I need to calculate the power needed to electrolyse 1 mole of water for a project. We will either use Polymer Electrolyte Membrane Electrolysis or Alkaline Water Electrolysis, they operate on 1.75V-2.20V and 1.8V-2.4V respectively. And each has a maximum efficiency of about 69%. So, I have done further calculations assuming that the potential difference between the electrodes is 1.8V and the efficiency is 70%. As per my requirements, I have also considered that I have 8 minutes for conducting electrolysis.

Here are the calculations :

The reaction : 2H2O 2H2 + O2
Half Reactions:
Reduction at cathode: 2 H+(aq) + 2e−  H2(g)
Oxidation at anode: 2 H2O(l)  O2(g) + 4 H+(aq) + 4e−

1 mole of hydrogen gas will be produced from 1 mole of water ie 18 g of water or 18 cc of water.
Time available for electrolysis = 8 minutes ie 480 seconds
Number of moles of electrons = 2
Charge = 2 x 96,485 C/F = 192,970 C
Current = Charge / Time = 192,970 / 480 = 400 Amperes (approx.)

Considering 70% efficiency, Current required = 570 Amperes
Potential Difference = 1.8V (common to both PEM electrolysis and Alkaline water electrolysis)

So, Power Required = 1026 Watts ≈ 1 kW (for 8 minutes per mole)

I took reference of this webpage to learn how to calculate it.

However, then I came across this figure on the internet - per mole water splitting requires at least 241.8 kJ of energy Source: The first answer to this question.

So I am a little confused. I guess, this 241KJ energy figure came out because the person might be considering 1.23V potential difference (which is considered ideal), and 100% efficiency. I can't say anything because he has stated, that the values of enthalpy of formation (which he has used) are experimental, but hasn't provided the voltage and neither other conditions for the reaction. He has just used the values of enthalpy of formation and not considered the efficiency of the method in use, in fact, that figure seems to be so general for all methods. As, PEM Electrolysis and Alkaline Water Electrolysis are the most efficient methods and both don't operate on that "ideal" 1.23V potential difference. Also, I think my reasoning is correct because if I redo the calculations considering 1.23V and 100% efficiency, the answer comes out to be the same as his.

Is my reasoning correct? and if not what is the reason for having different answers.

  • $\begingroup$ The essential topic for electrolytic efficiency is overpotential, especially overpotential of oxygen. $\endgroup$
    – Poutnik
    Jun 8, 2019 at 5:26

1 Answer 1


They said it right, "... requires at least 241.8 kJ ..." . You just found that one can simply calculate the molar formation enthalpy from the cell voltage, and vice versa, those two informations are redundant.

Of course you cannot run the electrolysis at the minimum voltage, because it will be extremely slow! You would just balance the electrochemical potential of the cell (created by the reverse reaction as soon as you make a small amount of hydogen and oxygen).

To make any actual electrolysis, you need a higher voltage, and the difference is converted into heat at the internal ohmic resistance of your cell. The minimum additional voltage is called the cell overvoltage, below that you don't get electrolysis. It is necessary to be able to break through the polarised electrolyte layers at the electrodes. So you cannot get towards 100% efficiency by running very slow.

  • $\begingroup$ Thanks. I had a question : you said that more voltage means speed of the reaction will increase. But according to what I have calculated (and according to that perdue web page: chem.purdue.edu/gchelp/howtosolveit/Electrochem/…) voltage is independent of the speed. So, could you please check my calculations and tell where have I gone wrong and why isnt my answer coming out to be the same as 241KJ. $\endgroup$ Jan 2, 2019 at 10:23
  • $\begingroup$ Only to a very small extent: The diffusion speed of ions in your electrolyte is limited (by diffusion). So there is a small voltage range in which the elecrolysis rate increases, below it is practically zero, above the resistance of the cell simply increases, because diffusion limits the current. $\endgroup$
    – Karl
    Jan 2, 2019 at 17:10

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