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I have a question about what happens in following experiment: Assume we have a battery with two Galvanic half cell, eg a magnesium and a copper cell. Since copper has higher electrode potential that magnesium, the electrode with magnesium oxidizes and therefore becomes the negative terminal und the electrode with copper reduces and becomes positive terminal.

I want to find out what happens at the moment when we connect the earth? (the earth is here considered as electrostatically neutral uncharged source with constant potential, which can absorb und donate a huge amount of electrons without changing it's potential) See also this image:

enter image description here

Question: What happens at the moment we connect the negative terminal (magnesium) having excess of electrons with earth und leave simultaneously the positive terminal unconnected?

Clearly, since the positive terminal is unconnected, there cannot be a steady current because there is no closed circuit between positive and negative terminals.

But I'm primary interested in electrostatic effects and want to know if there happens a electrostatic discharge at the contact moment within a very very short moment sending the excess electrons from negative terminal to earth or does there really literally 'nothing' happen?

Some ideas: Why I conjecture that such elecrostatic discharge should happen: At the moment when we connect the earth to the magnesium electrode seemingly the unconnected copper electrode isn't noticed by the system earth-magnesium electrode since we assumed that there is no connection to the copper electrode:

enter image description here

And since the earth is neatrally charged and the magnesium electrode contains a lot of electrons on it, it seems reasonable that all this electrons 'want' to run into the neutral earth at the moment we connect the earth to the magnesium electrode. Does it sound reasonable.

That seems to be quite similar to the question what happens immediately after we connect the positive to negative terminals of two Galvanic half cells but remove the salt bridge:

enter image description here

Again, clearly without salt bridge there can no consistent current happen because the salt bridge closes the circuit as ion conductor.

But nevertheless, at the moment we connect the two terminals/electrods whith a wire, shouldn't the excessed electrons at magnesium electrode not nevertheless flow to copper electode and then stay there to compensate the charge/potential difference? And only then because due to lack of salt bridge everything stops?

Are my reasonings correct or if not what is my error? (I asked the same question in physicsstack

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  • $\begingroup$ Please cite the source of your first diagram. $\endgroup$
    – Bob D
    Aug 30, 2021 at 21:26
  • $\begingroup$ chemguide.co.uk/physical/redoxeqia/introduction.html#top $\endgroup$
    – user267839
    Aug 31, 2021 at 21:51
  • $\begingroup$ ...and thanks to insightful discussions below I came to conclusion that the linked introduction to redox equilibria is very misleading, since the pictures of electrode bars and electrons there could lead to wrong thoughts that electrons in electrode bar are distributed there like in a statically charged metal bar (what essentially caused my confusion) $\endgroup$
    – user267839
    Aug 31, 2021 at 22:00
  • $\begingroup$ I think you got it. But I’m not sure any of the other answers completely considered that upon connecting the magnesium electrode to ground the capacitance between the positive electrolyte and earth could result in a transient current. I would have offered an answer along those lines but I’m not conversant enough in battery chemistry $\endgroup$
    – Bob D
    Aug 31, 2021 at 22:34
  • $\begingroup$ For sake of completeness I found here ( semanticscholar.org/paper/…) after long search a paper where the voltage is explained finally in a satisfactory way instead of this weird oversimplified and also wrong school book explanantion that allegedly the potential difference between the two terminals of a battery "arise due to excess of electrons in anode and lack of electrons in cathode", which can be found in first 50 or 100 entries if you google "potential difference in battery". $\endgroup$
    – user267839
    Sep 1, 2021 at 18:27

5 Answers 5

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The situation of the extra electrons in the magnesium bar is different of a bar with an excess of electrons due to friction for example. In this second case, any connection with the earth tends to neutralize the bar.

But the system magnesium bar + electrolyte must be taken as a whole. The excess of electrons in the bar are not free only because they are far from the positive ions in the solution. So, connecting it to the earth is like breaking the glass of an aquarium with coral reefs and fishes, connecting it to the ocean. The fishes have nothing to actract them out from its source of food.

The same happens if the bars are connected without also connecting the solutions. We can not think of the electrons in each bar forggeting its interactions with the ions.

Only after bars and solutions are connected we have another system, composed by the bars and the electrolyte, and a system out of equilibrium while the chemical reaction is running.

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  • $\begingroup$ So your argument is that if we consider a general half-cell this consists of two components, the metal bar, a lattice of Me-atoms and a electrolyte solution, in our example with magnesium the electrolyte solution is MgSO4 and depending on the chemical equilibrium Mg <-> Mg2+ + 2e some magnesium atoms give their electrons up "into the bar" and dissociate as positive Mg2+ ions into the solution. That's the background, now the punchline (if I correctly understood you): $\endgroup$
    – user267839
    Aug 29, 2021 at 20:32
  • $\begingroup$ You mean that the released electrons in the magnesium bar and not really "free" in the sense that that if we connect to the magnesium bar to a conductor which is less neagative charged, then the electrons will NOT be able without any obstacles to move there, beacause they are also attracted to the surface by the positive Mg2+ ions which bustle at the phase between the bar surface and the solution, that the point? So due to the opposed charges these ions prevent the negative electrons to "leave" the magnesium bar if we connect it to neutrally charged earth. $\endgroup$
    – user267839
    Aug 29, 2021 at 20:32
  • $\begingroup$ So in summary the punshline is that the presence of these Mg2+ at the surface-solution contact phase which with their positive charge fixing the electrons in the magnesium bar preventing them from flying away to eg neutrally charnged earth? Did I reproduce correct your point? $\endgroup$
    – user267839
    Aug 29, 2021 at 20:33
  • $\begingroup$ Yes, that is. The bar with excess of electrons and the solution with an excess of positive ions is a configuration energetically more favorable than both parts neutral. It is different from a metallic bar alone with excess of electrons. $\endgroup$ Aug 29, 2021 at 20:51
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I don't have a firm grasp on the chemistry, but the reaction is limited by the voltage difference between the two electrodes. The action of the cell will tend to keep that difference constant. So, if there's a transient flow of electrons from the negative terminal to ground that raises the voltage of that terminal with respect to ground by some amount,* then I would expect an equivalent amount of positive charge (i.e., half as many "+2" ions) to flow across the salt bridge, and raise the voltage of the positive terminal by the same amount.


* That is, assuming that the initial voltage of the negative terminal W.R.T. ground was less than zero. Making the electrical connection to ground in that case "raises" the voltage to zero.

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  • $\begingroup$ You wrote in your answer 'if there's a transient flow of electrons...' .then as you said as a consequence there should flow an equivalent amount of positive charge flow across the salt bridge to compansate the charge loss at negative potential. So how I understand you here is that if such flow hypothetically (!) would happen, then the consequences would be... $\endgroup$
    – user267839
    Aug 24, 2021 at 19:06
  • $\begingroup$ But my question is exactly if such transient flow of electrons from the negative terminal to ground really happens in nature (so if it detectable or only hypothetical). $\endgroup$
    – user267839
    Aug 24, 2021 at 19:27
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No current will occur since there is no potential difference between the battery clemma an the Earth.

On the other hand, lightnings are happening pricisely due to existing such a potential difference.

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  • $\begingroup$ Yes I argree with you that of course there could no steady current happen, simply because there is nowhere a closed circuit, that's what I already remarked in the question. $\endgroup$
    – user267839
    Aug 24, 2021 at 19:21
  • $\begingroup$ In the second sentence you refer to a 'potential difference' between negative terminal and earth, right? (maybe a stupid question but since there is also a 'potential difference' between the two terminals of the battery involved, I wanted to assure that the 'potential difference' you are talking in your answer is that one between the negative terminal and earth) $\endgroup$
    – user267839
    Aug 24, 2021 at 19:26
  • $\begingroup$ If I understand you correctly, then after connecting the negative terminal with the earth there should indeed happen for very short moment a electrostatic discharge by absorbation of electrons from negative terminal by earth. Followed by the movement of ions flowing across the salt bridge to compensate the lost charge in order to keep the potential difference between both battery terminals constant, right? And only then nothing more happens in this system? $\endgroup$
    – user267839
    Aug 24, 2021 at 20:20
  • $\begingroup$ No even electrostatic discharge may happen. For the Earth the battery is neutral as any other body consisting of equal number of positively and negatively charges. $\endgroup$ Aug 25, 2021 at 3:38
  • $\begingroup$ I'm still confused. Of course the battery as total system is neutral, but we assumed that we connect the earth only to the negative terminal chunk inhabited with all theese electrons (=magnesium electrode); which are "given up" by the magnesium atoms swimming now as possitively charged $2+$-ions. So seemingly as long as we keep the positive terminal (=copper electrode) disconnected, the earth 'sees' only this negatively charged magnesium chunk inhabited with all theese electrons. What prevents then the electrons from beeing absorbed electrostatically by the earth? $\endgroup$
    – user267839
    Aug 27, 2021 at 19:12
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The electrochemical reaction tries to maintain a constant potential difference between the magnesium and the copper with the copper being at a higher potential than the magnesium.
You can think of the electrochemical reaction a being like a pump which has the effect of producing a net movement of negative charges from one terminal (which becomes the positive terminal of the cell - copper) to the other terminal (which becomes the negative terminal of the cell - magnesium).
The excess charges on the two electrodes set up an electric field inside the cell in opposition to the migration of the charges and eventually the magnitude of that electric field is sufficient to stop the migration of charges between the two electrodes. There is then a constant potential difference between the two terminals.

What you do not know is the potential of the earth relative to either of the two electrodes.

If the potential of the earth is the same as that of the magnesium then connecting the two together has no effect.

If the potential of the earth is higher than that of the magnesium electrode before a connection is made, then the potential of the copper must rise to maintain the constant potential difference between the copper and the magnesium.
There will therefore be a net migration of negative charges from the copper, through the cell, to the earth until the potential difference between copper and magnesium is the same as it was before the connection was made.

If the potential of the earth is lower than that of the magnesium electrode before a connection is made, then the potential of the copper must fall to maintain the constant potential difference between the copper and the magnesium.
There will therefore be a net migration of negative charges from the earth, through the cell, to the copper until the potential difference between copper and magnesium is the same as it was before the connection was made.

The statement since the earth is neutrally charged that was made is possibly the reason why the question was asked. The important parameter is not the quantity of excess charge resident on the earth rather it is the potential difference between the earth and the copper/magnesium.

The removal of the salt bridge whilst the copper and magnesium are connected does indeed lead to a net migration of negative charges from the magnesium to the copper until the potential difference between the two is zero.

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  • $\begingroup$ You wrote about the case if the potential of earth is higher than that of the magnesium electrode. There you wrote that in this case as a consequence of the battery's effort to maintain the constant potential difference between the copper and the magnesium electrodes, there will be a net migration of negative charges from the copper throgh the cell to the earth. $\endgroup$
    – user267839
    Aug 29, 2021 at 19:42
  • $\begingroup$ How is it possible? I assume you are talking there about the first experiment from my opener where we connect the negative terminal to the earth and leave the positive terminal unconnected. How can in this case concretely negative charges from the copper (!) migrate through the cell? Since there is no connection to positive terminal (=copper), seemingly the electrons from there have "no way" to migrate to earth, or do I miss some important aspect? $\endgroup$
    – user267839
    Aug 29, 2021 at 19:42
  • $\begingroup$ If you have a potential difference across a piece of wire charges will flow until the potential difference is zero which will occur when the potential of the magnesium reaches the potential of the earth. That cannot be the final state as there is now no potential difference between the magnesium and the copper so there must also be a redistribution of charge within the cell to achieve the final steady state potential difference across the cell. $\endgroup$
    – Farcher
    Aug 29, 2021 at 22:27
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I have a question about what happens in following experiment: Assume we have a battery with two Galvanic half cell, eg a magnesium and a copper cell. Since copper has higher electrode potential that magnesium, the electrode with magnesium oxidizes and therefore becomes the negative terminal und the electrode with copper reduces and becomes positive terminal.

You first need to put some electrolyte in water. The left vessel will need magnesium sulfate and the right vessel copper (II) sulfate. They dissolve in water and we'll have Zn++ and SO₄-- ions floating in the left and Cu++ and SO₄-- in the right vessel.

Magnesium is more reactive and will lose electrons more readily than copper so the magnesium bar will become the negative terminal (anode). Doubly ionised Mg++ ions from the bar will join the other Mg++ ions in the solution. They will pull Cl- ions from the salt bridge. The Na+ ions from the salt bridge will dissolve in the right vessel and the Cu++ ions will stick to the copper cathode bar, which attracts it using the electrons it took from the magnesium bar. Presently, everything is neutral.

None of that happens of course until you connect the cathode to the anode with a conductor.

Question: What happens at the moment we connect the negative terminal (magnesium) having excess of electrons with earth und leave simultaneously the positive terminal unconnected?

If the copper terminal is unconnected, the magnesium bar does not have excess electrons. It has as many electrons as it has protons, it's electrically neutral. Connecting magnesium to the earth will have no effect. You might as well not connect it to anything. Also, there is neither positive nor negative terminal in a cell until you connect the terminals. A brand-new battery with a plus and a minus sign printed on it does not really have these terminals until you close that circuit; they become terminals in that instant. And because one metal is more eager to lose electrons than the other.

Nothing happens in the electrolyte either, the ions continue floating unaware of the earth having been plugged in to the magnesium bar.

That seems to be quite similar to the question what happens immediately after we connect the positive to negative terminals of two Galvanic half cells but remove the salt bridge

Without the salt bridge, some electrons will be exchanged between magnesium and copper but not many because the loss/gain of electrons will make magnesium slighly positive, preventing it from donating any more electrons and the copper will become slightly negative and therefore less eager to accept electrons. Nothing will happen. You needed that salt bridge to maintain the reaction but now it will stop before it really begins.

And the magnesium bar will not remain slighly positively charged just because it's easier for it to lose electrons. Copper can lose electrons for magnesium as well just somewhat less often.

So basically you're under the incorrect impresion that there are excess electrons in the magnesium bar. I think this is the source of all the confusion.

Some additional remarks. NOTHING travels across the bridge. Neither Mg nor Cu can travel across the bridge. The bridge has the job of keeping the electrolytes from mixing. However, it's filled with Na+Ch- which can leave the bridge and enter the vessels.

Next thing, metals have free electrons that are in constant motion and they are free to travel anywhere throughout the volume of the conductor. If you bring two pieces of metal in physical contact, their free electrons will now be able to cross from one to another, in both ways so there will be no net flow.

The Earth doesn't have the same propensity because it doesn't have as many free electrons as a metal, plus it's conductivity may be partially due to moisture. Earth is not objectively speaking as good a conductor as a metal, however it does a good job at conducting electricy dues to it's 3-dimensional geometry, whereas wires for example are one dimensional.

So there is a difference between metals and earth. Take also under consideration, the vastly varying orders of magnidute involved in electricity. There are about $6\cdot 10^{15}$ electrons per second in one miliampere, but there may be microamperes, picoamperes, at some point we would stop caring that there is a current if it's too small to detect. One electron is already a current. And absent electron generates electric field that will pull it back so we again have a current but in the opposite direction. This happens all the time but we won't measure it with any devices because it's quick and net is zero.

You need that salt bridge. You need all the parts for anything to happen.

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  • $\begingroup$ About the "Magnesium is more..." part: Probably you want to say that because magnesium is more reactive than copper and therefore tends stronger to lose it's electrons the the magnesium bar becomes negative terminal? (See the "definitions" part here: en.wikipedia.org/wiki/…) $\endgroup$
    – user267839
    Aug 28, 2021 at 18:59
  • $\begingroup$ Moreover when you say "They will pull Cl- ions from the salt bridge" probably you mean the SO₄-- ions from the right vessel which are attracted by Mg++ ions and can pass through the salt bridge. What do you mean by "The Na+ ions from the salt bridge will dissolve in the right vessel"? $\endgroup$
    – user267839
    Aug 28, 2021 at 18:59
  • $\begingroup$ Also I not completely understand the following part. On one hand you say that if the the copper terminal is unconnected, then if we connect only the magnesium bar to earth then no electrons will flow to earth because the half cell/terminal consisting of magnesium bar sitting in the electrolyte is electrically neutral: The number of electrons in the bar is the same as the number of protons in water/electrolyte solution, right? $\endgroup$
    – user267839
    Aug 28, 2021 at 18:59
  • $\begingroup$ On the other hand consider the second story where we connect the two terminals but remove the salt bridge. You say that here at the moment of connection there will short, but not permanent flow of electrons between two terminals. Why in this case there will be a electron flow, but in case of connecting the nagative terminal to earth there is no short flow (note, when I say "short flow" I only mean only short electrostatic effect; of course in both cases no permanent current is possible)? $\endgroup$
    – user267839
    Aug 28, 2021 at 19:00
  • $\begingroup$ Indeed applying the same logic as for earth-negative terminal case, when we connect the two terminal and let the salt bridge be removed, then each of the two terminals has the same number of electrons in the bar as the number of protons in water/electrolyte solution. So each terminal is due to this argumentation neutral and applying this "logic" as above there should be no electron flow between the two terminals. $\endgroup$
    – user267839
    Aug 28, 2021 at 19:00

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