I've heard from a few sources that over-voltage in a electrolysis cell for water will cause a greater heat buildup with no yield. Is this true, and if it is, why is it true? From my electronics background, voltage can be though of as electric potential, or thinking of it as water in a pipe for an analogy it is the strength of the substance flowing through the pipe, or as the strength of each electron, so why does it cause extra heating with no appreciable/no yield?


To first order, the power dissipated in an electrochemical cell is equal to that of an electrical circuit: $P=IV$. Thus, as the overpotential necessary to achieve a given current throughput increases, the power consumption increases also.

In the fluid flow analogy, think of an elevated overpotential as a throttling orifice inserted into one of the pipes. In order to keep the water moving through the pipe at the same volume flow rate, you have to increase the pressure drop across the system. A greater pressure drop to achieve the same flow rate requires a pump with a greater power rating.

  • $\begingroup$ Wow, that was really a duh type moment, I didn't even think to go to the power function. $\endgroup$ – Sarah Szabo Mar 21 '16 at 2:14
  • $\begingroup$ @SarahSzabo <shrug> It wasn't obvious to me, when I first started working in earnest with electrochemistry. Why should it be that an electrochemical boundary layer behaves at all like a circuit component, in terms of power dissipation? But, it apparently does. $\endgroup$ – hBy2Py Mar 21 '16 at 2:30

You regard water in pipe as electrons in conductor, it's well. No yield means water pressure is not enough to wash the barrier away, increasing the water pressure is needed, the increment of water pressure is overvoltage. The barrier usually comes from polarization of electrode reaction (electrolysis of water).

  • $\begingroup$ This looks like a bunch of random Markov chains. At least NAA $\endgroup$ – M.A.R. Mar 22 '16 at 10:44
  • $\begingroup$ @IͶΔ No, the underlying thread here is pretty accurate, actually, though mostly duplicative of what I wrote. The last sentence, actually, is important for someone to have mentioned, that polarization of the interface is part of the physical explanation for the overvoltage. Definitely needs clarification. $\endgroup$ – hBy2Py Mar 22 '16 at 12:08

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