Why does a voltaic cell induce an electric force proportional to voltage?

Question

I'm trying to understand the intuition behind Ohm's law (that current is proportional to voltage).

I understand the basics of the Drude model: if the electrons frequently collide and on average reset their velocities, then they won't accumulate much momentum and the average velocity will be proportional to the force from the constant electric field $E.$

What I'm missing is why we are allowed to substitute the voltage $V$ for $E$ in the Drude model. In particular, why does the voltage induce an electric field whose strength is proportional to $V?$

Normally we start with an electric field, and this naturally induces an electric potential (eg moving an electron against the field increases its potential). But when you start with the purely chemical fact that a reaction could take place and consume electrons at some unknown rate and release free energy in the process, I don't see how this defines the magnitude of an electric field.

My attempt at digging deeper into what the voltage actually means: The voltage of a galvanic cell is $V = -\Delta G/(nF)$ where $n$ is the number of electrons transferred in the reaction and $F$ is Faraday's constant. This makes sense to me: it is roughly the energy released when one Coulomb moves through the conductor and enables the chemical reaction, which matches the definition of voltage (amount of work required per unit of charge to move a test charge from one point to another).

But why is there an electric field proportional to the Gibbs free energy of the reaction? I've seen people informally use language like: "the chemical reaction causes a build-up of electrons at the anode, this is what causes the electric field." Is this the full story? If so, why is the amount of built up electrons at the anode proportional to the Gibbs free energy of the reaction?

Zooming out a bit, it seems strange to me that current should be proportional to Gibbs free energy, because current is the speed of electron flow, and Gibbs free energy is supposed to have nothing to do with the actual rate of the reaction...

Answer 1

I try to explain a little with informal language; the deep truth is: it's very complicated. If you manage to understand what really the electrons do, well you can get immediately a job in the huge battery industry/R&D. Sorry I'm no battery expert, I write the following from general electromagnetism knowledge.

1. Yes in the galvanic cell we can somehow say that the electrons accumulate at the anode and cause the electric field; it gets exactly proportional to the Gibbs energy because they accumulate until the anode is charged enough to repel more electrons, so there is equilibrium between chemical and electric potential. Electrons really like the anode and ammass there until it is crowded enough, that is the same for an electron to be in the crowd (high electric potential) or bound at the catode (Gibbs chemical energy). No net current here!
2. About current, well, the galvanic model works for "very small" or zero current actually. If you allow the battery to discharge and current to flow, yes you should look at the actual reaction rate and ions/electrons velocities in the medium. But, in normal cases, it is assumend that the reaction rate is MUCH FASTER than the electrons outflow, so the battery remains in almost chemical-electrostatic equilibrium and can keep the voltage. If you instead let a large current to flow, you may incur into the limitation of reaction rate ("internal resistance"?) but this is outside the scope of the simple galvanic cell model you are looking at.
3. So, for small current (high resistance of the external circuit), the current gets actually proportional to the Gibbs potential. At low current, the electrons/ions have negligible resistance at moving inside the battery, so they keep the anode fully crowded of electrons at all time and the voltage is stable, basically identical as in the stationary condition. The chemicals get slowly consumed and the Gibbs energy decrease, thus slowly decreasing the potential until the battery is fully discharged. This is the operation for very low currents; the Gibbs-Voltage relationship breaks down for higher currents, and the "max current achievable" will depend on many other parameters, like chemistry, temperature, battery geometry and size.
4. About the $V$ and $E$, I will say that the voltage is defined as integral of the electric field; and in a Drude model conductor, there is an external fixed voltage and this define the electric field; the travelling electrons (supplied in arbitrary amount by some source!) adapt their speed to the electric field, giving rise to a current; while overall the conductor is neutral (does not change the electric field, as instead in the galvanic cell).