyze water and energy released by hydrogen combustion?
 A: Re your second question: have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html. The 237kJ/mol is the Gibbs free energy required for electrolysis, while the 286kJ/mol normally quoted for combustion is the enthalpy. To compare the two you need to take into account heat exchange with the environment and even the work done by the escaping gas.
Re the first question: the energy density you quote is not a useful quantity in this context. To do thermodynamic calculations you normally only consider changes between initial and final states. The article I linked to above does the calculation in this way.
The efficiciency of electrolysis is always below 100% for various reasons. This can simply be resistive/heating losses, but a problem specific to electrolysis is overpotetials. These cause inefficiencies because if the overpotential is $V$ volts then $V$ electronvolts of energy are lost for each electron.
A: Regarding the first question, I believe that you missed a couple of sentences in the Efficiency section of the Wikipedia article:

It also requires energy to overcome the change in entropy of the reaction. Therefore, the process cannot proceed below 286 kJ per mol if no external heat/energy is added.

If you use 286 kJ/mol, you "don't get something for nothin."
A: You can get an efficiency over 100%.  For a fuel cell efficiency is delta G / delta H and this is 83%, where the loss is due to entropy.  Since an electrolyzer is opposite a fuel cell, its efficiency will be delta H / delta G or thus 119% in theory.  However to reach this efficiency you need to gain entropy?  How could you ever do that?  It's simple-  Your electrolysis device will start cooling.  If it totally isolated it will cool all the way to 100% efficiency.  However it is not isolated, but set in the environment.  What happens is the electrolysis cools, and the surrounding environment dumps heat (i.e. entropy) into the electrolysis device. (Hot temperature transfers to cold temperature).  So this is very weird case in that entropy actually helps you out.  Now in reality the catalytic barriers are very big and this prevents greater than 100% efficiency at all but the smallest currents.
A: Sorry for entering this thread late, but my question is directly related.
I went through most of this for a project last year and at that time I recall articles that the Gibbs free energy of 237 kJ/mol varies with the pressure of the water.
Now it's been a year and I'm not finding the references I recall, but I do keep finding occasional articles claiming this is the case.
Is the Gibbs free energy significantly affected by water pressure? As in is the difference more than ~ 1%?
Second question.
If the water is under very high pressure, such as High Pressure Electrolysis, where the volume of the gas is greatly reduced due to V = n R T / P, does this change things such as bubble formation or methods of collection? Assume P < 40,000,000 Pascals (400 atm).
