# How does current start to flow when galvanic cell is first connected?

Using the Daniell Cell as an example, where in the steady state we expect zinc to be oxidised and copper to be reduced spontaneously due to the overall reduction in Gibbs free energy, how does the system "know" to start moving electrons from the anode to the cathode? What is the underlying physical mechanism?

When the two half cells are first connected via the external circuit, it seems that they would not yet "know" anything about each other; each half cell would not already "know" it is anode or cathode (e.g. zinc could be the cathode in a different galvanic cell) and so how would the system immediately start driving the electrons in the right direction?

In other answers on this site (e.g. https://chemistry.stackexchange.com/a/28638/98580) I've seen statements like "the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell".

However, as a physicist this is somewhat unsatisfying, as the phrase "induce reduction" seems to bundle up a whole lot of physical processes in the half cells and the external circuit into a very high level concept. Fundamentally, it must come down to electrical forces and thermodynamics to explain the overall behaviour of the system.

My understanding is that prior to connecting the external circuit, each electrode is in equilibrium with its electrolyte, undergoing both oxidation and reduction at the same time; so each electrode could both push and pull electrons to/from a connected wire. The moment the external circuit is connected, do electrons start flowing from the zinc electrode to the copper electrode because zinc has a greater "tendency" (than copper) to push electrons onto a connected wire, and copper has a greater "tendency" (than zinc) to pull electrons from a connected wire? That would then sound like the galvanic cell gets started via a statistical process and is sustained by electrical charging of the electrodes thereafter.

• If you connected a single cell to a very sensitive chargemeter, you would know wether the zero current equillibrium is reductive or oxidative. But who cares? Ions in an electrolyte are by many orders of magnitude less mobile than electrons in a wire.
– Karl
Sep 6, 2020 at 22:16
• @poutnik why do the redox reactions "try" to maintain a particular potential? Is there some underlying physics governing that? And could you elaborate on the origin of the current that disbalances the potentials and what that means from a microscopic perspective? Sep 6, 2020 at 23:09
• My comments are moved to an answer. Sep 7, 2020 at 5:38

The electrode redox reactions of the Daniell cell:

$$\ce{Zn(s) <=> Zn^2+ + 2 e- }$$ $$\ce{Cu(s) <=> Cu^2+ + 2 e- }$$

maintain at the respective electrode a particular potential where both opposite reactions have the same rate, implying there are no external galvanic causes that affect this potential. Rates of oxidation electrode reactions exponentially grow with growing potential and vice versa for reduction. This applies on potentials near the equilibrium one, when the reaction rate is not limited by the diffusion yet. Diffusion cuts off this exponential growth. The faster of the opposite reactions either pushes electrons to an electrode either pulls electrons from it until the electrode potential reaches the equilibrium potential. At this potential, rates of both reactions are equal. Even before electrodes are ever connected, these electrode potentials are in place. Electrodes do not need any external galvanic circuit.

A current flowing between 2 galvanically connected places of different potentials is not part of electrochemistry, but a part of physics which you are already familiar with. The connected external circuit is then just following the Ohm law $$I=\frac UR$$ The current is shifting electrode potentials by moving electrons, what is the same process as in electrostatics, where potentials of capacitor electrodes are changing, when galvanically connected.

As the resulting potentials are not the equilibrium ones any more, one of the electrode opposite redox reactions takes the upper hand, leading to oxidation at the anode and reduction at cathode. Particularly, the potential of the $$\ce{Zn}$$ electrode with the lower equilibrium potential gets higher and starts oxidation, making the electrode the anode.

$$\ce{Zn(s) <=>> Zn^2+ + 2 e- }$$

Similarly, the potential of the $$\ce{Cu}$$ electrode with the higher equilibrium potential gets lower and starts reduction, making the electrode the cathode.

$$\ce{Cu(s) <<=> Cu^2+ + 2 e- }$$

This leads to a kind of race on the zinc electrode in providing electrons on the chemistry side and taking them by the circuitry side. And vice versa for the copper electrode. The actual electrode potential under the load is a kind of the score of this race.

Electronic chips are known for the internal use of electronic charge pumps to get different voltage levels than externally provided. E.g. once famous 8bit CPU Z80 uses external voltage +5V and 0V, internally generating also +12V,-12V,-5V. Electrodes in electrochemical cells can be considered in some sense as chemically powered charge pumps with different target potentials.

Great basic knowledge of chemistry ( and mainly of physics ) is Hyperphysics-Chemistry-electrochemistry, in form of linked cheat sheets. I would recommend electrochemistry chapters of textbooks of a physical chemistry. I have once studied a translation of Moore, EN natives would suggest much more. More advanced sources are here: SE Chemistry resources-for-learning-chemistry

• Thanks @Poutnik, I am slowly getting there. So would it be correct to say that the microscopic situation, when the external wire is first connected, is that the Zn electrode is truly at a lower electric potential than the Cu electrode, and hence there are literally electric forces acting on the electrons as per the usual electrodynamical theory? Sep 7, 2020 at 11:47
• As another question, can Cu ever be the anode? E.g. if one half cell had a F electrode, would Cu then be oxidised whilst F is reduced? Sep 7, 2020 at 11:55
• F is not good example, as fluorine would blow up near any electrolyte and solvent out of the container being extremely oxidative. But if Cu is used against Ag/Ag+, or Hg/Hg2^2+, or against a half-cell with an inert electrode and all-in-solution redox system, they yes, Cu/Cu^2+ half cell can be an anode. Note that a half-cell is the whole redox system around 1 electrode, while an electrode in wider sense is the same as a half-cell, in narrower sense it is just that conductive stick, that may(Zn) or may not ( Pt, glassy C ) be involved in redox reactions. Sep 7, 2020 at 12:00
• Thanks again. My chemistry knowledge is sorely lacking; would you recommend any particular texts that have a good treatment of this subject? I'd be interested in both introductory and more advanced treatments, especially from a fundamental physical perspective. Sep 7, 2020 at 12:08
• See the A update Sep 7, 2020 at 12:43

BTW, electrons don't actually actually "flow" very fast. The drift velocity of electrons is incredibly slow, on the order of a millimeter every 5 or ten seconds in a room-temperature conductor. The propagation of current is like the particles in Newton's Cradle, where the electric field bumps each electron down the line, a much faster process. Furthermore, electrons are swept out of the conductor by the electromagnetic force, which propagates on the surface of the conductor at a large fraction of the speed of light in vacuuo, C. The fraction is the velocity factor, perhaps 0.5 to 0.75 C.

So the answer to your question is not as clear-cut as it might seem. You could say the reaction propagates at the speed of the EMF, C times the velocity factor in the connecting wire. However, one then must take into account the buildup of slow-moving ions, creating a space charge slowing things down.