This answer to this post addresses "what happens if you dip a single zinc electrode into some electrolyte solution." It explains that

However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode.

This seems very counter-intuitive to me. The word "can" seems important. Perhaps it is a point of minor importance that the author did not think through. In the absence of an applied external potential can or does immersing metallic zinc in water result in the surface retaining electrons while ions diffuse into the water? Why are neutral zinc metal atoms insoluble in water, or are they soluble, just very reactive? The questions are partly addressed in comments below the answer and links to articles are provided, but this seems worthy of a separate post (it may have been asked already I realize).

What actually happens to zinc when immersed in pure water?

Edit: assume the initial conditions as anaerobic pure water (no electrolyte), room temperature and pressure.

  • 7
    $\begingroup$ Have a look at Gouy-Chapman-Stern (and related) theories/methods for interface at metals/solutions, in short there is a structured area which is diffusion-limited for ions near the surface and then a continuum as some distance thereafter. Shells or planes can be defined, I haven't time to elaborate but your answer might be found with some targeted Googling. $\endgroup$ Commented Mar 19 at 14:52
  • $\begingroup$ Thanks @ToddMinehardt There's a lot happening from the moment you dip the metal in water. My question is perhaps like asking what happened seconds after the Big Bang. $\endgroup$
    – Buck Thorn
    Commented Mar 20 at 16:47
  • $\begingroup$ This study describes what happens in the presence of air and electrolyte over time: sciencedirect.com/science/article/pii/S2452199X18300240 $\endgroup$
    – Buck Thorn
    Commented Mar 20 at 17:57
  • $\begingroup$ I have a related question about self-diffusion of metal atoms/ions at this interface: Let's say we have the natural abundance of Zinc isotopes in the solution in the form of a soluble salt. The metal however is 99% enriched with the 68Zn isotope, which usually has a natural abundance of 4%. If we let the metal rest in the solution for a long time, are we going to see an exchange of the metal 68Zinc atoms and solution Zinc ions? So basically the self-diffusion of neutral 68Zinc and ionized Zinc in the solution? Or should there be a huge energy barrier for this kind of self-diffusion? $\endgroup$
    – Timo
    Commented Mar 26 at 7:59
  • $\begingroup$ @Timo If you have a question please create a new post rather than post the question as an answer to an existing question. The site does not function as a forum. Each page corresponds to one question and potentially multiple answers to the same question. $\endgroup$
    – Buck Thorn
    Commented Mar 26 at 8:51

3 Answers 3


We can expect one of three things to happen:

  1. The electrons also diffuse out from the electrode, becoming solvated electrons.

  2. If the electrons cannot diffuse, their negative charge creates an electric field that opposes the diffusion of the zinc ions, so the latter are reined in.

  3. The electrons react with the water to evolve hydrogen and leave behind hydroxide ions, which then do (1) or (2) plus maybe undergoing secondary reactions with the zinc ions (precipitation, complexation).

With zinc in water we don't see evidence of (1) or (3), but (2) requires no visible change and thus is an experimentally plausible option. Magnesium just barely, and alkali and heavier alkaline earth metals more prominently, give (3). Alkali metals in ammonia solvent readily give (1) combined slowly with (3).

  • $\begingroup$ Thanks, this begins to address the mechanistic problem. Do you have suggestions for supporting references? After spending some time googling it seems that eventually water will react with Zn to form passivating oxide and release H2. I suspect Zn never leaves the electrodes. $\endgroup$
    – Buck Thorn
    Commented Mar 20 at 16:51

If a zinc plate is dipped into pure water, maybe a small number of zinc atoms produce $\ce{Zn^{2+}}$ ions while the corresponding electrons are staying back in the metal. But this is only valid for the first few zinc atoms. When additional zinc atoms will do the same, they are prevented by the presence of positive $\ce{Zn^{2+}}$ in water and the negative charges in the metal. So additional zinc atoms will not be ionized and dissolved. So few ions are generated that the concentration of zinc ions in solution is totally negligible. But it is sufficient to prevent further ionization of zinc atoms plus dissolution of the corresponding $\ce{Zn^{2+}}$ ions in the future.

  • $\begingroup$ So is it about the interionic forces in water? $\endgroup$ Commented Mar 19 at 12:51
  • 7
    $\begingroup$ 2 electrons would not make so big difference. Consider the capacitance of the electrode and the equation dE = dq/C. Very roughly some 10 millions of electrons just for Delta E = 1 V for C = 1 pF. More at a much higher capacitance. Finally, potential gets so low electrons start to get hydrated and quickly reducing water or H+(aq) to hydrogen. $\endgroup$
    – Poutnik
    Commented Mar 19 at 13:26
  • $\begingroup$ @Poutnik I guess that explains the non-reactivity of Zn towards cold/hot water and reactivity only towards steam? $\endgroup$ Commented Mar 19 at 15:17
  • 4
    $\begingroup$ @Poutnik. I expected your answer. I perfectly know that one and only one atom cannot be ionized and dissolved. But it was an image. I think instead of one, I should have said "the first millions". But this amount is not known with accuracy. And it would start a long series of remarks about its real values, which is not important for the present problem. It is better to choose a simple number like "one", even if it is probably not one. $\endgroup$
    – Maurice
    Commented Mar 19 at 16:44
  • 5
    $\begingroup$ But it is severely misleading, and without much justification. "Many" is enough. $\endgroup$
    – Poutnik
    Commented Mar 19 at 17:58

This a question that offers an answer with electrochemical equilibrium. Our experiment is as follows:

  • We have a typical beaker of volume $V = \pu{200 mL}$ of water. On top of it we have a volume $V_\mathrm{a} = \pu{1 dm3}$ of air, at ambient conditions of $p = \pu{1 bar}$ and $T = \pu{25 °C}$.
  • We add a mass $m = \pu{1 g}$ of zinc.
  • Determine the final state after a long period of time.

First attempt

One of the most important issues with these type of problems is to determine which types and how many reactions will take place. Let's consider for simplicity that only the oxidation of zinc takes place: $$\ce{Zn(s) -> Zn^2+(aq) + 2e^- \tag{1}}$$ However, if only this happens we have the following problems:

  • The electrolyte solution becomes positively charged with $\ce{Zn^2+}$ without any counterbalance of anions. This is a violation of charge conservation taking place at the macroscopic scale in the liquid phase.
  • The solid zinc becomes negatively charge with $\ce{e^-}$. This is a violation of charge conservation taking place at the macroscopic scale in the electronic phase.

Thus, we reject that only Eq. (1) takes place.

A more rapid way to show that this is impossible is the known phrase taught at at general chemistry: 'an oxidation needs a reduction or vice versa', so we cannot have a single oxidation.

Second attempt

We know that self-ionisation of pure water brings negligible but some amount of protons and hydroxides to the solution. This raises the interesting question: can the two electrons from Eq. (1) be grabbed by the hydrogen ions, and produce hydrogen gas? This situation will not violate any law of charge conservation, so we see whether this is possible by thermodynamics.

The system has two electrochemical reactions and one chemical reaction \begin{align} \ce{Zn(s) &-> Zn^2+(aq) + 2e^-} \tag2 \\ \ce{2H+(aq) + 2e^- &-> H2(g)} \tag3 \\ \ce{H2O(l) &-> H+(aq) + OH^-(aq)} \tag4 \\ \end{align} Eqs. (2) and (3) lead to two electrochemical equilibrium, while Eq. (4) to a chemical equilibrium \begin{align} E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag5 \\ E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln \left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right) \tag6 \\ K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag7 \end{align} and we count 5 unknowns: $E$, $[\ce{Zn^2+}]$, $p_\ce{H2}$, $[\ce{H+}]$, and $[\ce{OH-}]$. We need two more equations. The first one is charge conservation \begin{align} [\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag8 \\ \end{align} The other one is a relation between the partial pressure of hydrogen and other species. Let $\xi_2$ and $\xi_3$ be the extents of reactions 2 and 3. Then, the amount of hydrogen, protons, and hydroxides in terms of these extents are \begin{align} n_\ce{H2} &= \xi_2 \tag{9} \\ n_\ce{H+} &= n_\ce{H+}^0 - 2\xi_2 + \xi_3 \tag{10} \\ n_\ce{OH-} &= n_\ce{OH-}^0 + \xi_3 \tag{11} \end{align} By Eq. (11) we can write $\xi_3 = n_\ce{OH-} - n_\ce{OH-}^0$. Introducing this expression and Eq. (9) in Eq. (10) gives \begin{align} n_\ce{H+} &= n_\ce{H+}^0 + 2n_\ce{H2} + n_\ce{OH-} - n_\ce{OH-}^0 \\ 2n_\ce{H2} &= n_\ce{H+}^0 - n_\ce{H+} + n_\ce{OH-} - n_\ce{OH-}^0 \\ 2n_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)V \\ p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0) V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{12} \\ \end{align} and we note that the volume $V$ refers to the species in the liquid phase, which is not the same as the volume $V_\mathrm{a}$ for the gas phase.

And we have a set of 5 non-linear equations for the system, which we repeat again \begin{align} E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag{13} \\ E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln \left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right) \tag{14} \\ K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag{15} \\ [\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag{16} \\ p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0) V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{17} \\ \end{align} It is possible to combine this equations to a single unknown, but this will unecessary extend the answer. We tabulate the solutions: \begin{array}{|c|c|} [\ce{H+}] \; \pu{(mol dm^-3)} & \pu{9.0632E-13} \\ [\ce{OH-}] \; \pu{(mol dm^-3)} & \pu{5.5168E-2} \\ [\ce{Zn^2+}] \; \pu{(mol dm^-3)} & \pu{1.1034E-3} \\ p_\ce{H2} \; \pu{(bar)} & \pu{0.27352} \\ E \; (\pu{V}) & \pu{-0.69598} \\ \end{array} The results indicate:

  • Indeed, the hydrogen ions were consumed and a bit of zinc entered the solution.
  • The electrode potential $E = \phi^\mathrm{M} - \phi^\mathrm{S} < 0$, thus the electric potential of the liquid phase became more negative.

By mass conservation, we can finally see how much the mass of zinc solution changed (remember we put $\pu{1 g}$ initially) \begin{equation} m_\ce{Zn} = m_\ce{Zn}^0 - [\ce{Zn^2+}]VM(\ce{Zn}) \to \boxed{m_\ce{Zn} = \pu{0.99278 g}} \end{equation} where $M(\ce{Zn}) = \pu{65.41 g mol^-1}$ is the molar mass of zinc.

  • $\begingroup$ It's good that you kept oxygen out of the analysis. I should have been clearer and stated the initial conditions as anaerobic, RT and pressure. $\endgroup$
    – Buck Thorn
    Commented Mar 20 at 17:52
  • $\begingroup$ @BuckThorn The system can get more complicated as you wish, as more reactions keep being added... Oxygen may enter the liquid phase and react with water or $\ce{H+}$, which I did not consider... $\endgroup$ Commented Mar 20 at 18:04
  • $\begingroup$ Yes, and oxide can eventually form, at least according to a reference I placed in comments under the question. But that describes an experiment in the presence of electrolyte exposed to air. $\endgroup$
    – Buck Thorn
    Commented Mar 20 at 18:08
  • $\begingroup$ I will read it later. However, you were interested in the case of introducing zinc in pure water, right? $\endgroup$ Commented Mar 20 at 18:26
  • $\begingroup$ Yes, pure water (no salt). I could not find an article that addresses that problem. $\endgroup$
    – Buck Thorn
    Commented Mar 21 at 6:59

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