When building a simple salt water battery, consisting of a zinc-anode, copper-cathode and an NaCl-solution, I noticed the following:

When closing the circuit, the current decreases over time. (No surprise, since either charge adds up around the electrodes or $\ce{Na^+}$- and $\ce{Cl^-}$-ions get depleted.) When I gently slide my finger along the zinc-anode (under the water), the measured current increases very little (and continues to decrease over time). When I do the same at the copper-anode, the current increases much more than at the anode (and continues to decrease over time).

I understand that at the zinc-anode $\ce{Zn^2+}$ dissolves into the water. And I understand that at the copper-cathode the electrons will interact with the water, not the $\ce{Na^+}$.

Now, I have 3 questions, where no. 3 is my main question:

  1. Close to the anode: Will $\ce{Cl^-}$-ions just float next to the $\ce{Zn^2+}$-ions independently or will they form $\ce{ZnCl_2}$?

  2. Close to the cathode: What exactly is happening there? Since the voltage I measured is ~0.7V, it can't be electrolysis of water, since 1.23V are required for this to happen. I would assume due to autoprotolysis of water, $\ce{H_3^+O + e^- -> 1/2H_2 + H_2O}$. The autoprotolysis leaves behind an $\ce{^-OH}$. I assume it bonds with the Na, giving $\ce{NaOH}$.

  3. Why do I measure a much greater temporary increase in electric current when I gently stirr the water close to the copper-cathode, compared to stirring close to the zinc-anode? Shouldn't both water-volumes around the two electrodes be equally neutral? On one side, $\ce{Zn^2+}$ gets cancelled out by $\ce{2Cl^-}$, on the other side $\ce{2OH^-}$ get cancelled out by $\ce{2Na^+}$?

Here is the battery:

enter image description here

And here is where I got my ideas for the reactions happening. (I'm not a chemist.)

  • $\begingroup$ Be aware it is not really Zn/Cu galvanic cell, but Zn/H2-on-Cu cell. By fiddling with cathode, you remove hydrogen and improve access of electrolyte to copper. $\endgroup$
    – Poutnik
    Dec 8, 2021 at 14:48
  • $\begingroup$ Ah, that's interesting. So the hydrogen stays close to the cathode, forming kind of a shield around the cathode? I thought it would rise and leave the water eventually. I should mention that I was not able to see any bubbles around the cathode, even after waiting for a long time... But I thought the bubbles might be too small. $\endgroup$ Dec 8, 2021 at 15:14
  • $\begingroup$ Related: chemistry.stackexchange.com/q/159681/79678. $\endgroup$
    – Ed V
    Dec 8, 2021 at 16:54
  • $\begingroup$ I know, that was me. I am still trying to figure out some details I am not totally sure about (see my 3 questions). Thanks for helping earlier btw. $\endgroup$ Dec 8, 2021 at 17:15
  • $\begingroup$ copper-anode $->$ copper-cathode $\endgroup$ Dec 10, 2021 at 5:25

1 Answer 1


There are some mistakes in your questions.

  1. At the anode, $\ce{Zn}$ will form $\ce{Zn^{2+}}$ and not $\ce{ZnCl2}$. Simultaneously the anode attracts anions $\ce{Cl-}$ to compensate for the new positive charges created at the zinc plate. The $\ce{ZnCl2}$ compound does not exist in solution. but it can be obtained after evaporation of the aqueous solution.

  2. At the cathode, the only reaction is : $\ce{2 H2O + 2e^- -> H2 + 2 OH^-}$. Simultaneously, the copper plate will attract $\ce{Na^+}$ to compensate for the new $\ce{OH-}$ ions created on the copper plate. So the solution near the cathode is made of $\ce{Na+ + OH-}$ ions. But the solution does not contain $\ce{NaOH}$. This substance may be obtained after evaporation of the solution.

  3. When stirring the cathodic plate, you remove the hydrogen bubbles adsorbed on the cathode surface. Without stirring, the surface where $\ce{H2O}$ can produce $\ce{OH-}$ decreases slowly with the time, because the hydrogen bubble prevents further production of $\ce{OH-}$ ions. The current is proportional to the surface of the electrode. As the surface of the plate where $\ce{OH-}$ ions can be created decreases, the current decreases

  • $\begingroup$ Dear Maurice, awesome! Thank you so much. This is exactly what I was looking for. Now I understand it. $\endgroup$ Dec 8, 2021 at 22:26
  • $\begingroup$ Let me just add one more question: Shouldn't the electrolysis of water require at least 1.23V? I was only able to measure ~0.7V. $\endgroup$ Dec 8, 2021 at 22:31
  • $\begingroup$ @Rainer_Zoufal Electrolysis is the opposite process as a galvanic cell providing power. Furthermore, water electrolysis means creation of hydrogen AND oxygen, what is not the case. $\endgroup$
    – Poutnik
    Dec 9, 2021 at 8:59

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