Reading the wiki for electrolysis it reads that a potential of 1.23V is needed to separate hydrogen from the oxygen. With that being said, how was that calculated? Does it change with the water volume and conductivity?


1 Answer 1


I'm assuming you mean this Wikipedia article on the hydrolysis of water in which the method of determining that potential difference is described farther down the page.

I'll address your other questions - what might cause this value to change.


No, except see below on concentration. The reaction potential for this electrolysis is independent of the amount of water since it is actually a thermodynamic function. Electrolysis may occur at different rates and take varying time depending on the amount of water, but the potential needed remains the same.


Conductivity is a measure of the ease of transmitting electrical current (electrons) through a substance. So, increased conductivity will increase the rate of electrolysis, but since it does not change the original or final substances in the chemical equation, does not change any thermodynamic values like the reaction potential.

However, there is a danger here. We typically increase conductivity by increasing electrolyte content. Some of these electrolytes could be electrolyzed at a reduced potential compared to water. For example, if you use sodium iodide as your electrolyte, the iodide ion is oxidized to the triiodide ion at -0.53 V:

$$\ce{3I- -> I3- + 2e-}\ \ E^\circ = -0.53\ \pu{V}$$


The reaction potential -1.23 V is the standard reaction potential, which is determined when all applicable species are at 1 molar concentration and 1 bar pressure. The two half reactions in this case are:

$$\ce{2H2O -> O2 + 4H+ + 4e-}\\ \ce{4H+ + 4e- -> 2H2}$$

Thus, this potential is only valid at $[\ce{H+}] = 1\ \pu{M}$ or $\pu{pH}=1$. As the reaction proceeds, water is consumed and the volume decreases. The amount of protons remains constant, so the pH increases. As the pH increases, so too does the potential required and the electrolysis eventually reaches equilibrium and progresses no further unless you are providing an overpotential.

Note that -1.23 V is equally valid at $[\ce{OH-}] = 1\ \pu{M}$ or $\pu{pH}=14$ thanks to Hess's Law.

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    $\begingroup$ The -1.23 Volts also implies a very low current. Any appreciable current necessitates overpotential as well. // Current/Voltage also depends on size and type of electrodes. // Practical considerations imply at least a crude relationship between cell volume and surface area of electrodes. In other words it is hard to have 1 m^2 area electrodes with 1 ml of cell volume. $\endgroup$
    – MaxW
    Commented Jul 22, 2017 at 15:33
  • $\begingroup$ @MaxW Let's take for example the electrodes I'll be using which will be made out of aluminum, how can you theoretically calculate overpotential? Also, is it minus 1.23 Volts or close to that number? $\endgroup$ Commented Jul 22, 2017 at 19:36
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    $\begingroup$ @Tonakis2108 - Sorry, but you obviously are not getting the point. You will almost certainly need over-voltage. The needed over-voltage depends on a dozen nebulous factors. $\endgroup$
    – MaxW
    Commented Jul 22, 2017 at 21:26
  • $\begingroup$ @MaxW Sorry for my ignorance this is all new to me. I am trying to understand electrochemistry as much as I can since it's not the main subject in my college.I am sure I will need over voltage, the answer also stated that as pH increases. This far I know but can it be calculated also? Or do I give a random voltage over 1.23 V? $\endgroup$ Commented Jul 23, 2017 at 0:51

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