In a standard galvanic cell, I understand that electrons flow from the anode to the cathode. For example, in a Daniell cell, electrons flow from the Zn anode to the Cu cathode. Before the two half-cells are connected (open circuit), they are each in chemical equilibrium:

$$\require{mhchem} \ce{Zn(s) ⇌ Zn^2+(aq) + 2e^-}$$

$$\require{mhchem} \ce{Cu(s) ⇌ Cu^2+(aq) + 2e^-}$$

This means that in open circuit conditions each electrode will contain a surplus of negative charge. However, the zinc electrode will be more negatively charged because it oxidizes more easily. When the cells are connected via a wire and salt bridge, the electrons will flow from the zinc anode to the copper cathode as they attempt to re-establish new equilibrium positions.

Here is my uncertainty:

When connected, it appears that electrons are flowing from a more negatively charged area (Zn) to a less negatively charged area (Cu). Can the less negatively charged Zn be considered "positive" in relation to the more negative Zn? It seems that instead of a positively charged ion attracting them into the cathode, they travel there because the electron density is less. Is there an error in my logic?

I understand there is an electrochemical potential difference due to a difference in Gibbs free energy--which I have researched--but a quote by a blogger summed up my frustrations perfectly: "To me, this explanation seems to be invoking the "electrochemical potential" as a magic wand that pushes positive ions from the negative to the positive electrode without needing to invoke any electric fields."


1 Answer 1


One model for a solid metallic conductor is a lattice of singly-charged positive ions, interacting with a gas of conduction electrons, for overall neutrality. Apparently when a metal ion leaves the solid for the electrolyte, it leaves a second electron behind in the metal. If you look at a table of ionization energies, you find that the second ionization energy for zinc (1700 kJ/mole) is slightly less than the second ionization energy for copper (2000 kJ/mole). If you imagine a copper and a zinc electrode with the same shape, the dissolving of the conductor into the electrolyte makes the electrolyte positively charged and the metal negatively charged. There is an electric field pointing from the positive electrolyte towards the negative surface charges on the metal; this field is maintained by the chemical action of the electrolyte. But for the same amount of ionization work, you get more zinc ions than you do copper ions.

To complete the circuit, you must have paths for both the positive and the negative charges. The positive charges travel through the electrolyte, which you usually accomplish by putting your zinc and copper electrodes in the same bucket of electrolyte. The negative charges usually travel through a metallic conductor, like a wire. So the electrolyte is everywhere more positive than the wire is everywhere. Because the zinc is easier to remove, there are more electrons at the zinc end of the metal conductor, but there are also more zinc ions in the electrolyte near the zinc electrode than there are copper ions in the electrolyte near the copper electrode.

When you have completed the circuit, electrons will flow through the metal from the more-negative zinc electrode to the less-negative copper electrode. But in the electrolyte, there are more zinc ions than copper ions for the same amount of ionization work, so the positive charges equilibrate when zinc ions move towards the copper electrode. As the cell discharges, you end up with mass flow from the zinc electrode to the copper one, both in the wire and in the electrolyte.

The current will continue to flow until the copper electrode has been zinc-plated. Applying an external voltage to the electrodes reverses the ion flow in the electrolyte, and could eventually give you copper-plated zinc.

  • $\begingroup$ You cleared this up considerably. Thank you for giving me such a great answer. I was initially confused by electrons flowing to a less negative area. I had reasoned the attraction into the cathode must involve some type of positive charge like a cation. However, I now see that a "less negative" area can be represented as a positive charge (and will exert an attractive force on the electron?). It's sort of like moving from an electric potential of -5V to -3V. Would I be correct in saying that? $\endgroup$ Commented Jun 2, 2022 at 4:50
  • $\begingroup$ Yes, just like that. $\endgroup$
    – rob
    Commented Jun 2, 2022 at 6:09

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