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?
Best Answer
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.