# 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?

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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.