Pressure-dependence of electrolysis voltage

I'm trying to better understand electro-chemistry, so please explain why this second-kind perpetual motion device won't work. I guess it's safe to assume it won't work one way or another.

There is a cycle, which starts with low-pressure water.

• The water is pressurized by a compressor. This requires work, but since water is nearly incompressible, the required work is very low: the distance over which the pressurization force is applied is almost zero.
• Next, the high-pressure water is split into high-pressure hydrogen and oxygen gas. This requires a certain amount of electrical work.
• Next, the high-pressure hydrogen and oxygen gas are expanded through turbines, expanding cylinders or other devices that allow the expansion to apply output work. Heat is added to keep the temperature constant during expansion, to maximize the work output. Note that, unlike the compression stage, expansion deals with gases, which are very compressible; therefore, the amount of work produced is substantial.
• To close the cycle, the hydrogen and oxygen gases are recombined in a fuel cell, to end up with the same amount of water, at the same pressure as when we started the cycle. This delivers some electrical work.

Of course there are going to be energy losses; I expect the fuel cell will always yield less energy than is required in the electrolyzer. However, the difference between the compression work and the expansion work can be scaled up arbitrarily by using a higher pressure difference, so at some point this difference in work must yield more energy than is lost in the electrolysis/fuel cell combination. This energy comes from the heat flowing into the expander, but simply turning heat into excess work should not be possible.

So, what is wrong? The only way I see to prevent this to work is if the electrolyzer requires more work per reaction mass (essentially its voltage) than the fuel cell, even if they are both 100% efficient. This would imply that the reaction voltage is pressure-dependent, with a higher voltage at higher pressures. Is this the case? And is this only the case if the output material is more compressible (gas) than the input material (liquid)?

(The same should be true for evaporation/condensation enthalpies, as we could have designed a very similar cycle with evaporation/condensation instead of electrolysis/fuel cell.)

• I'm voting to close this question as off-topic because it's rather physics then chemistry. Mar 24 '19 at 21:45
• IF there are losses in the system (as you admit) then a perpetual motion machine is impossible. And, no, scaling up a mechanical process might get you a higherefficiency, but it can't get more energy out than you put in. There are always losses. Mar 24 '19 at 21:46
• @Mithoron my goal is to learn about the (physical) properties of a chemical process (electrolysis). The physical set-up with compression and expansion is just a tool to achieve this.
– cjp
Mar 24 '19 at 21:50
• 1) This is chemical thermodynamics; while it could (since it's thermodynamics) be asked on the physics site, I think this is a better home for it. 2) The OP isn't claiming a perpetual motion machine is possible. He understands it's impossible, and thus wants to understand what the particular flaw is in the above set of steps that reveal it to be impossible. Mar 24 '19 at 23:48
• Electrochemical potential is pressure dependent. It is just the free energy difference of the systems in equilibrium, i.e. if you produce high-pressure gasses from a liquid, you have to put this extra work in the system in form of higher electrochemical potential. The higher the pressure you use to gain that work, the more energy you will need to split the water, while you get back always the same energy when you use the fuel cell after decompressing the gas.
– Greg
Mar 25 '19 at 2:07

Of course, we're starting by acknowledging a perpetual motion machine is impossible. The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.

The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater than you get back from your step 4, even if we don't have losses. That's because, everything else being equal, the work needed to create gases against a high pressure is greater than that needed to create gases against a low pressure. I.e., $$|w_2| > |w_4|$$.

To understand why, let's simplify things by considering the pressure-dependence of the free energy of a pure substance at constant T: dG = VdP. In step 2, because the pressure is so, high, you have increased the free energies of both the reactants and the products relative to what they'd be at atmospheric pressure. To get the actual changes, you would need to calculate $$\int^{P_f}_{P_i}V(P) dP$$ for the reactants and products. Since the products are gases (oxygen and hydrogen), and the volume of gases is much larger than that of liquids, the free energies of the products would be raised far more than that of the reactant (a liquid). Thus the separation between the free energies of the reactants and products, (i.e., the magnitude of $$\Delta G_r$$) for step 2 (which is at high pressure), will be substantially greater than it is for step 4 (which is at low pressure).

You could also understand it using Le Chatelier's principle: The volume of the products is higher than that of the reactants, and thus higher pressure will favor the reactants over the products.

• This sounds informative, but I'm not that familiar (anymore) with free energies (especially in relation to chemical reactions). Does this translate into a higher required voltage for the electrolyzer, compared to the fuel cell, as I theorized in my question?
– cjp
Mar 24 '19 at 22:15
• It translates into more energy from the electrolyzer. Energy = voltage x charge, and charge = current x time, so energy = voltage x current x time. Thus, for constant current and time, it translates into more voltage. Mar 24 '19 at 23:38
• I think it might be easier to understand in terms of the Nernst equation, which directly relates the voltage to the concentration or pressure of the reactants and products. Higher pressures of product gases means higher voltage required. Mar 25 '19 at 17:14

As you noted, there will be some definite energy losses, however I think it's important to keep in mind as well, that electrolysis is a slow process, like charging an electric car. You can speed it up by applying more current, but ultimately there is a theoretical limit before your electrodes are toast (I nearly experienced that my first time experimenting with this stuff!).

I'm not sure what you mean by "the oxygen and hydrogen gases are expanded into turbines," as yes, they can be pressurized, but sooner or later they are going to find a way to escape the container. Do you mean simply let loose like air from an air compressor, or something a little more specialized? Either way, I'm excited for you! Keep up the good work and don't forget to keep your friends on Stack Overflow posted on your progress!!!

I arrived to your question and it seems that is an answer to another question I had. Let's examine the processes:

We can increase the pressure of water with very little work, while the volume of water will remain almost constant. This is because ... I will try to explain it naivelly ... there is no much available free space between molecules in order to allow for compression.

Then we divide water into oxygen an hydrogen keeping the volume constant. We have to understand that at high pressures, both hydrogen and oxygen behave as "water", this means that "there is still no space available".

If you look at the curves of compressibility, you will notice that gases are compressible between 0 and 200 bar even more than an ideal gas, while at 200 bar they arrive to be again as compressible as an ideal gas. Afterwards, they become compressible less than an ideal gas and with rising difficulty as the pressure rises.

So by compressing water you cannot store much energy, and similarly by compressing high pressure gases you cannot store much energy. Here the "store energy" is wrong but more on that later on.

So suppose you have water at 10.000 bar pretty a high pressure by the way, then you will get H2 and O2 at 10.000 bar or even more, but as water does not give back energy when uncompressed, similarly you will not be able to get energy from ultra high pressure H2 and O2 until you arrive to human pressures of 500bar or less and you start to be able to decompress and get energy.

Thanks a lot for the question as it gives ideas to work on.

The question has to do with physics and chemistry both.

When you increase pressure of e.g. aqueous vapor, you arrive to have water and solid water afterwards. Similarly when you increase the pressure of oxygen you start to have liquid oxygen and solid oxygen but that is valid when you are under the critical temperature ( that for oxygen is −118.6 °C) . When you are above the critical temperature, there is only a status of gas-liquid oxygen that is in the same phase irrespective of the pressure.