What induces electrons to leave the atoms around the part of the zinc metal in a solution to travel up along that zinc metal and above the solution, through the wire and to the copper metal and reduce the copper ions in that solution.

Is it a result of the positive metal ions in a 'sea of electrons' being attracted to the negative ions of the solution making the electrons surrounding them leave, are the electrons repelled by the electrons in the solution or do these electrons 'sense' the copper ions on the other side waiting to be reduced and flows towards them?

I'm just not sure what prompts them to leave in the first place if the positive copper ions are not directly in contact with the zinc.


3 Answers 3


First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}<=>Zn^2+_{(aq)} +2e-}$$

At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. 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 excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.

But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- <=> H_{2(g)} +2H2O}$$ The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.

To go back to the core of your question,

I'm just not sure what prompts them to leave in the first place if the positive copper      
ions are not directly in contact with the zinc.

If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a salt bridge.

The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.

This paper is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).

  • 1
    $\begingroup$ I think the big remaining question is why does "zinc adsorbs hydronium very poorly"? The paper seems to skip any explanations and just states that as a fact. What prevents the hydronium from stealing electrons from zinc the way it steals electrons from copper? $\endgroup$
    – akhmed
    Commented May 10, 2018 at 7:59
  • 1
    $\begingroup$ @akhmed Here is a good primer on how it works. In short, some metals can easily form an M-H intermediate, greatly lowering the energy barrier to making H2 versus briefly having a lone H atom in the vicinity of the electrode. $\endgroup$ Commented May 10, 2018 at 20:44
  • $\begingroup$ This is awesome! Basically, proton needs to detach from hydronium and attach to metal first or it can't steal electrons efficiently. So for zinc electrode to work, we rely crucially on the fact that $\ce{Zn-H}$ bond is weak. If it were not so weak, hydroniums would be stealing excess electrons and we would not be able to accumulate electrons inside the zinc electrode! $\endgroup$
    – akhmed
    Commented May 10, 2018 at 22:41
  • $\begingroup$ I'm actually a little confused regarding your last paragraph's statement: "Copper's electrons are not normally high enough [...] ". It seems like you are describing a scenario where the electrolyte solution being used in the two half-reactions with a salt bridge is, in fact, just water. If the electrolyte solution is just water, why would electrons travel to the copper electrode (from the zinc electrode) to ultimately bind to hydronium...couldn't they have just left the zinc electrode directly to access nearby hydronium? Or does hydronium have better adsorption properties on copper than zinc? $\endgroup$
    – S.C.
    Commented Mar 13, 2020 at 23:16
  • $\begingroup$ @S.Cramer pure water is not a very good electrolyte; you still need more ions in there to make the cell work because otherwise the ionic conductivity will be very low between the two electrodes. You are on to something with your last question though: zinc can react directly with the water, making Zn2+ and H2, but hydronium adsorbs very poorly to zinc, so this happens rather slowly unless the pH is low. $\endgroup$ Commented Mar 15, 2020 at 5:34

If you have studied semiconductors and the method of conductivity by them, mainly extrinsic semiconductors, you would recall that the movement of holes is relative and equivalently considered as the hole moving but actually the electrons are moving, in the opposite direction.

Consider the solution, copper reduces via the neighboring electrons, a very rare but favorable probability that happens similarly with zinc but with the difference that is looses some electrons in its surroundings.slowly the process starts involving electrons at far distances and suddenly when both processes approach each other, a kind of link is created which is mutually favorable and the process proceeds further more easily and now involving almost all atoms which was earlier limited to a very few atoms.

The whole process is also a particular cause of the random motion of solvent and other particles like Brownian motion. Note that random always means something that involves very small factors (do you know the butterfly effect? ) so that it is almost impossible but not exactly impossible to predict the outcome, not does it means something that is luck.


First of all, both electrodes (the copper and the zinc, in this case) are dissolving in the electrolyte. However, normally the dissolved electrode material comes out of solution just as fast. It's like trying to dig a hole in the dirt when someone else is filling it up at the exact same speed; nothing happens. Nothing happening is also called an equilibrium.

Second of all, when it dissolves, it becomes an ion, that is, it either gains or loses an electron. Now, some materials want to ionize different amounts than others. This "want" is measured in volts.

Taking into account these two factors, when you put any two electrodes together in an electrolyte, you can tell from a "Standard Electrode Potential Table" what will happen, if anything. In the case of a lemon battery, the ionic zinc wants to lose two electrons more than the copper wants to gain them. Note that one electrode must gain the electrons and one must lose them. So we have these two electrodes giving off ions into solution and also taking them back in. However, when a copper ion by chance touches the zinc metal, it transfers two electrons to the copper making metallic copper and ionic zinc. After the copper adheres to the zinc metal, it can't dissolve anymore because if it, by chance, does dissolve, the zinc metal around it instantly changes it back to metallic copper by giving it two more electrons and dissolving another zinc atom in the copper's place. Likewise, the zinc can't adhere to the copper because it would have to steal electrons from the copper, but the copper doesn't want to.

Some things to keep in mind:

  • This transfer of electrons is what generates the electricity in the first place.
  • Not all electrode pairs react.
  • If you want more in depth information, look up "standard reduction potential table" in Google, as well as "how to calculate the voltage of a galvanic cell".
  • $\begingroup$ This isn't really right because of some common misconceptions of this type of cell. The half-reaction at the copper electrode has nothing to do with copper, in a lemon battery, because there is no significant quantity of copper ions to reduce in the solution. The reduction reaction at the copper electrode is simply the production of hydrogen from the water. What happens if there is copper in the solution is that some electrons will flow through the circuit to reduce copper, but a lot of copper will simply plate onto the zinc electrode, doing no useful work in the process. $\endgroup$ Commented Dec 29, 2014 at 2:19
  • $\begingroup$ Welcome to chemistry.SE! You seem to be well aware of the rules, but I anyhow, guide you to the help center or taking a tour of the website. $\endgroup$
    – M.A.R.
    Commented Dec 29, 2014 at 2:46
  • $\begingroup$ @MichaelDMDryden Do zinc ions oxidate, because of the acid? You said that in neutral solutions, zinc would do nothing $\endgroup$
    – Petr L.
    Commented Mar 12, 2023 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.