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I've been trying to visualise what is going on inside the batteries when you connect, lets say two 1,5V batteries, and why the voltage is added to 3V.

Several good explanations have tried to explain it in different ways, batteries as pumps etc. But I have never actually got a good picture of what the electrons are doing that creates the additive voltage effect (do you like my jargon). Until now. Introducing: the super explanatory theory of electron behaviour in serial connected batteries. Well maybe not, but I would like some feed back to see if I am on the right tract to understanding this mystical phenomenon. Here goes:

Lets this battery for example:

anode: Zn(s) + 2OH−(aq) -> ZnO(s) + H2O(l) + 2e−

cathode: 2MnO2(s) + H2O(l) + 2e− -> Mn2O3(s) + 2OH−(aq)

When the battery is unconnected, some of the redox reaction above will take place, creating free electrons in the anode and removing electrons from the cathode, until the energy needed to dislodge electrons from the cathode is to great, and the reaction stops. This creates the voltage (electric potential) for the battery. More electrons in the anode the more they want to go to the cathode, meaning higher Voltage .

When you complete a circuit each electron moved from the anode to the cathode facilitates a new chemical reaction replacing the electron at the anode, keeping the reaction going with even Voltage until the materials are spent.

If you connect two batteries in series without completing the circuit the electrons in the anode of battery 1 rushes to the cathode of battery 2, which in turn pushes the chemical reaction in battery 2. Thus you have twice as many electron is the anode of battery 2 and the original amount of electrons in the cathode of battery 1. Hey presto double voltage.

Am I at all correct in my impression of what is going on?

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  • $\begingroup$ What happens when you connect 2 charged capacitors in a serie ? Aside of very fast balancing of potentials of connected electrodes - nothing. The same for electrochemical cells. Note that it is important to distinguish the electrostatic and electrochemical definition of zero potentials. $\endgroup$
    – Poutnik
    Sep 23 at 16:50
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Voltage is analogous to setting the chemical potential of the electron in the material. The analogy is not about electrons "rushing" around to create voltage. The analogy is about combining electrochemical potentials that electrons have to climb up or can fall downward through.

Imagine that you two batteries of exactly the same voltage in series using the sequence (+ -)(+ -). To pass through this connection, the electrons must traverse two electrochemical potential wells (or two potential hills). These wells or hills exist regardless of whether the electrons move through them or not.

The electrochemical potential energy difference is the difference between the absolute chemical potentials of the products and reactants. It is sensed by the electrons; it is not "imparted to the system".

Consider that the work needed for a charge $Q$ to go through a voltage difference $\Delta V$ is $Q\ \Delta V$ (eV/electron or J/electron). One electron going through one battery requires (or gives off the energy equivalent to) work $w = q\ \Delta V$. Going through a second battery gives off or requires that same energy.

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  • $\begingroup$ Thank you for the answer. If voltage is the the potential energy that's the batteries impart to the system through the chemical reaction, what is the voltage between the two batteries? It seems to me like the voltage must be the same 3V in the circuit and between the batteries, but that clashes a bit in my mind when I think of it as potential hills, where the voltage over the second battery would be 1,5V. $\endgroup$ Sep 23 at 21:32
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Am I at all correct in my impression of what is going on?

You are not too far off, but you are too focused on the number of electrons and not focused enough on the electrolyte. A battery has two electrodes and each electrode has both a metal and an electrolyte. You cannot understand batteries without both.

The electrons cannot move in the electrolyte and the ions cannot move in the metal. But at the surface of the electrode the redox reaction occurs which creates or consumes charge carriers on each side. This changes the charge carriers from electrons to ions and vice versa, so that current can go across the interface even though the individual charge carriers cannot.

Like any chemical reaction, as it proceeds it consumes reactants and produces products. With most chemical reactions as the concentration of products increases the reaction slows and eventually halts, and this could give the impression that a certain number of electrons remaining at the interface would halt the reaction.

However, in the ranges of concentrations that occur in normal batteries, that is not the case. It is not the concentration of electrons on the electrode that matters, but the electrical potential difference across the interface. The reaction may proceed even with a high number of electrons in the electrode if there is also an excess of negative ions in the electrolyte. Conversely the reaction may cease even with few electrons in the electrode if there are also few negative ions in the electrolyte.

So the function of the electrode is to produce an electrical potential across the interface. The chemical reaction is energetically favorable, so it can drive the reaction against an electrical potential (I.e. in the direction opposite what you would expect from Ohm’s law). This only proceeds up to a certain point, where the opposing electrical potential difference matches the driving chemical potential.

It is this potential difference that determines if the reaction proceeds or not. It doesn’t matter if the potential difference is due to a few electrons on one side and a few cations on the other or if it is due to many electrons on one side and a few anions on the other.

So, you can ground any point of the circuit, and adding more batteries will not change the number of electrons at that point. But elsewhere you will have metal/electrolyte interfaces that increase the electrical potential due to the additional chemical reactions, regardless of the number of electrons present.

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  • $\begingroup$ "The chemical reaction is energetically favorable, so it can drive the reaction against an electrical potential". How do you mean? Is the electric potential of the electrodes creating an electric field inside the battery which pushes against the negatively charged ions when they move from the cathode to the anode? I know Volt is J/C and, know now, that the reaction creates the potential difference, but how does a higher potential manifest for the electron traveling through the circuit? In mechanical physics potential energy would become kinetic and friction when released. Thank you! $\endgroup$ Sep 24 at 8:14
  • $\begingroup$ @ErikEriksson again, don't forget about the electrolyte. The potential is created at the interface of the metal and the electrolyte. You wind up with a bilayer of charge carriers that produces an electric field at the interface. This electric field opposes the current, and makes a potential difference across the interface. The chemical reaction being energetically favorable means that even with the electric field opposing it the reaction still occurs, up to a point. $\endgroup$
    – Dale
    Sep 24 at 14:20
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    $\begingroup$ Thank you for the answers. I am studying structural mechanics as an engineer, which is way more intuitive for me. Electricity always seems like magic and formula. But now I am trying to force my self to tackel that monster and I feel very dumb at times. But thanks the the site and people like you I can get a little less confused. $\endgroup$ Sep 24 at 14:27

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