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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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  • $\begingroup$ Check your numbers for energy of electrolysis. In physical chemistry textbooks, there are very small "over voltages" required to start electrolysis, and those over voltages add to the energy of electrolysis. However, there are two elements in the periodic chart that have abnormally large over voltages, and as you probably guessed, those elements are hydrogen and oxygen. Such high over voltages add substantially to the electrical energy required for electrolysis, and it would be wise to verify the number that you used. $\endgroup$ Mar 11, 2021 at 1:14

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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.

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  • $\begingroup$ Can you clarify to me - are the calculations that I gave correct in IDEAL conditions (no loses, absolute efficiency)? To put it in another way, if someone is able to perform electrolysis with no loses would he be able create more energy from resulting gases than he used? $\endgroup$
    – Dusan
    Apr 30, 2012 at 14:43
  • $\begingroup$ The simple answer in no, unfortunately there are no perpetual motion machines. The more complicated answer is that if you just consider the gas, and not the water that it came from, you can get more energy from burning the gas that you put in as electricity. However this is misleading because you need to include the change in free energy of the water bath. $\endgroup$ Apr 30, 2012 at 14:57
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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."

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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.

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  • $\begingroup$ I disagree. NO practical process produces an overall efficiency of more than 100%, or it would be a perpetual motion machine which violates the 1st Law of Thermodynamics. $\endgroup$ Mar 11, 2021 at 1:20
  • $\begingroup$ @DavidWhite This does not violate the 1st law because the extra energy you are getting is heat from the environment. dU = dW + dQ. eff = dU/dW can be greater that 100% if dQ is positive. However, this does not allow for perpetual motion because the flow of heat is defined by the entropy change, and will switch sign when running a fuel cell. In a fuel cell you must give heat to the environment, leading to an efficiency less than 1. $\endgroup$ May 26, 2021 at 4:19
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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).

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If the electrolysis is conducted at a high temperature, e.g. 800 or 900 C, which requires some attention to materials, then the T * delta S term can push the apparent electrical efficiency beyond 100%. The resulting energy storage captures some of the energy from the thermal reservoir.

This approach is relevant for solar thermal and nuclear installations where high-grade process heat is available for much less than the cost of burning coal or hydrocarbons. Currently being commercialized by Bloom Energy, probably with backing from Bill Gates.

The researchers claimed 15 +/- years ago that the electrical efficiency is greater than unity, but the claims from Bloom address efficiency relative to competitive technologies. Beyond the benefit of high temperature on the entropic boost, the high temperature also benefits the ionic conductivity of the solid oxide material.

Disclaimer: I am not a paid shill.

The World’s Largest and Most Efficient Solid Oxide Electrolyzer https://www.bloomenergy.com/bloomelectrolyzer/

Bloom Energy has begun generating hydrogen from the world’s largest solid oxide electrolyzer installation at NASA’s Ames Research Center, the historic Moffett Field research facility in Mountain View, Calif. This high-temperature, high-efficiency unit produces 20-25% more hydrogen per megawatt (MW) than commercially demonstrated lower temperature electrolyzers such as proton electrolyte membrane (PEM) or alkaline.

This electrolyzer demonstration showcases the maturity, efficiency and commercial readiness of Bloom’s solid oxide technology for large-scale, clean hydrogen production. The 4 MW Bloom Electrolyzer™, delivering the equivalent of over 2.4 metric tonnes per day of hydrogen output, was built, installed and operationalized in a span of two months to demonstrate the speed and ease of deployment. I don't think that is mentioned in the marketing materials. ...

Running at high temperatures and high availability, the pilot results reveal the Bloom Electrolyzer is producing hydrogen at 37.7 kWh per kilogram of hydrogen. Alternative electrolyzer technologies, such as PEM or Alkaline, consume as much as 52 – 54 kWh per kilogram of hydrogen produced. ...

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