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I am trying to have a more fundamental understanding of electricity and specifically what voltage is.

My memory of highschool physics was that a battery has an excess of electron on one terminal, and a shortage on the other. This explanation caused some confusion when I thought about batteries in series; why does two 1,5V batteries add to 3V.

If it is excess and shortage of electrons, created from a chemical reaction, that attract the electrons from cathode to anode, wouldn't the volt (J/C) be the same for a circuit with two batteris; every electron in the cathode has one void to fill in the anode?

Some reading lead me to see battery voltage as a measurement of how much energy per electron (J/C) the chemical reaction produces, and the number of reactions per second as the current (C/s). I know 1 coulomb is not 1 electron, and one reaction doesn't necessarily = 1 free electron, this is a rough picture in my mind.

Does this mean that it is the energy released from the reaction that is accelerating the electron a certain amount depending on the amount of energy, and how is it that two batteries accelerate the electron twice as much?

This is my first attempt at articulation my confusion, so I'm sorry that this is not as eloquent as it could be

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Voltage (up to a factor) is the work one electron can produce when travelling from anode ($-$) to cathode ($+$). Current is the number of electrons flowing.

The energy needed to do the work comes from the energy released when electron participates in the reaction happening at cathode. This is chemistry-specific, so one cell of a battery has usually a very specific voltage.

Now imagine what happens if you have two cells in series. When, electron travels from A2 to C1 it produces work $W$ and when it reaches $C1$ it releases energy $U$. But at the same time, it allows another electron to travel from $A1$ to $C2$ and release another chunk of energy $U$. So in the end, $W=U+U=2U$ and we say that two cells in series have twice as large voltage, meaning for every electron participating in outer current, there is now twice as many reactions.

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// Of course, in reality, single electron does not travel the whole distance from anode to cathode and the energy is not magically released and transferred. There is a sea of electrons in the wire and they inhabit different energy levels, and chemical reactions in different metals of the cell produce level shifts, which in turn produce an electric field in the wire creating electron driving force, etc. But it doesn't change the general picture of why the voltage is doubling

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  • $\begingroup$ As so often happens, once I wrote the question down I got a new idea how it all works. I began to think of it as follows: No connection between batteries or circuit - electrons build up from reactions in both A1 and A2. An equilibrium is reached and reactions stop. Connect batteries but not circuit - Electrons from A1 flow to C2 which facilitates more reaction in the second battery which doubles the electrons in A2. You now have the normal amount of positive charge in C1 and double Negative in A2, creating double Volt. Am I right in thinking like this? $\endgroup$
    – Erik Eriksson
    Commented Sep 23, 2021 at 13:24
  • $\begingroup$ Although, in general you are correct, there is spontaneous reaction of exchange between $\text{Zn}$ and $\text{Zn}^{2+} + 2e^{-}$, you are safe to ignore it. Until something removes electrons from the anode, you won't see a noticeable build up of zinc ions (the same is true for cathode). The same I believe is true when you connect A1 to C2. The small initial flow of electrons will be immediately balanced by electrostatic field because of the charge mismatch. I doubt one could measure it at all. $\endgroup$
    – Vasily Mitch
    Commented Sep 23, 2021 at 16:23
  • $\begingroup$ If you connected a multimeter between the two batteries, what voltage would it show? $\endgroup$
    – Erik Eriksson
    Commented Sep 23, 2021 at 18:18

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