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I'm not at all an expert in electrochemistry and want just understand a little what is going on:

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I have a NaCl solution in water, connected most simply by plain copper conductors. My observation is: When the switch is closed, the electrolyte is connected to 5V. After opening the switch, I measure a slight voltage of ~100mV. What could be the reason for this voltage? It seems, that the arrangement is slightly (!) charged like a battery and inherits a voltage source - could that be and if yes, what are the reactions going on?

BTW, when I switch polarity of the battery, also the remaining voltage will invert to a negative voltage of about ~100mV. So it is not just some offset voltage of the voltmeter.

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    $\begingroup$ Consider the electrolytic cell as a very very inferior rechargeable cell. The residual voltage is caused by differences of composition of electroactive particles at both electrodes, created during electrolysis.. $\endgroup$
    – Poutnik
    Mar 7, 2022 at 14:28
  • $\begingroup$ So what I observe is quite normal...? $\endgroup$
    – MichaelW
    Mar 7, 2022 at 14:30
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    $\begingroup$ It will serve you better if you try to answer it yourself. $\endgroup$
    – Poutnik
    Mar 7, 2022 at 14:31
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    $\begingroup$ In fact, you need not any knowledge of chemistry for that, just read carefully the 1st comment. Said by other words, a perfectly symmetric cell creates zero voltage output. If some process like electrolysis breaks this symmetry, it produces nonzero voltage. $\endgroup$
    – Poutnik
    Mar 7, 2022 at 14:46
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    $\begingroup$ @MichaelW I encourage you to ask this question over on electronics.stackexchange.com. Because we discourage cross-posting, you'll need to delete the question here. I have no doubt you will get several satisfactory explanations from the electronics/electrical engineering crowd. $\endgroup$ Mar 7, 2022 at 15:31

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The electrolysis of $\ce{NaCl}$ solutions is not the best way to handle this problem, because of secondary reaction occurring during the electrolysis between $\ce{Cl2}$ produced at the anode, and $\ce{OH-}$ ions produced at the cathode.

Let me first describe a simple case : electrolyzing an aqueous solution of sulfuric acid $\ce{H2SO4}$. The following reactions are carried out on the electrodes : $$\mathrm{Cathode, (-) } : \ce{2H+ + 2 e- -> H2}$$ $$\mathrm {Anode, (+)}: \ce{2 H2O -> 4 H+ + O2 + 4 e-}$$ At the end, a couple of bubbles of $\ce{H2}$ remain adsorbed on the cathode and $\ce{O2}$ on the anode. Of course, much more of these gases have been produced previously, which have been lost by bubbling and getting out of the solutions around the electrodes. But the few remaining bubbles on the electrodes make up a weak galvanic cell. When the switch is off, this cell is able to produce a small current for at least a couple of seconds, according to the equations $$\mathrm{Anode, (-)} : \ce{H2 -> 2 H+ + 2 e-}$$ $$\mathrm{Cathode, (+)}: \ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$ This behavior does not last long, because of the ridiculous small amount of $\ce{H2}$ and $\ce{O2}$ available on the electrodes.

It is worth while pointing out that $\ce{H2}$ is produced by electrolysis at the cathode (negative pole). But at the end of the electrolysis, when the two electrodes are directly connected, the system is a galvanic cell. The electrode holding the $\ce{H2}$ gas is still the negative pole, but it is the anode now. Anode is always the electrode where reduction happens.

With solutions of $\ce{NaCl}$, the behavior is similar, although the chemistry is more complicated, because of secondary reactions occurring during the electrolysis between $\ce{Cl2}$ produced at the anode, and $\ce{OH-}$ ions produced at the cathode, like : $$\ce{Cl2 + 2 OH- -> Cl- + ClO- + H2O}$$ $$\mathrm{and, when } \mathrm{hot } : \ce{3 ClO- -> ClO3^- + 2 Cl-}$$

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