Being really fascinated by how electricity actually works, I started studying batteries.
I initilally understimated the complexity of this apparently-simple objects, and thorugh studies and researches I have understood how they are the result of complex chemical and physical discoveries, blendend together in a triumph of engineering.
Anyway there are some things that I still have to figure out completely, I made some pictures that aim to clarify my doubts and the question objects.
This is the picture of a zinc-copper cell with a non-specificated electrolyte:
enter image description here
It represents essentially how I imagine a cell of this kind.
Zinc and Copper have both reached a dynamic equilibirum with the electrolyte, so charges leave electrodes at the same rate as charges enter.
This creates an electrode's electric field, and the elctrode potential at the same time. The net electric field represents the net force driving the elctrons from the zinc to the copper, resulting from the superoposition of the elctrodes fields.
Given that this analysis is right ( If it's not, please help me understand where is flawed), I don't understand how this electric field ( and so the potential) changes as time passes.
If electrons are leaving zinc to reach copper (and eventually react with some positive ions in the electrolyte,right?), how the electric field is mantained so that the flow can go on?
As electrons leave, how the zinc electrode field is affected?
Are positive ions leaving the zinc at a faster rate in order to compensate for th electrons deficency? (That would explain the zinc corrosion)
Is there an electrode's size limit (corrosion limit) where the electrons that make up the elctrode aren't enough to keep the same voltage(the chemistry's votlage)?
Is it at the point that we start to see a decrease in the voltage of the cell?
Of course, all the considerations presume the precence of a salt bridge capable of mantaing the elctrolyte charge stable.


closed as too broad by Mithoron, M.A.R. ಠ_ಠ, pentavalentcarbon, airhuff, Todd Minehardt Mar 27 '18 at 23:11

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi, if anyone who put the question on hold has any suggestion on how to make it more concise I'll be happy to follow it. I think it would be bad to close this quesiton since it touches some useful point that may be unclear to a lot of people. Moreover I spent a lot of time, trying to make it good and clear :) $\endgroup$ – Gabriele Scarlatti Mar 28 '18 at 14:24

I'll assume you are someone less than 16 years of age, since usually it's around 16-17 years (11th or 12th grade) that you'll probably learn how batteries work.

You're assumptions about how electrons are taken from the zinc cell and moved to the copper cell are correct. But considering your setup, as the charge moves, an electric field and a potential difference would build up, won't it? In such case a battery would eventually stop functioning.

While batteries do stop functioning after a while, it's due to the reaction being completed (all of the zinc converted to $\ce{Zn^{2+}}$ and/or all of the $\ce{Cu^{2+}}$ deposited as copper onto the copper electrode. Charge separation does not occur. Then how do the charges remain balanced, you might ask. And that's where the salt bridge comes into the picture.

A Salt Bridge is a tube that connects the two half-cells to maintain electrical neutrality. This tube makes sure that the ionic solutions on both sides are electrically connected, but at the same time prevents diffusion of both solutions to become one (see the link to know exactly how this works).

Feel free to comment if you need any help or clarifications.

  • $\begingroup$ Thank you for your help Pritt! :) I'm 20 unfortunately but still very curious even if I don't have a strong chemistry background! I will try to explain my understanding with a simplified version :) Let'say the dynamic equilibrium in the zinc cell it's reached when there are always 7 "unpaired" electrons on the surface and 7 positive ions in the electrolyte, while in the copper cell it's reached when there are 3 electrons on the surface and 3 positive ions in the electroyte. The net electric field favours the copper cell. $\endgroup$ – Gabriele Scarlatti Mar 28 '18 at 15:02
  • $\begingroup$ Two electrons will then travel from zinc to copper in order to balance charge, they will reach the electrode and join the positive ions in the copper cell. The salt bridge will then release 2 positive ions in the solution in order to keep the ions concetration unaffected.In the meantime, in the zinc cell, as soon as the 2 electrons leave, 2 zinc ions will be released by the zinc in the electrolyte in order to mantain the dynamic chemical equilibrium, so that there are always 7 "unpaired" electrons on the surface. $\endgroup$ – Gabriele Scarlatti Mar 28 '18 at 15:08
  • $\begingroup$ The salt bridge will then release 2 negative ions in order to mantain the electrolyte ions concentration unaffcted. This cycle repeats. My guess was that voltage start to decrease when Zinc electrode's size start to be so small that the chemical equilibirm can't be mantained anymore. I hoep my analysis is clearer and it's now easier for you to address any misconception. $\endgroup$ – Gabriele Scarlatti Mar 28 '18 at 15:09
  • $\begingroup$ @GabrieleScarlatti To an extent yes, but it's still an approximation. Let's consider the "unpaired electrons" in terms of a more easily measurable term called "emf", which is simply the work done in moving a unit charge across the electrodes. Emf is also the potential difference across the electrodes. The emf is not always constant (for routine purposes it appears to be) and actually depends on the concentration of the ions in the solution. For your copper-zinc case, the reaction is $$\ce{Zn + Cu^{2+} <=> Cu + Zn^{2+}}$$ The equation for emf is infact, continued in next comment... $\endgroup$ – Pritt Balagopal Mar 28 '18 at 15:20
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    $\begingroup$ Thank you for all your time and patience, I think that this formulas are a little advanced for my current level. I hope that my understanding even if a little naive, it's not flawed and reflects the general process even if it abstracts from the formulas. If there is something you wish to correct that you omitted, I would be glad to hear it, or if you have some suggestions. Thank you very much, I'm in debt. $\endgroup$ – Gabriele Scarlatti Mar 28 '18 at 15:51

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