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In a zinc-copper voltaic cell,
What causes the initial electron movement from the zinc to the copper?

I understand that copper would pull the electrons more strongly than zinc, but only if it is electron deficient. Both half cells of the voltaic cell are electrically neutral (same amount of electrons and protons), so I don't understand what would cause the electron transfer / oxidation.

Also, I understand that $\ce{CuSO4}$ solution in the cathode half cell is needed so that the $\ce{Cu++}$ will accept the electrons, but what's the function of the $\ce{ZnSO4}$ in the anode half cell?

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  • $\begingroup$ The fact that the chemical potential of the two separate half is not the same. I don't claim it is simple. But you can ask the same for every reaction not only for redox ones in an electrochemical apparatus. $\endgroup$ – Alchimista Jan 19 at 11:52
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For both electrodes, regardless of their galvanic connection, are ongoing the opposite processes

$$\ce{M(s) <=> M^n+(aq) + n e-}$$.

The forward process charges the electrode negatively, but slows down at more negative potentials.

The backward process charges the electrode positively, but slows down at more positive potentials.

As the resulting net effect, an electrode establishes such an equilibrium potential, where both processes have the equal rate and the zero net effect. These equilibrium potentials are different for different combinations of electrode material and solution composition.


When you connect 2 electrodes ( connected also internally by a free ion passage ) by a galvanic circuit, different electrostatic potentials lead to the electric current. The current disbalances the equilibrium state on electrodes and one of the respective processes takes the upper hand.

That's the core of my question. Where is the difference in the electrostatic potential? If we count the electrons and protons in each half cell, they're the same, so what would still cause the electrons to move?

They are not exactly same and there are differences within the half cells, between electrodes and electrolyte.

We started with neutral electrodes and neutral electrolyte, where did the difference in potential came from? I'm sorry if I wasn't clear in the original question, but that is exactly what's confusing me. If I'm able to see how the electrostatic potential is created, all the other pieces will fall into place for me.

Dissolving a metal electrode charges it negatively and solution positively. The opposite process - metal depositioning - does the opposite. The metal with the greater tendency to dissolve causes the electrolyte to be more positive than the one in the other cell, cause positive ions ions to migrate there and negative here. As result, the net charge of half cells is not zero anymore and neither the charge of the electrodes.


The "spectator" ions like $\ce{SO4^2-}$ do not take part in electrode reactions. They electromigrate according to the local solution electrostatic potential gradient away from the $\ce{Cu}$ side, where they are becoming excessive and forming slightly negative solution charge. They pass the salt bridge ( or a porous membrane or a diaphragm) to the $\ce{Zn}$ side, where they are needed to counterpart being created $\ce{Zn^2+(aq)}$ to cancel slighly positive solution charge there.

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  • $\begingroup$ When you connect 2 electrodes by a galvanic circuit, different electrostatic potentials lead to the electric current. - That's the core of my question. Where is the difference in the electrostatic potential? If we count the electrons and protons in each half cell, they're the same, so what would still cause the electrons to move? $\endgroup$ – Toi Yiot Jan 21 at 9:42
  • $\begingroup$ We started with neutral electrodes and neutral electrolyte, where did the difference in potential came from? I'm sorry if I wasn't clear in the original question, but that is exactly what's confusing me. If I'm able to see how the electrostatic potential is created, all the other pieces will fall into place for me. $\endgroup$ – Toi Yiot Jan 21 at 9:52
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You may prefer a "human-like" picture. You may prefer to interpret the chemistry of the cell in saying that the zinc atom "feels like" losing its electrons to become a zinc cation. On the other side, the copper electrode has not at all the same tendency. It is even ready to accept foreign electrons to discharge its own cations. I know this explanation is not very scientific, and I beg your pardon for stating it. From my experience in teaching chemistry to unmotivated high school students, I know it helps.

Now the function of the $\ce{ZnSO4}$ in the anodic solution is to help electrically speaking, because pure water is an insulator. If the anodic compartment were just pure water, the first $\ce{Zn^{2+}}$ ions created in the water from zinc metal would prevent the next ones from appearing in solution, and the cell would soon stop working. But if there are already some $\ce{Zn^{2+}}$ in the solution, the newly created $\ce{Zn^{2+}}$ ions would simply push the already dissolved cations through the solution up to the bridge where the sulfate ions are coming from the cathodic region.

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  • $\begingroup$ the zinc atom "feels like" losing its electrons to become a zinc cation - but why that happens? If we count the electrons and protons, they're the same amount. Is it caused by something that doesn't boil down to electrostatic energy? If it is, then what is it? $\endgroup$ – Toi Yiot Jan 21 at 9:38
  • $\begingroup$ @Toi Yiot This happens because redox potential is negative $\endgroup$ – Maurice Jan 21 at 10:17

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