Re: How efficient is the electrolysis of water?

The efficiency depends fairly critically on how fast, and on what scale you 
are trying to generate the gases.

A relevant discussion that gives data that enable efficiency for one 
typical set of conditions to be worked out is contained in some lecture notes 
published on the web by Dr Joseph McNeil of Chatham College. The relevant 
section is headed 'Electrochemistry and Overvoltage' and is found about 2/3 
way down the very long page.

For platinum electrodes, we are told, at a current density of 0.01 amp/cm2, 
an overvoltage of 0.07 volt is required for evolution of hydrogen, and 0.40 
volt for oxygen evolution.

In a cell with an electrolyte of 0.5 M sulfuric acid, the standard cell 
potentials are

2 H+ + 2 e- ---> H2    Eo = 0.00 volt
O2 + 4 H+ + 4 e- ---> 2 H2O    Eo = 1.23 volt

so a reaction that requires only 1.23 volt to drive it as a reversible 
reaction in thermodynamic terms needs 1.23 + 0.07 + 0.40 = 1.70 volt in 
practice at these current densities. The efficiency is therefore 1.23/1.70 
=72%.

What do the conditions mean? A current density of 0.01 amp/cm2 corresponds 
to 100 amp/m2. So a cell with two platinum electrodes, each of 1 m2 
effective surface area, and drawing 100 amps of current, will have the 
appropriate current density. Such a cell would take 965 seconds = 16 
minutes to generate 0.5 mole = 12 litres of hydrogen gas and half the 
volume of oxygen gas.

Its total power consumption would be 100 amp * 1.7 volt = 170 watt, of 
which 28% or 47 watt would be dissipated as low grade heat at the 
electrode/electrolyte interface. And of course all of this assumes a design 
such that the internal resistance of the cell and the resistance in the 
rest of the electrical circuit are negligibly small.