I find a similar question in stack exchange (What will happen if the electrolysis of Brine takes place for a long time?), yet it does NOT seem to respond to my question.

Suppose I am electrolysing a very dilute sodium chloride solution. At the anode, the hydroxide ions will get oxidised into oxygen gas, while at the cathode, the hydrogen ions will get reduced into hydrogen gas.

Now, if I leave the set-up ongoing for a long time, is it true that all the water will be consumed and I will obtain a beaker of molten sodium chloride (given that heat is constantly provided)?

My lecturer pointed out that this is impossible to happen. When a lot of water is consumed, then I will obtain concentrated sodium chloride solution, namely brine. So, at the anode, the chloride ions will get oxidised to give chlorine gas, while at the cathode, the hydrogen ions are still reduced to give hydrogen gas.

Can anyone provide me with the whole picture of the case provided by me? Please be noted that I am talking about very diluted NaCl solution instead of brine.

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    $\begingroup$ Well, your scenario becomes this earlier scenario with time, so it kinda does answer. $\endgroup$
    – Mithoron
    Commented Nov 28, 2023 at 13:31

2 Answers 2


The redox potentials cited by Harikrishnan are all valid for ionic concentrations equal to $1$ M. As soon as these concentrations are different from $1$ M, these potentials are modified according to Nernst's law. This is the case here, because in $\ce{NaCl}$ solution, the concentration, for example, of $\ce{H+}$ ion is not $\ce{1}$ M. It is around $\ce{10^{-7}}$ M. The same sort of deviation may occur for all ions in this electrolysis, including the oxidation of chloride ion to chlorine, as the present solution is "very dilute". As a consequence, all potentials should be recalculated to be adapted to a "rather dilute" and nearly neutral solution of $\ce{NaCl}$. Unfortunately this cannot be done here, as the concentration of $\ce{NaCl}$ is not known.

The potentials will change, but the only useful result is that $\ce{H2}$ is still produced at the cathode according to : $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ Depending on the concentrations, both $\ce{Cl2}$ and $\ce{O2}$ are produced at the anode, according to the following equations : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ $$\ce{2 Cl- -> Cl2 + 2 e-}$$ Note that the ions $\ce{H+}$ from the anode react with the $\ce{OH-}$ from the cathode, and produce new water molecules. If the discharge of $\ce{Cl-}$ is momentarily forgotten, electrolysis by $4$ electrons consumes $6$ $\ce{H2O}$ and produce $4$ $\ce{H+}$ plus $4$ $\ce{OH-}$ that recombine to form $4$ $\ce{H2O}$. As a result, water is consumed. As the amount of $\ce{Na+}$ ion is not destroyed, the solution becomes more concentrated in $\ce{NaCl}$. It also contains a little bit of $\ce{NaOH}$, as there are two possible reactions at the anode, and only the one starting with $\ce{H2O}$ produces $\ce{H+}$ ions. So the solution becomes slowly basic. After a rather long time, all water is destroyed, and solid $\ce{NaCl}$ plus $\ce{NaOH}$ remained in the electrolysis cell, as Freeby Freeby mentioned it. The trouble is that, at the end, you cannot prevent $\ce{Cl2}$ from reacting with $\ce{NaOH}$ according to : $$\ce{Cl2 + 2 NaOH -> NaClO + NaCl}$$ so that the solution is a bleach, and the evaporation of bleach gets a disproportionation according to : $$\ce{3 NaClO -> NaClO3 + 2 NaCl}$$

As a final result the powder obtained on the long range, out of an electrolyzed solution of $\ce{NaCl}$, is made of $\ce{NaCl}$ plus a few percent of $\ce{NaOH}$ and $\ce{NaClO3}$

  • $\begingroup$ I understand the complexities of the situation now Maurice! Thank you. Further, it's nice that our answers both complement each other. $\endgroup$ Commented Nov 29, 2023 at 13:41
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    $\begingroup$ Thank you Harikrishnan. You should also know that, in the last three lines of your initial answer, you have interchanged the words ANODE and CATHODE. The anode must produce electrons. The cathode mut consume electrons. $\endgroup$
    – Maurice
    Commented Nov 29, 2023 at 13:58
  • $\begingroup$ Yes Maurice I will edit it. $\endgroup$ Commented Nov 29, 2023 at 13:59
  • $\begingroup$ Looking at your answer I get an exciting application. If this process was made in brine it would result in the formation of bleach in higher yield is it not? $\endgroup$ Commented Nov 29, 2023 at 14:02
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    $\begingroup$ @Harikrishnan. The electrolysis of NaCl solutions is the standard way to produce bleach in industry. $\endgroup$
    – Maurice
    Commented Nov 29, 2023 at 21:58

During the electrolysis of aqueous sodium chloride solution, the products are $\ce{NaOH}$, $\ce{Cl2}$ and $\ce{H2}$ . In this case besides $\ce{Na+}$ and $\ce{Cl–}$ ions we also have $\ce{H+}$ and $\ce{OH–}$ ions along with the solvent molecules, $\ce{H2O}$.

At the cathode there is competition between the following reduction reactions:

$\ce{Na+ (aq) + e– → Na (s) : E = – 2.71 V}$

$\ce{H+ (aq) + e– → ½ H2(g): E = 0.00 V}$

The reaction with higher value of $E^⊖$ is preferred and therefore, the reaction at the cathode during electrolysis is the 2nd one.

But H+ (aq) is produced by the dissociation of H2O, i.e.,

$\ce{H2O (l) → H+ (aq) + OH– (aq)}$

Therefore, the net reaction at the cathode may be written as the sum of above 2 reactions:

$\ce{H2O (l) + e– → ½H2(g) + OH–}$

At the anode the following oxidation reactions are possible:

$\ce{Cl– (aq) → ½ Cl2 (g) + e– : E = 1.36 V}$

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

The reaction at anode with lower value of $E^⊖$ is preferred and therefore, water should get oxidised in preference to $\ce{Cl–}$ (aq). However, on account of overpotential of oxygen, the 1st reaction is preferred. Thus, the net reactions may be summarised as:

$\ce{NaCl (aq) ->[H2O] Na+ (aq) + Cl– (aq)}$

CATHODE: $\ce{H2O (l) + e– → ½H2(g) + OH–}$

ANODE: $\ce{Cl– (aq) → ½ Cl2 (g) + e–}$

NET REACTION: $\ce{NaCl(aq) + H2O(l) -> Na+(aq) + OH-(aq) + 1/2H2(g) + 1/2Cl2(g)}$

Source: NCERT [Refer page 85 / 23rd page in the PDF.]


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