# Why are H+ and OH- ions preferentially discharged in the electrolysis of NaCl(aq)?

There are 4 types of ions in an $\ce{NaCl(aq)}$ solution: $\ce{H+}$, $\ce{OH-}$, $\ce{Na+}$ and $\ce{Cl-}$. According to the E.C.S. table, $\ce{H+}$ and $\ce{OH-}$ should preferentially discharge at cathode and anode, respectively.

However, according to the Wikipedia page on self-ionization of water, water only dissociates slightly into $\ce{H+}$ and $\ce{OH-}$ in neutral solution. (This is the reason why pure water conducts electricity poorly). Therefore, in the $\ce{NaCl(aq)}$ solution, the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely small, and by comparison those of $\ce{Na+}$ and $\ce{Cl-}$ are much higher. As a result, it seems to me that $\ce{Na+}$ and $\ce{Cl-}$ should be preferentially discharged because of the concentration effect.

However, my chemistry textbook says that in fact $\ce{H+}$ and $\ce{OH-}$ are preferentially discharged, and hence the main products of the electrolysis are hydrogen and oxygen. So apparently my understanding is wrong. But where?

• – Mithoron Mar 12 '18 at 20:04
• @Mithoron Related, yes, but not duplicate. – hBy2Py Mar 12 '18 at 20:27

Water does not need to be dissociated into ions in order to participate in electrochemical reactions. While it is common to refer to electrode reactions as "discharging" this or that species, there is no requirement that the reactant of an electrode reaction must always be an ion.

In neutral solution, it's quite common for the water electrolysis reactions to be written as:

\begin{align} \text{Cathode:} &\quad \ce{4H2O + 4e- -> 2H2 + 2OH-} \\ \text{Anode:} &\quad \ce{2H2O -> O2 + 4H+ + 4e-} \end{align}

It's absolutely the case that the electrochemistry written using the ions as reactants in many cases is likely to be more facile:

\begin{align} \text{Cathode:} &\quad \ce{4H+ + 4e- -> 2H2} \\ \text{Anode:} &\quad \ce{4OH- -> O2 + 2H2O + 4e-} \end{align}

But it's still far (far) more favorable for $\ce{H2O}$ to be reduced than $\ce{Na+}$, and in many cases more favorable for $\ce{H2O}$ to be oxidized than $\ce{Cl-}$. You're quite right that the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely low in neutral salt solutions, but that fact is mostly irrelevant to the electrochemistry.

• So, water is actually discharged in the form of complete molecules, rather than in the form of H+ and OH- ions? Also, according to your first two equations, water is simultaneously reduced and oxidized, so this is actually an example of disproportionation? – user161070 Mar 13 '18 at 7:03

It isn't true. In "chlor-alkali" electrolytic processes it is chloride ions that get discharged at the anode, not water. See here for an introductory look.

• Is it still false if the solution is in very dilute condition? Also, how about sodium ions? – user161070 Mar 12 '18 at 15:06
• Well ... there is even a setup where sodium is reduced instead of hydrogen, although this cell, where they form an amalgam, is on its way out. Industrially the brine has to be concentrated so nothing is said in the article about very dilute solutions; but there is such a thingcas running out of chloride ions thus forcing water oxidation. – Oscar Lanzi Mar 12 '18 at 15:36
• Thanks, but actually my doubt is about the "claim" of the question. Maybe I should ask the question in this way: Is the following claim true: Water only dissociate very slightly into H+ and OH− by itself (which is the same reason why pure water almost doesn't conduct electricity). Therefore, in any aqueous solution, the concentration of other ions will be a lot higher than that of ions of water. If we take this aqueous solution into an electrolysis process, the ions of water will never get preferential discharged, because of the way higher concentration of other ions. – user161070 Mar 12 '18 at 16:11