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This is the reaction:

$$\ce{2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g)}$$

If the above reaction goes on for long enough, we will eventually run out of Cl- ions. There will only remain an aqueous solution of NaOH in the pot as the hydrogen and chlorine gases have floated away.

Then the electrolysis of NaOH solution will start, and H2 gas will be produced at the cathode and O2 gas will be produced at the anode. Now, if this electrolysis also goes on, at some point, only solid NaOH will remain in the pot.

Is my analysis correct?

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If the electrolysis is done without a separation membrane, the chlorine gas will partly be dissolved in the solution and react with NaOH according to : $\ce{2 NaOH + Cl2 -> NaClO + NaCl + H2O}$. So the composition of the solution changes gradually. The solution becomes bleach.

If a membrane separates both electrode zones, you may be nearly right. After a very, very long time, the water will be nearly totally consumed. But you will not obtain NaOH solid, because the contact between both electrodes will be interrupted when the NaOH is highly concentrated. The remaining $\ce{NaOH}$ drops will be more and more separated, and the contact between them become poor. The inner resistance of the system will increase a lot, so that the current will become smaller and smaller, to come to zero before total consumption of water.

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  • $\begingroup$ I've put more thought into this, and have realized that this is a complicated matter. Let me run the whole process, as I understand it, by you. Let us assume that we have a 1L 1M NaCl aqueous solution. So, we have 1 mole Na+ and 1 mole Cl- ions roaming around in the water, and we have 55.55 moles of H2O. Now, 10^-7 mole of that 55.55 mole will dissociate into H+ and OH- ions. So, number of mole of H+ ions = number of mole of OH- ions=10^-7 mole. Now let us dip two graphite electrodes in the solution. We have two positive ions: H+ and Na+, and we have two negative ions: Cl- and OH-. (...) $\endgroup$
    – tryingtobeastoic
    Commented May 30, 2022 at 6:43
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    $\begingroup$ @tryingtobeastoic There is the important note that cathodic/anodic reactions are not limited to cations/anions. Water can be reduced at cathodes to hydrogen or oxidized at anodes to oxygen as well. Electrolysis of concentrated KOH solution can hardly rely on negligible content of H+(aq). Similarly, for H2SO4 solutions, oxidation of water is much easier than of sulphate. $\endgroup$
    – Poutnik
    Commented May 30, 2022 at 8:09
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    $\begingroup$ @tryingtobeastoic OH- does not seem to me like water, neither does H+(aq). // Consider rather including you later thoughts and analysis into the question elaboration, (or own answer if it answers your question), instead of in extensive comments. It is not good if sum of your comments is much longer than the question itself. $\endgroup$
    – Poutnik
    Commented May 30, 2022 at 8:21
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    $\begingroup$ @tryingtobeastoic $\ce{2 H2O + 2e- -> H2 + 2 OH-}$ // $\ce{2 H2O -> O2 + 4 H+ + 4 e-}$ $\endgroup$
    – Poutnik
    Commented May 30, 2022 at 8:24
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    $\begingroup$ @tryingtobeastoic I have not advised posting a new question. Focus on improving quality of your existing questions. $\endgroup$
    – Poutnik
    Commented May 30, 2022 at 9:25

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