Applying an electric potential to impure water splits water into its constituents H/O.

Does the temperature of the electrolyte change during this electrolysis?

  • 2
    $\begingroup$ ""Applying an electric potential to impure water splits water into it's constituent H/O."" This is wrong, just "potential" does change nothing. One needs to have a current flowing to achieve electrolysis. And if you read a physics book first, you may learn about currents and heat. $\endgroup$
    – Georg
    Aug 25, 2013 at 18:53
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    $\begingroup$ Because the electrolyte has a finite resistance, heat will be generated based on the power, $I^2R$, the time of electrolysis (since $power=\frac{energy}{time}$) and the heat capacity $C=\frac{energy_{heat}}{\Delta T}$. This would be a very rough estimate since it assumes the solution is the greatest source of resistance. $\endgroup$ Aug 26, 2013 at 1:33
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    $\begingroup$ However, Wikipedia states that there are cases where the temperature drops because of the electrolysis (“in the electrolysis of steam into hydrogen and oxygen at high temperature, […] heat is absorbed from the surroundings”). So it may not be as easy as Joule effect… $\endgroup$
    – F'x
    Aug 26, 2013 at 8:05
  • $\begingroup$ @F'x Interesting point, although this is a case where a reference on the Wikipedia page would be beneficial since this is a fairly unexpected phenomenon. $\endgroup$ Aug 26, 2013 at 11:20
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    $\begingroup$ @AlexanderGuyer The comments are mixing a variety of concepts. A temperature drop during electrolysis of water is unexpected. If we now complicate the problem to incorporate gas phase electrolysis with phase changes, then all bets are off. My comment was in reference to the lack of decent citations in the Wikipedia page (9 years ago). $\endgroup$ Sep 3, 2022 at 13:56

1 Answer 1


Since you have not given the voltage you are applying to the system I assume that there is excess voltage meaning to spilt water needs about 1.23 V (quick wiki)


$$\ce{ 2H2O(l) → O2(g) + 4H+(aq) + 4e- -1.23 V} $$


$$\ce{2H+(aq) + 2e- → H2(g) 0.00 V}$$

To get more current to flow more exactly to get more products you will need to supply more voltage thus the excess voltage and current for that voltage will convert into heat according to the fundamental law of nature the energy conservation law.

Total Electric Energy Input gets converted to Chemical Energy + Heat (if exceeds voltage is appied)

$$\ce{V_{input} * I = (1.23 V) * I + Heat}$$

If you take the excess heat energy out of the system (I.E heat sink or what ever means) keep the system temperature constant then $\pu{1.23 V}$ will remain constant otherwise the potential will change according to the Nernst Equation.

Please also note that $\pu{1.23 V}$ comes due to standard state $\pu{ T= 298.15 K }$

  • 1
    $\begingroup$ +1 Also, even if you do nail the reaction voltage, that voltage just refers to the potential difference across the interface between the electrode and the electrolyte. You'll necessarily have to apply a little extra voltage -- to compensate for resistance i.e. voltage drop in the rest of the circuit -- to reach the required interface potential. Some of that circuit will be outside the electrolyte but most will be submerged conductors and also the electrolyte itself. The heat dissipation is given by $P=I^2R$ where $I$ is the current and $R$ is the resistance of the rest of the circuit. $\endgroup$
    – Jason C
    Nov 15, 2021 at 18:23
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    $\begingroup$ (Other forms of the dissipation equation can be readily derived from $V=IR$ and $P=IV$, too). $\endgroup$
    – Jason C
    Nov 15, 2021 at 18:27

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