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Are the plates of a battery really charged?
Naive Question About BatteriesBattery and current confusion?How do electrons distribute themselves along a wire attached to one pole of a battery?What is the cause of the time-varying electric and magnetic near-fields around a wire with a 60Hz AC current?Why does not a battery kept in empty space discharge on its own? If space between the terminals is empty, what prevents the electrons to flow?Electron gun; potentials around charged platesIf there is a potential difference between the ends of a battery, does it mean that there is always an electric field?Mechanism by which a capacitor is charged, using surface chargesWhy is the electric field inside a charged conductor zero?How does the electric field produced by a battery look like and behave?Using a significantly large metallic sphere, could we draw out the electrons from the negative terminal of a battery?
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$begingroup$
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 !
electromagnetism electricity electric-circuits electrons electric-fields
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
$endgroup$
|
show 8 more comments
$begingroup$
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 !
electromagnetism electricity electric-circuits electrons electric-fields
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
$endgroup$
$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
1
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
1
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
4
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
1
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51
|
show 8 more comments
$begingroup$
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 !
electromagnetism electricity electric-circuits electrons electric-fields
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
$endgroup$
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 !
electromagnetism electricity electric-circuits electrons electric-fields
electromagnetism electricity electric-circuits electrons electric-fields
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
edited Jun 30 at 12:37
mohamed azaiez
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
asked Jun 30 at 11:21
mohamed azaiezmohamed azaiez
601 silver badge11 bronze badges
601 silver badge11 bronze badges
$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
1
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
1
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
4
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
1
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51
|
show 8 more comments
$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
1
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
1
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
4
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
1
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51
$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
1
1
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
1
1
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
4
4
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
1
1
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51
|
show 8 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
$endgroup$
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
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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.
$endgroup$
– Poutnik
Jun 30 at 17:22
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@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
|
show 6 more comments
$begingroup$
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.
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
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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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
|
show 10 more comments
$begingroup$
Metals of electrodes in the open circuit undergo these electrochemical reactions:
$$requiremhchem ceZn(s) <=> Zn^2+(aq) + 2 e-$$
$$ceCu(s) <=> Cu^2+(aq) + 2 e-$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^circ_M + fracRTnFln a_M^n+$$
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.
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
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it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
|
show 11 more comments
$begingroup$
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 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
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$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
add a comment |
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4 Answers
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4 Answers
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$begingroup$
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.
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
$endgroup$
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
|
show 6 more comments
$begingroup$
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.
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
$endgroup$
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
|
show 6 more comments
$begingroup$
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.
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
$endgroup$
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.
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
edited Jun 30 at 20:06
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
answered Jun 30 at 14:46
DaleDale
8,4691 gold badge11 silver badges37 bronze badges
8,4691 gold badge11 silver badges37 bronze badges
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
|
show 6 more comments
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 15:56
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
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.
$endgroup$
– Poutnik
Jun 30 at 17:22
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
$begingroup$
@mohamed azaiez no, not all are removed. I will add a bit to the answer
$endgroup$
– Dale
Jun 30 at 18:24
2
2
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
$begingroup$
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.
$endgroup$
– Dale
Jul 1 at 16:18
1
1
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
$begingroup$
@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 :)
$endgroup$
– Luaan
Jul 2 at 7:35
|
show 6 more comments
$begingroup$
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.
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
|
show 10 more comments
$begingroup$
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.
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
|
show 10 more comments
$begingroup$
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.
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
$endgroup$
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.
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
edited Jun 30 at 20:24
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
answered Jun 30 at 15:41
Alfred CentauriAlfred Centauri
49.8k3 gold badges52 silver badges158 bronze badges
49.8k3 gold badges52 silver badges158 bronze badges
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
|
show 10 more comments
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:54
1
1
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
In fact, it does not halt the reactions, it just makes the opposite reactions to have equal rate, forming thermodynamic equilibrium.
$endgroup$
– Poutnik
Jun 30 at 17:17
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
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
$endgroup$
– mohamed azaiez
Jun 30 at 17:20
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
$begingroup$
@Poutnik, thus the scare quotes around halts and the parenthetical (net)
$endgroup$
– Alfred Centauri
Jun 30 at 17:33
1
1
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
$begingroup$
@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.
$endgroup$
– Alfred Centauri
Jun 30 at 17:46
|
show 10 more comments
$begingroup$
Metals of electrodes in the open circuit undergo these electrochemical reactions:
$$requiremhchem ceZn(s) <=> Zn^2+(aq) + 2 e-$$
$$ceCu(s) <=> Cu^2+(aq) + 2 e-$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^circ_M + fracRTnFln a_M^n+$$
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.
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
|
show 11 more comments
$begingroup$
Metals of electrodes in the open circuit undergo these electrochemical reactions:
$$requiremhchem ceZn(s) <=> Zn^2+(aq) + 2 e-$$
$$ceCu(s) <=> Cu^2+(aq) + 2 e-$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^circ_M + fracRTnFln a_M^n+$$
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.
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
$endgroup$
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
|
show 11 more comments
$begingroup$
Metals of electrodes in the open circuit undergo these electrochemical reactions:
$$requiremhchem ceZn(s) <=> Zn^2+(aq) + 2 e-$$
$$ceCu(s) <=> Cu^2+(aq) + 2 e-$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^circ_M + fracRTnFln a_M^n+$$
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.
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
$endgroup$
Metals of electrodes in the open circuit undergo these electrochemical reactions:
$$requiremhchem ceZn(s) <=> Zn^2+(aq) + 2 e-$$
$$ceCu(s) <=> Cu^2+(aq) + 2 e-$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^circ_M + fracRTnFln a_M^n+$$
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.
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
edited Jun 30 at 17:53
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
answered Jun 30 at 13:13
PoutnikPoutnik
9105 silver badges10 bronze badges
9105 silver badges10 bronze badges
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
|
show 11 more comments
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
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 ?
$endgroup$
– mohamed azaiez
Jun 30 at 13:49
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
$begingroup$
Who says zinc stays neutral ?
$endgroup$
– Poutnik
Jun 30 at 14:01
1
1
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
Zinc sends electrons to circuit, not to copper. No charges on electrodes = no current from battery.
$endgroup$
– Poutnik
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
it transforms into zn2+ but the zinc electrode is still neutral
$endgroup$
– mohamed azaiez
Jun 30 at 14:06
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
$begingroup$
Wrong. Try again. The electrode is not neutral.
$endgroup$
– Poutnik
Jun 30 at 14:08
|
show 11 more comments
$begingroup$
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 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
$endgroup$
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
add a comment |
$begingroup$
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 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
$endgroup$
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
add a comment |
$begingroup$
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 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
$endgroup$
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 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
answered Jun 30 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
answered Jun 30 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
answered Jun 30 at 14:01
Stanislav BashkyrtsevStanislav Bashkyrtsev
4302 silver badges14 bronze badges
4302 silver badges14 bronze badges
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
add a comment |
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:57
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
But Why would zn electrode accept the extra electrons when it wants to get rid of them ?
$endgroup$
– mohamed azaiez
Jun 30 at 15:58
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
$begingroup$
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.
$endgroup$
– Stanislav Bashkyrtsev
Jun 30 at 16:21
1
1
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
$begingroup$
@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.
$endgroup$
– Luaan
Jul 2 at 7:40
add a comment |
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$begingroup$
Is this en.m.wikipedia.org/wiki/Lemon_battery the type of battery that you are referring to?
$endgroup$
– my2cts
Jun 30 at 11:32
1
$begingroup$
C a n you update your question with more details?
$endgroup$
– my2cts
Jun 30 at 11:44
1
$begingroup$
i will try to change my question Thanks for trying anyway !
$endgroup$
– mohamed azaiez
Jun 30 at 12:30
4
$begingroup$
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.
$endgroup$
– Dawood ibn Kareem
Jul 1 at 6:15
1
$begingroup$
"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.
$endgroup$
– Loong
Jul 1 at 19:51