If my understanding is correct, the chemical reaction in a battery is basically a conveyor belt that moves electrons from the positive to the negative terminal of the battery until the electric field between the terminals slows the rate of the reaction to zero.

If this is true, would the electric potential difference between the terminals then be given by the electric field between the terminals multiplied by the distance between them (assuming the electric field between them is constant)?

And if so, then would separating the terminals further increase the potential difference?

I feel like this is incorrect though because it would increase the potential energy stored in the battery which I thought was a constant value determined by the chemicals in the reaction, but if the electric field is the only thing slowing the reaction I don't see why separating the terminals should have any effect since it wouldn't change the electric field (again assuming the battery was designed similar to a capacitor in that the electric field between the terminals remained constant at every point between them).


1 Answer 1


Your first two paragraphs are correct. [One could quibble and say that the pd is the integral of the electric field due to charge separation with respect to distance as one goes on any path from one terminal to the other. This takes into account any non-uniformity of the field.]

The answer to the question in your third paragraph is "no". The pd stays the same but the mean value of the electric field decreases. This is equally in accord with your first two paragraphs. It is the pd that is fixed by the chemical processes inside the cell, the processes that give rise to the separation of charges. Using your analogy, the conveyor belt can only lift goods to a certain height, that is the chemical reactions can only give the charges enough energy per charge to produce a certain potential difference (the cell's emf).

  • $\begingroup$ So does this mean that it isn't the electric field that slows the reaction down, but the potential difference? And if so, then how exactly does it do that, since wouldn't the electric field (times charge) be the only "force" acting against the reaction? $\endgroup$ Dec 12, 2020 at 23:38
  • $\begingroup$ I'd say that your first sentence is correct. As for your second... If you move the electrodes further apart, the mean electric field strength will decrease but you don't significantly weaken the electric field force over the small regions near the electrodes where the reactions take place as there will be very little potential drop through the electrolyte. There is also the argument that chemical reactions are not the province of classical Physics and we shouldn't try to discuss them in terms of force! $\endgroup$ Dec 13, 2020 at 0:04
  • $\begingroup$ Oh yeah, I was forgetting that reaction rates are based on energy gained/lost by a reaction and that they go both ways (forward/reverse), thanks! $\endgroup$ Dec 13, 2020 at 0:17

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