Are the plates of a battery really charged?

Question

In a zinc/copper Daniell cell correct me if I am wrong :

Zinc has 2 valence electrons. So it wants to get rid of them. To do so it sends them to the copper which needs 2 to complete its valence shell.
There needs to be a wire between the zinc and the copper for this reaction to happen.
So technically the plates are not charged. It's just the charges flowing out that create the electric field.

TLDR : Are the plates of a battery more like a capacitor with excess charges on the plates? Or do they simply throw in and out electrons near their terminals and the individual plates of zinc and copper are neutral?

My confusion is this : I understand that the zinc wants to get rid of electrons, and the copper wants more electrons, but : The zinc and copper atom are "neutral". it's only the defecit of electrons on the conductor near the positive terminal and the excess of electrons near the negative terminal, That for me would make an electric field .

Or Maybe it's the "wanting to get rid" and the "wanting to get more" electrons that create an electric field, if it's indeed that please confirm !

Thanks !

Is this https://en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to? — my2cts, Jun 30 '19 at 11:32
Then I don't understand what you are asking, so I deleted my answer. Good luck. — my2cts, Jun 30 '19 at 12:21
It seems the OP is completely confused about basic electrochemistry. Better than a limited answer, I suggest studying the basic concepts from suitable textbook, or at least Wikipedia articles. — Poutnik, Jun 30 '19 at 12:50
relevant: https://physics.stackexchange.com/questions/31923/naive-question-about-batteries — WillO, Jun 30 '19 at 14:09
It's almost always wrong to use words like "wanting to" to explain results in physics and chemistry. Zinc and copper are inanimate, and devoid of wants and needs. They never "want to" lose or gain electrons. To understand what's really going on, it's a good idea to abandon this notion. — Dawood ibn Kareem, Jul 01 '19 at 06:15
"it's the "wanting to get rid" and the "wanting to get more" electrons that create an electric field, if it's indeed that please confirm !" - Confirmed :) — JimmyB, Jul 01 '19 at 13:47
"Zinc has 2 valence electrons. So it wants to get rid of them." – No. The ionization energy of neutral atoms is always positive. Thus, the process M -> M+ + e- is endothermic. No element wants that. — , Jul 01 '19 at 19:51
We know that God isn't a computer scientist, because then chemical cells would implement late binding: the charge would appear only when needed to avoid the upfront cost of allocating it for batteries that just sit on the shelf without being put to use. — Kaz, Jul 04 '19 at 18:00

Dale · Accepted Answer · 2019-06-30T20:06:02.407

35

I think that most confusion about batteries comes from ignoring the electrolyte. For example:

Zinc has 2 valence electrons. So it wants to get rid of them. To do so it sends them to the copper which needs 2 to complete its valence shell.

The actual reaction is between the metallic zinc and the dissolved zinc. Zinc wants to get rid of two electrons and it does so by becoming an ion and going into solution.

This reaction is energetically favorable and can occur even if there is a small electric field opposing it at the surface of the electrode. However, the reaction products near the electrode surface, zinc ions and electrons, are highly charged and quickly produce a strongly opposing field which overcomes the energetic favorability and halts the reaction. For the reaction to proceed the reaction products must be removed from the region near the electrode surface.

The electrons can be removed from the electrode surface by transport through the wire, and the ions can be removed from the surface by transport through the fluid. The transport of electrons requires the complementary reaction at the copper electrode, and the transport of the ions requires a complementary transport of the solute ion in the electrolyte. Understanding the electrolyte is essential for understanding batteries, and is the usual neglected piece.

There needs to be a wire between the zinc and the copper for this reaction to happen.

The purpose of the wire is not to make the reaction happen. The reaction is energetically favorable, so it briefly happens regardless. The purpose of the wire is to remove the reaction products so it can continue to happen.

Note that in this process not all of the excess electrons on the plate are normally removed. As soon as a few are removed the reaction proceeds and replenishes those few. The reaction thus proceeds at the rate that the products are removed from the immediate vicinity of the electrode surface. In abnormal situations, like a short circuit, a substantial fraction of the excess charges at the surface can be depleted and the current is limited by the reaction kinetics. This manifests as an “internal resistance” for the cell.

edited Jun 30 '19 at 20:06

answered Jun 30 '19 at 14:46

Dale

99,825

ok my confusion is this when the wire removes the reaction products does it remove all the excess charges that the zinc had , thus resulting in the zinc plate becoming neutral again – mohamed azaiez Jun 30 '19 at 15:56
A wire does not remove products, all it does is allowing electrons to reach higher potential. And no, it does not remove all excess charge. Not even when the cell voltage is zero and no current flows any more. – Poutnik Jun 30 '19 at 17:22
@mohamed azaiez no, not all are removed. I will add a bit to the answer – Dale Jun 30 '19 at 18:24
So the number of excess charges in the electrode before the reaction coming to a halt is the voltage – mohamed azaiez Jun 30 '19 at 19:36
Roughly speaking yes, or at least it is proportional to the voltage. – Dale Jun 30 '19 at 19:54
Note the zinc got rid of its electrons in context of the metallic bonds with delicalized valance electrons. – Poutnik Jun 30 '19 at 20:31
So in a short circuit does the electric field become weaker due to less excess charge on the surface of the electrodes? – mohamed azaiez Jul 01 '19 at 15:07
2

Yes, it becomes substantially weaker. On the electrical side this is usually modeled as an “internal resistance”, but the effect is to reduce the voltage for high currents. – Dale Jul 01 '19 at 16:18
1

Zink does not want to get rid of two electrons. This part of the reaction is not energetically favourable. The first ionization energy of Zn is +9.394199 eV; the second is +17.96439 eV. I.e. this part is endothermic. – Jul 01 '19 at 20:02
1

@Loong hmm, good point. Perhaps I should say “Zinc wants to become an ion and go into solution and it does so by getting rid of two electrons” – Dale Jul 01 '19 at 21:45
2

@Loong Only if you ignore the electrolyte (as Dale noted is often done). Zinc does dissolve readily, losing the expected two electrons in the process (which of course need to be used to reduce the other half of the battery for the reaction to continue). Of course, this quickly stops if you don't put the battery in a circuit, which is why the battery can maintain "charge" over time, rather than spontaneously discharging before it gets to the consumer :) – Luaan Jul 02 '19 at 07:35
@Dale By energetically favorable do you mean exothermic or do you mean that the $\Delta G<0$? – tryingtobeastoic Apr 27 '22 at 05:42
I have understood from your answer how the Zn electrode is negatively charged. I'm still a bit confused as to how the Cu electrode is positively charged electrostatically. Is it appropriate to say that the Cu electrode too is also negatively charged, but it is less negatively charged than the Zn electrode? – tryingtobeastoic Apr 29 '22 at 05:35
1

@tryingtobeastoic by “energetically favorable” I mean that there is an electrochemical potential difference that drives the reaction “forwards”. Your follow up question should be asked as its own question (after looking for duplicates). It is a little much for comments – Dale Apr 29 '22 at 10:57
@Dale Posted the question. You can check it out if you want! Thank you so much for your response by the way. – tryingtobeastoic Apr 29 '22 at 12:10

Alfred Centauri · Answer 2 · 2019-06-30T20:24:50.677

11

TLDR : Are the plates of a battery more like a capacitor with excess charges on the plates?

Yes. This is why a (charged) cell has a non-zero open-circuit (no external circuit) voltage across. Despite the fact that there is no external circuit through which charge can flow, the reactions of the plates with the electrolyte result in one plate having a deficit of electrons (positively charged) and the other plate having an excess of electrons (negatively charged).

Essentially, it is this separation of charge (and the associated electric field) that 'halts' the (net) chemical reactions at the plates and produces the constant open-circuit voltage.

If electrons are allowed to flow through an external circuit (load) from the negative plate to the positive plate, the reactions proceed and the cell discharges.

In the case of a rechargeable cell, an external source can move electrons from the positive plate to the negative plate essentially reversing the chemical reactions at the plates and the cell charges.

(From the comments)

OK but when we connect both terminals with a circuit wouldn't all the excess charge simply flow in the wire and terminals become neutral again ?

Unlike a capacitor, the energy stored in a cell is in the form of chemical energy and not electrostatic energy. Only a tiny fraction of the stored chemical energy is needed to establish the open-circuit voltage.

Like a capacitor, the plates have equal and opposite electric charge but unlike a capacitor, chemical reactions 'try to' maintain the charge separation on the plates even as excess electrons flow from the negative plate, through the load, an onto the positive plate.

Of course, there is a limit and cells have a so-called short-circuit current that is the maximum current that can be supplied when the plates are connected together with a wire (I should point out that this is generally not a safe test to perform).

There is of course, lots more to all of this, and I do believe there's some good Q & A here on this subject.

edited Jun 30 '19 at 20:24

answered Jun 30 '19 at 15:41

Alfred Centauri

59,560

OK but when we connect both terminals with a circuit wouldn't all the excess charge simply flow in the wire and terminals become neutral again ? – mohamed azaiez Jun 30 '19 at 15:54
1

In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium. – Poutnik Jun 30 '19 at 17:17
Yeah but the negative terminal won t be charged anymore due to all excess charges going in the circuit it wont be like a capacitor anymore more like an electron pumping device – mohamed azaiez Jun 30 '19 at 17:20
@Poutnik, thus the scare quotes around halts and the parenthetical (net) – Alfred Centauri Jun 30 '19 at 17:33
I see, I must have misread it somehow. – Poutnik Jun 30 '19 at 17:35
@Poutnik What do you mean by to have equal rates and thermo dynamic equilibrium , thanks – mohamed azaiez Jun 30 '19 at 17:36
Ohhhhh i think i got it – mohamed azaiez Jun 30 '19 at 17:39
@mohamedazaiez, why do you think this way (referring to 1st comment)? Are you thinking that the chemical reactants are exhausted and so cannot continue when the external circuit is connected? In my answer, I explicitly write that when the external circuit is connected, the reactions continue. – Alfred Centauri Jun 30 '19 at 17:39
Equal rates mean motion of ions from electrode to solution and vice versa is ongoing by the same speed. If solution and metal were made from different isotopes, it would be observed they intermixed. Thermodynamic equilibrium = mechanical + thermal + chemical + radiation equilibrium, with dG=0 ( G is Gibbs energy or also called free anthalpy ). See TD equilibrium – Poutnik Jun 30 '19 at 17:42
1

@mohamedazaiez, that is, only a tiny fraction of the energy stored in the reactants is used to initially charge the plates to the open-circuit voltage. Unlike a capacitor, the energy stored in the cell is in the form of chemical energy rather than electrostatic energy. Now that I better understand what you've asked, I think I'll edit my answer to explicitly address this. – Alfred Centauri Jun 30 '19 at 17:46
Thanks guys you are really helpful so if the charges go out faster (short circuit) the cell chemical reaction will 'speed up' zinc will more rapidly disolve and the battery life team will be reduced but voltage will stay the same at a high current – mohamed azaiez Jun 30 '19 at 18:15
And can we say that the reaction still undergoes "short duration halts even when in a circuit – mohamed azaiez Jun 30 '19 at 18:16
@Poutnik and Alfred sorry i cannot like your posts ( 13 rep points) but thanks for making me understand .and last question is how many charges that both electrode can hold before the reactions coming to a halt that determines the voltage ? – mohamed azaiez Jun 30 '19 at 18:31
Voltage is current dependent due limited rate of kinetics of chemical reactions and ion diffusion. This affects formation of several internal cell potential differences, acting against the cell voltage. It is effectively manifested as the (variable) cell internal resistance. – Poutnik Jun 30 '19 at 18:37
The answer to your last question depends on the electrode capacitance what is not trivial to determine. – Poutnik Jun 30 '19 at 18:39
I have understood from Dale's answer how the Zn electrode is negatively charged. I'm still a bit confused as to how the Cu electrode is positively charged electrostatically. Is it appropriate to say that the Cu electrode too is also negatively charged, but it is less negatively charged than the Zn electrode? – tryingtobeastoic Apr 29 '22 at 05:31

Poutnik · Answer 3 · 2019-06-30T17:53:43.600

2

Metals of electrodes in the open circuit undergo these electrochemical reactions:

$$\require{mhchem} \ce{Zn(s) <=> Zn^2+(aq) + 2 e-}$$ $$\ce{Cu(s) <=> Cu^2+(aq) + 2 e-}$$

The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.

$$E_M=E^\circ_M + \frac{RT}{nF}\ln a_M^{n+}$$

Zinc ions have higher tendency to dissolve in solution, compared to copper, leading to more positive potential on the copper electrode.

When the circuit is closed, the potential difference causes the current flowing.

As consequence, both electrode potentials are disbalanced from their equilibrium values.

Zinc electrode potential gets higher, what leads to dissolving of zinc and pumping electrons to the circuit.

Copper electrode potential gets lower, what leads to copper ion reduction to metsl and draining electrons from the circuit.

Within the cell, a ion flow forms between electrodes, via a diaphragm or ionic bridge, following potential and concentration gradients.

About capacitors, there are at least 3 type of capacitance:

bulk metal of electrodes
dielectric layer of ions adsorbed at electrodes
solution itself due bulk volume charge displacement.

edited Jun 30 '19 at 17:53

answered Jun 30 '19 at 13:13

Poutnik

1,902

I get why zinc sends electrons to copper , but my confusion is this , Zinc SENDS electrons to copper But ZINC stays neutral in charge so no electric field can occur , My question is this , is it the zinc electronegativity and the copper electro positivity that create this electric field ? – mohamed azaiez Jun 30 '19 at 13:49
Who says zinc stays neutral ? – Poutnik Jun 30 '19 at 14:01
1

Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery. – Poutnik Jun 30 '19 at 14:06
it transforms into zn2+ but the zinc electrode is still neutral – mohamed azaiez Jun 30 '19 at 14:06
Wrong. Try again. The electrode is not neutral. – Poutnik Jun 30 '19 at 14:08
But the electrode never had excess charge it's just that zinc has 2 electrons in it's valence shell isn't it ? – mohamed azaiez Jun 30 '19 at 14:09
or do the electrons that it sends stay on the zinc electrode ? – mohamed azaiez Jun 30 '19 at 14:10
Neither, neither. – Poutnik Jun 30 '19 at 14:11
ok Last try do the electrodes have some sort of capacitence to build up an excess charge ? like a capacitor ? – mohamed azaiez Jun 30 '19 at 14:12
They do have capacitance, but it is rather unimportant, side feature. In fact, there is capacitance of electrode itself, and capacitance of dielectric layer. – Poutnik Jun 30 '19 at 14:13
Ok wait is the electric field between the electrodes or the excess charges near them ? – mohamed azaiez Jun 30 '19 at 14:16
either, either. Chemical processes create extra charges, what creates the field. At the electrodes, in the circuit, inside the cell. – Poutnik Jun 30 '19 at 14:17
Wait i think i got it the nernst equation is a balance between when the chemical reaction starts and the potential of the plate so when it reaches this it will start giving electrons out ? and it needs to reach it again to give electrons again ??? also really thanks for the help – mohamed azaiez Jun 30 '19 at 14:20
The E. equation tells the potential when both opposite reactions are equally fast, therefore there is thermodynamic equilibrium. If an electrode is forced to more positive potential , the net reaction is metal dissolution. If to more negative one, the net reaction is metal deposition. – Poutnik Jun 30 '19 at 14:23
So the zinc would more be like a full glass of water (water molecules are electrons) and when you try to put in more water , water overflows out of the glass (electrons leave the zinc) and the potential of this 1/2 cell is how much water can the glass hold ? if i am totally wrong just give me a book / a course or even a subject for more investigations . Thanks , – mohamed azaiez Jun 30 '19 at 14:32
Start with metal conductivity band and metallic bond Good are also electrochemistry chapters in Physical Chemistry textbooks. There are many, e.g. Moore. – Poutnik Jun 30 '19 at 14:37
I have understood from Dale's answer how the Zn electrode is negatively charged. I'm still a bit confused as to how the Cu electrode is positively charged electrostatically. Is it appropriate to say that the Cu electrode too is also negatively charged, but it is less negatively charged than the Zn electrode? – tryingtobeastoic Apr 29 '22 at 05:31
@tryingtobeastoic Well, both Zn and Cu electrodes have in fact positive potential, as the standard hydrogen electrode (SHE) has "absolute" potential wrt free electron +4.44+-0.02 V. – Poutnik Apr 29 '22 at 05:41
The standard reduction potential of Zn is negative and that of Cu is positive. So, I'm a bit confused as to what you are saying. Could you please elaborate? – tryingtobeastoic Apr 29 '22 at 06:14
@tryingtobeastoic That is because their potential is conventionally referenced to the potential of SHE which is conventional zero in electrochemistry context. – Poutnik Apr 29 '22 at 06:30

score 2 · Answer 4 · answered Jun 30 '19 at 14:01

2

Both Cu and Zn are dissolving from the respective metal pieces and return back at some rate. In order to be dissolved in water they need to be ions (because water is a polar solvent). So who has a greater tendency to dissolve? The element that's easier to ionize. In this case it's Zn because it has lower electronegativity and can give away its electrons (to Zn electrode) easier.

So a bit more of Zn will be dissolved than Cu and thus Zn electrode will have some build up of electrons. That will create an electric field in the conductor which will increase the density of electrons nearby which in turn will create an electric field down the wire.

Though this would halt very quickly because electrons will quickly build up on the other end. For this reason you also need a salt bridge. AFAIU, it will either contain additional ions itself that would dissolve in water or simply allow Zn- to flow to the Cu cell and vice versa.

PS: I'm not sure what role SO4 plays in this though, but it seems like it has even more affinity to Zn than just water which makes the dissolution better.

answered Jun 30 '19 at 14:01

Stanislav Bashkyrtsev

624
4
16

But Why would zn electrode accept the extra electrons when it wants to get rid of them ? – mohamed azaiez Jun 30 '19 at 15:57
But Why would zn electrode accept the extra electrons when it wants to get rid of them ? – mohamed azaiez Jun 30 '19 at 15:58
Electrode was neutral from the beginning. When Zn atom leaves the electrode it doesn't take its electrons (gets rid of them) leaving them on the electrode. Thus minus 1 Zn atom, but 2 of its electrons staid making the electrode slightly negative. – Stanislav Bashkyrtsev Jun 30 '19 at 16:21
1

@mohamedazaiez There's an equilibrium formed. A few zinc ions leave the electrode (each leaving two electrons behind), and producing a slight negative charge on the electrode. This is energetically favourable. If you don't close the circuit, this is maintained over time (batteries do degrade, but it takes a lot of time, especially with modern primary batteries). As many zinc ions get dissolved in the solution as get reabsorbed on the electrode. When you close the circuit, the electrode can get rid of the negative charge, and more zinc ions dissolve in the electrolyte, continuing the reaction. – Luaan Jul 02 '19 at 07:40
@Luaan I have understood from Dale's answer how the Zn electrode is negatively charged. I'm still a bit confused as to how the Cu electrode is positively charged electrostatically. Is it appropriate to say that the Cu electrode too is also negatively charged, but it is less negatively charged than the Zn electrode? – tryingtobeastoic Apr 29 '22 at 05:33

Are the plates of a battery really charged?

4 Answers4

Linked