Understanding section on electromotive force in Purcell and Morin's Electricity and Magnetism

Question

I am following the book Electricity and Magnetism by Purcell and Morin.

There is a section on the electromotive force that I would like to understand better. I will write out the snippets from the book and my interpretation of what they are saying.

First there is an explanation about how a lead-sulfuric acid cell works, that we can visualize below

They then say

The cell will remain indefinitely in this condition if there is no external circuit connecting its terminals. The potential difference between its terminals will be close to 2.1 volts. This open-circuit potential difference is established “automatically” by the chemical interaction of the constituents. This is the electromotive force of the cell, for which the symbol $\epsilon$ will be used. Its value depends on the concentration of sulfuric acid in the electrolyte, but not at all on the size, number, or separation of the plates.

Open-circuit means there is no wire connecting the terminals of the battery. The cell in the picture above is charged, meaning there is charge on the terminals, hence an electric field between them, and hence a potential difference between the terminals.

Electromotive force (emf) is not actually a force. It is actually this potential difference between the terminals and it represents work done by a force per unit charge that has magnitude equal to the electric field between the terminals (at least inside the battery near the terminals).

Question: What does it mean that the potential difference between the terminals is established automatically by the chemical interaction of the constituents?

We have an electrolyte (liquid or gel containing ions) between the terminals. My interpretation of the charged state depicted above is that there is some kind of "equilibrium" (where this is not well-defined by me) where no net charge is being added to the terminals. I don't know exactly why this is, but is seems it might be because if we were to write out the chemical reaction equation at each terminal it would be a reaction happening in both directions simultaneously at the same rate. Ie, something being added to the positive plate and removed from the positive plate at the same rate. Not sure if this is correct.

Now let's close the circuit.

We can think of charge moving between the battery terminals through the circuit external to the battery. We can think of negative charge moving from the negative (lower potential) terminal to the positive (higher potential) terminal. If nothing else were to happen, then the potential difference (and hence the current, by Ohm's law) would fall until it reached zero.

However, something does happen: the battery keeps the potential difference constant, or at least tries to because it can't do this forever. It does this by taking electrons flowing into the positive terminal and having them participate in a chemical reaction with the lead dioxide on the positive plate, and increasing the concentration of lead sulfate in the electrolyte a a result. This lead sulfate participates in another chemical reaction, this time on the negative terminal, which releases electrons into the external circuit.

About this whole situation of closing the circuit, the book says

Now connect the cell’s terminals through an external circuit with resistance $R$. If $R$ is not too small, the potential difference $V$ between the cell terminals will drop only a little below its open-circuit value $\epsilon$, and a current $I = V/R$ will flow around the circuit.

I don't really understand this. As far as I do understand, we have the relationship

$$V_a-V_b=IR=\epsilon$$

Let $V_b=0$. Then

$$V_a=\epsilon=IR$$

The smaller $R$ is, the more current will flow. The larger $R$ is, the less current will flow. As far as I can see $V_a$ is fixed given $\epsilon$ which I think is determined by the characteristics of the battery we are using.

Question: So why does the book say "the potential difference $V$ between cell terminals will drop only a little below it's open-circuit value $\epsilon$"?

Answer 1

Electromotive force (emf) is not actually a force. It is actually this potential difference between the terminals

Correct, and it represents a work done if a charge is brought from one end to the other.

Question: What does it mean that the potential difference between the terminals is established automatically by the chemical interaction of the constituents?

We do not need to do anything to the system, it will maintain this voltage by itself. Automatically. Just putting these objects together will be enough.

there is some kind of "equilibrium" (where this is not well-defined by me) where no net charge is being added to the terminals. I don't know exactly why this is, but is seems it might be because if we were to write out the chemical reaction equation at each terminal it would be a reaction happening in both directions simultaneously at the same rate.

It does not have to start out at the same rate, but because the whole system will end up at thermal equilibrium (and in the case of flow, a steady state), the final rates must be the same, and then it will obviously no longer change when that happens.

Look, each electrode only cares about the electrolyte nearby it, and all of these interactions are happening at multiples of Avogadro's number! It is all probability and statistical, and at equilibrium, the tendency of the acid to eat away at the electrodes must be balanced by the already accumulated charges repelling more of the reaction, calculated by the voltages that have accrued. That is why the open circuit voltages would be what they are—just enough to cancel out the tendency for the electrolyte to be used up.

This lead sulfate participates in another chemical reaction, this time on the negative terminal, which releases electrons into the external circuit.

IIRC, it is some other chemistry. Not the same thing. The acid should become weaker as the battery charge level is used up, and the acid should become stronger as you charge the battery back up. But this is a detail that you definitely can just look up.

Remember that the ``equilibrium" is due to probability and statistics, that the net rate of chemical reactions depends upon the voltages. If you continue to supply precisely the open-circuit voltage $\epsilon,$ then the rate of chemical reaction ought to be precisely zero, which means zero current. In order to sustain a certain current $I,$ the voltage must thus drop a little to let the rate of chemical reaction be equal to the circuit current $I.$

It is this that we crudely model as an internal resistance, so $$\epsilon=I\left(R+r\right)\qquad\implies\qquad V=IR=\epsilon-Ir$$