How a Battery creates a Potential difference in an electrical current

[The Tortoise-Man]

Hello,
I'm seriously confused on several things around how concretely
batteries create potential difference in order to force the electrons
circulate trought the circuit wire.

Almost all explanations I found (wikipedia, diverse tutorials, intro scripts, etc.) explain it in nearly
the same way, namely they say something like: There are two electrochemical cells with different potentials,
and at that cell with lower electrode potential, the negative terminal, the electrons are going chemically
removed from atoms and are stored at this negative terminal of the battery, while at positive terminal the
positive ions attract and absorb the oncoming electrons.

So essentially the reason for the creation of the potential difference in the battery
between two poles which keeps the electric current circulating, is seemingly based
due to the explanations I found on surplus of electrons at negative terminal and
lack of electrons at positive terminal.

Therefore seemingly the potential difference in batteries inducing the electrical current
is based purely on exactly the same mechanism as electrostatic discharges like known from lightning
phenomena, that the excessive electrons from negative terminal want to compensate the electron
deficiency on the other side.

First of all, did I rephrased this usual explanation for reason of electric circuits for batteries correctly,
or did I misunderstood it already here?

Now we come to my actual concern. I think that if we use this way to explain the currency
in a battery circuit via surplus of electrons on one pole and deficiency of electrons
on other pole, following construction/ thought experiment leads us to wrong prediction:

Assume we have our circuit consisting of a battery, circuit wire and resistor und
we connect to the wire somewhere the connection to Earth. The Earth is considered here
as idealized model of a reservoir with everywhere same potential, infinite
amount of electrons which it could donate or absorb without changing it's potential.

Empirically we know, that if we connect the Earth to our battery circuit, nothing
happens. The circuit still runs. But on the other hand, if we adhere to our model
how the battery creates the potential difference between two poles I explained above seemingly
we running here into troubles.

Namely if we assume that in our circuit there is a surplus of electrons at negative terminal and lack of electrons
at positive terminal, then at the moment when we connect the Earth to the circuit theoretically we should
expect electrostatic discharge: the electrons from negative terminal going be absorbed by the Earth and
simultaneously the Earth provides enough electrons to fill the electron deficiency in positive terminal.
Then the the battery current should collapse since there is no more potential difference between two poles.What is the error in my reasonings here?

 

[.Scott]

Voltage is relative.
When a battery is not connected to a circuit, it creates a voltage difference between its terminals - pushing electrons towards the negative and and attempting to draw them from the positive end.

If you connect one end of a battery to an electrostatic source, leaving the other end open (unconnected), it will still maintain this voltage difference - even if the result is a surplus or deficiency at both terminals.

Strictly speaking, batteries do not "force the electrons to circulate". When the terminals are left open, electrons do not move. The battery provides a steady push - like the water behind a faucet. When you turn the faucet on (equivalent to providing a electric circuit for the battery), water moves.

 

[ergospherical]

Upon inserting an electrode into a solution, the difference between the Fermi energy level in the metal and the ionic energy levels in the solution provides the "driving force" for the transfer of electrons between the metal and the solution. As a consequence, a potential difference $\Delta \phi = \phi_M - \phi_S$ arises at the interface. Consider an example half-cell \begin{align*}
\mathrm{e}^{-}(\mathrm{metal}) + \sum_{A \in \mathcal{R}} \nu_A A \quad \leftrightharpoons \quad \sum_{B \in \mathcal{P}} \nu_B B
\end{align*}where $\mathcal{R}$ is the set of reactants and $\mathcal{P}$ is the set of products. Given that this reaction is in equilibrium, one can equate the electrochemical potentials $\bar{\mu} := \mu + qF\phi$ of the reactants and the products to obtain\begin{align*}
(\mu_{\mathrm{e}^{-}} - F\phi_M) + \sum_{A \in \mathcal{R}} \nu_A (\mu_A + q_A F \phi_S) = \sum_{B \in\mathcal{P}} \nu_B (\mu_B + q_B F \phi_S)
\end{align*}Charge conservation implies that $-1 + \displaystyle{\sum_{A \in \mathcal{R}}} \nu_A q_A = \sum_{B \in\mathcal{P}} \nu_B q_B$ therefore\begin{align*}
F\phi_S\left( \sum_{A \in \mathcal{R}} \nu_A q_A - \sum_{B \in \mathcal{P}} \nu_B q_B \right) - F\phi_M &= -\mu_{\mathrm{e}^{-}} + \sum_{B \in \mathcal{P}} \nu_B \mu_B - \sum_{A \in \mathcal{R}} \nu_A \mu_A \\ \\

F(\phi_M - \phi_S) &= \mu_{\mathrm{e}^{-}} - \sum_{B \in \mathcal{P}} \nu_B \mu_B + \sum_{A \in \mathcal{R}} \nu_A \mu_A
\end{align*}furthermore $\mu_A = \mu^{o}_A + RT \ln{a_A}$ hence\begin{align*}
\phi_M - \phi_S &= \Delta \phi^{o} + \dfrac{RT}{F} \left[ \sum_{B \in \mathcal{P}} \nu_B \mathrm{ln}(a_B) - \sum_{A \in \mathcal{R}} \nu_A \mathrm{ln}(a_A) \right] \\ \\

\Delta \phi &= \Delta \phi^o + \dfrac{RT}{F} \ln \left[ \dfrac{\displaystyle{\prod_{B \in \mathcal{P}}} a_B^{\nu_B}}{\displaystyle{\prod_{A \in \mathcal{R}}} a_A^{\nu_A}}\right]
\end{align*}where $\Delta \phi^o = \dfrac{1}{F} \left[ \mu_{\mathrm{e}^{-}} + \displaystyle{\sum_{A \in \mathcal{R}}} \nu_A \mu_{A}^o - \displaystyle{\sum_{B \in \mathcal{P}}} \nu_B \mu_{B}^o \right]$. This is called the Nernst equation for a half-cell. Due to the impossibility of measuring absolute voltages between the metal and the solution, the electrode potential of an arbitrary redox couple $\mathrm{W}/\mathrm{X}$ is defined w.r.t. an agreed reference - e.g. a $\mathrm{H}^+/\mathrm{H}_2$ couple:$$\mathrm{E}_{\mathrm{W}/\mathrm{X}} := \Delta \phi_{\mathrm{W} / \mathrm{X}} - \Delta \phi_{\mathrm{H}^+/\mathrm{H}_2}$$If two half-cells are combined to produce a cell $W | X \, || \, Z | Y$, the cell potential difference is nothing but\begin{align*}
\mathcal{E} = E_{\mathrm{Y/Z}} - E_{\mathrm{\mathrm{W/X}}}
\end{align*}As for how this relates to thermodynamics, notice that under reversible conditions we have$$dU = TdS + (-pdV + dw^*)$$where $dw^*$ is the work done to transfer a charge $-F dn$ across a potential difference of $\mathcal{E}$, i.e. $dw^* = -F\mathcal{E} dn$. Therefore, at constant temperature and pressure,\begin{align*}
G &= U + pV - TS \\ \\
dG &= dU + pdV + Vdp - TdS - SdT \\
(dG)_{T,p} &= dU + pdV - TdS \\ \\
(dG)_{T,p} &=dw^* = -F \mathcal{E} dn\end{align*}which means that the free energy changes by $\Delta G = -F\mathcal{E}$ for each mole of charge moved across the cell. If $\Delta G <0$ (i.e. $\mathcal{E} > 0$) then the reaction as written is spontaneous in the forward direction, and the configuration acts as a galvanic cell.

 

[hutchphd]

If I may, the most important point in that very complete, correct and concise description is the concept of the electrochemical potential. This is what distinguishes batteries from other sources electrical energy. They can be successfully modeled (for electronic circuits) as a fixed emf in series (or occasionally in parallel) with a small (large) resistor. They are "like" no other source...that is why they are treated as a fundamental circuit element. Avoid all attempts think of them as "sort of like" anything else

 

[vanhees71]

See, e.g.,

https://doi.org/10.1119/1.2826658

 

[willem2]

Namely if we assume that in our circuit there is a surplus of electrons at negative terminal and lack of electrons
at positive terminal, then at the moment when we connect the Earth to the circuit theoretically we should
expect electrostatic discharge: the electrons from negative terminal going be absorbed by the Earth and
simultaneously the Earth provides enough electrons to fill the electron deficiency in positive terminal.
Then the the battery current should collapse since there is no more potential difference between two poles.

If you connect only one side of the battery to the earth, no current can flow. The battery has to stay neutral as a whole, Removing/adding electrons to one pole will produce a local charge at this pole that will have to be removed by a current of ions towards the other pole of the battery, and then a current from the other pole to the earth.

If you connect both sides of the battery to the earth, that will shorten the battery. There will be no potential across the battery, and no current through the resistor, but there will be current at both poles to the earth, until the battery is empty.
(this supposes connection to the Earth with no resistance)

 

[The Tortoise-Man]

If you connect one end of a battery to an electrostatic source, leaving the other end open (unconnected), it will still maintain this voltage difference - even if the result is a surplus or deficiency at both terminals.

Why this is true? As you said if we not connect the battery to circuit, at the nagative terminal
there is a bunch of free electrons and at the other positive terminal the kations are "waiting"
to absorb electrons.

Assume we connect to say negative end of the battery a electrostatic source. For example a charged one
or uncharged like Earth from my example. And we leave the other terminal (the positive in this example)
unconnected.

Why there happen nothing? For example say our electrostatic source is positively charged.
Shouldn't it attract electrostatically the electrons from the negative end?

Or say the electrostatic source is neutrally charged like Earth. The electrons are negatively charged, and therefore
repulse themselves staying in negative terminal.

Why they not going absorbed by Earth to avoid this self repulsion?Also what is the precise meaning of this "steady push" in electronics? I can imagine
this analogy to water behind a faucet, but I not understand how this potential is
concretely formed by the batteries. And is this a electrostatic effect or
should one think about it as as kind of black box effect having no analogies in
electrostatic world?

 

[The Tortoise-Man]

If you connect only one side of the battery to the earth, no current can flow. The battery has to stay neutral as a whole, Removing/adding electrons to one pole will produce a local charge at this pole that will have to be removed by a current of ions towards the other pole of the battery, and then a current from the other pole to the earth.

If you connect both sides of the battery to the earth, that will shorten the battery. There will be no potential across the battery, and no current through the resistor, but there will be current at both poles to the earth, until the battery is empty.
(this supposes connection to the Earth with no resistance)

On the "If you connect only one side of the battery to the earth" part:
So do I understand it correctly, that altough there are indeed there free electrons at the negative end present, and if we connect an electrostratical source only to the negative end, which is able to steal electrons electrostatically (for example if it is positively charged), then the chemical equilibrium condition induces for every electron passing away a ion movement such that the constant potential in the battery is still preserved? Or did I misunderstood you point?

 

[hutchphd]

If one attaches any chunk of conductor to the "electrostatic source" at high potential then electrons will quickly flow to that source until the chunk of conductor is at high potential. The same will be true for a battery but this has nothing to do with chemical equilibrium but rather with its conductivity. The battery will continue to be a battery...you may not want to touch it though!
Is there a particular system you are trying to understand?

 

[.Scott]

Why this is true? As you said if we not connect the battery to circuit, at the negative terminal
there is a bunch of free electrons and at the other positive terminal the kations are "waiting"
to absorb electrons.

Yes - and the higher the voltage the less patiently they wait.

Assume we connect to say negative end of the battery a electrostatic source. For example a charged one
or uncharged like Earth from my example. And we leave the other terminal (the positive in this example)
unconnected.

Why there happen nothing? For example say our electrostatic source is positively charged.
Shouldn't it attract electrostatically the electrons from the negative end?

When you say "Earth", what you mean is an enormous electrical conductor - so large that any charge from our battery (either negative or positive) will have a negligible effect on its potential. So here's what happens when you connect such an "Earth" to a battery terminal.

1) We set up the experiment with both battery terminals open. There is no flow of electricity - but there is a charge on each terminal. Since the terminals are relatively small, it doesn't take much of a charge to create the voltage difference between those terminals. Let's say that the charge (measured in electrons) is +1 billion at the negative terminal and -1 billion at the positive terminal. The battery will remain in a steady state - with no further flow of charge - until we do something else with it.

2) We now connect our negative terminal to our neutral (uncharged) "Earth". This Earth is so large, that it will immediately absorb those billion electrons with no measurable change in potential. So, unless the battery does something, the voltage difference across its terminals is only half the rated battery voltage. So what happens next is that the battery pumps another 1 billion electron from the positive terminal to the negative terminal and into the "Earth". This will still have negligible effect on the "Earth", but it will leave the positive terminal with a charge of -2 billion electrons - just enough to restore the voltage difference between the terminals to the rated battery voltage.

So when you connect one battery terminal to "Earth", you will get a momentary flow. But it will be nanoseconds in duration - limited by battery resistance, circuit impedance, and the speed of light.

 

[rcgldr]

So what happens in a sequence where alternating poles of a battery are connected one at a time to one plate of a capacitor, with the other plate grounded, after thousands of cycles?

 

[.Scott]

So what happens in a sequence where alternating poles of a battery are connected one at a time to one plate of a capacitor, with the other plate grounded, after thousands of cycles?

Normally "ground" is part of a circuit. But since we are only connecting one battery terminal at a time, we will be treating "ground" as a planetary-sized charge carrier.
Since the amount of current will be limited by the battery voltage and the charge-carrying capacity of the unconnected battery terminal, the total current will be minuscule and any practical capacitor will act as a conductor.
So you will get some tiny amount of momentary current each time you connect, but it won't be much.

 

[The Tortoise-Man]

When you say "Earth", what you mean is an enormous electrical conductor - so large that any charge from our battery (either negative or positive) will have a negligible effect on its potential. So here's what happens when you connect such an "Earth" to a battery terminal.

1) We set up the experiment with both battery terminals open. There is no flow of electricity - but there is a charge on each terminal. Since the terminals are relatively small, it doesn't take much of a charge to create the voltage difference between those terminals. Let's say that the charge (measured in electrons) is +1 billion at the negative terminal and -1 billion at the positive terminal. The battery will remain in a steady state - with no further flow of charge - until we do something else with it.

2) We now connect our negative terminal to our neutral (uncharged) "Earth". This Earth is so large, that it will immediately absorb those billion electrons with no measurable change in potential. So, unless the battery does something, the voltage difference across its terminals is only half the rated battery voltage. So what happens next is that the battery pumps another 1 billion electron from the positive terminal to the negative terminal and into the "Earth". This will still have negligible effect on the "Earth", but it will leave the positive terminal with a charge of -2 billion electrons - just enough to restore the voltage difference between the terminals to the rated battery voltage.

So when you connect one battery terminal to "Earth", you will get a momentary flow. But it will be nanoseconds in duration - limited by battery resistance, circuit impedance, and the speed of light.

So in summary if we connect the negative terminal with surplus of electrons to "Earth" there will be indeed a ultrashort discharge current flow of these electrons into the Earth immediately after the connection take place, but the chemical equilibrium compensates it by subtracting the same amount of these lost electrons from positive terminal.
So the final steady result is then - which becomes stationary if we not do something else - that the potential
at "negative" terminal and "Earth" is the same now (all surplus electrons there are absorbed now by the Earth), but the positivity at other terminal also raised, therefore the compensation mechanism you have explaned ensures that the potential difference between the two terminals/poles stays still constant, that's the punchline, right? Did I understood you correctly?And generally in cases where we connect any electrostatic source to only one terminal und leave the other terminal unconnected in behaves simmilary:

There is an ultrashort discharge current phase between the electrostatic source and the terminal connected to it, ensued by the potential shift at the unconnected terminal due to mechanisms preserving the
electrochemical equilibrium making the potential difference between the two terminals again the same as before.

The second interesting case is what happens if connect the two terminals to each other via the most simple circuit consisting of a conductive wire and a resistor. Now there should be a flow of electrons from negative to positive terminal just due to potential difference at the terminals as one might naively expect.

This gives rise to following two interesting questions:

1) We know that the there is a potential difference between the terminals intrinsically given by the electrode potentials. And during a treveling electron along the wire comes closer to the positive terminal the potential difference descreases and is zero at the positive terminal.

Question: Can we completely reconstruct the progression of the potential function knowing it's values only at negative terminal (maximal value) und positive terminal (minimal value)?

Clearly, the potential is determined up to a constant, that's fine, so we can assume that it is zero at positive terminal. Futhermore it's continuous and descreases monotonous from negative to positive terminal.

Is that sufficient the reconstruct the complete course of the potential function of the battery circuit?
Motivation: The potential function determines completely the electric field E inside the wire pushing the electrons from negative to positive terminal.

2) Can the behavior of curcuit divided also in several characteristic phases if we consider the following experiment:

We start with battery- wire- resistor circuit as in Question 1 and connect to the wire somewhere the "Earth" (considered again as before as idealized enormous electrical conductor). By experience there should nothing happen. Or does it happen like the case you discussed above that at the moment we connect the Earth to the wire, there happen an ultrashort discharge current flow, but then the circuit stabilizes again?

 

[.Scott]

So in summary if we connect the negative terminal with surplus of electrons to "Earth" there will be indeed a ultrashort discharge current flow of these electrons into the Earth immediately after the connection take place, but the chemical equilibrium compensates it by subtracting the same amount of these lost electrons from positive terminal.
So the final steady result is then - which becomes stationary if we not do something else - that the potential
at "negative" terminal and "Earth" is the same now (all surplus electrons there are absorbed now by the Earth), but the positivity at other terminal also raised, therefore the compensation mechanism you have explained ensures that the potential difference between the two terminals/poles stays still constant, that's the punchline, right? Did I understood you correctly?

Yes.

The second interesting case is what happens if we connect the two terminals to each other via the most simple circuit consisting of a conductive wire and a resistor. Now there should be a flow of electrons from negative to positive terminal just due to potential difference at the terminals as one might naively expect.

This gives rise to following two interesting questions:

1) We know that the there is a potential difference between the terminals intrinsically given by the electrode potentials. And during a traveling electron along the wire comes closer to the positive terminal the potential difference decreases and is zero at the positive terminal.

Let's be careful about how "easy" we make this easy case.
The simplest model of a charged battery is simply a voltage source. But since your applying a resistive load, let's bear in mind that the battery itself has an internal resistance.

Let's use a 1-volt battery with an internal resistance of 100 mOhms. With no load, the terminals will show 1 volt and no current. But if you short the terminals together, you will get 0 volts and 10 amps (and a very toasty battery).

Question: Can we completely reconstruct the progression of the potential function knowing it's values only at negative terminal (maximal value) and positive terminal (minimal value)?

If I understand what you are asking:
1) By convention, potentials are measured as increasing towards the positive terminal. It was established long before the electron was identified. Let's keep to that convention.
2) Let's take the negative terminal as a potential of zero - and all other potentials will be relative to that negative terminal.
3) Let's put a 1 Ohm resistor across this battery.
4) We will assume that the battery terminal itself and the leads to the resistor are superconductors - with no resistance. In practice, they would be of negligible resistance.
5) So the potential at the negative terminal and all non-resistive parts attached to that terminal (including the resistor lead) will be at a potential of zero.
6) The current through our circuit will be 1Volt/(1ohm+100mOhm) = 909mAmps.
7) So the potential at the positive end of the resistor will be 909mA x 1 Ohm = 909mV.
8) So the positive battery terminal and the positive resistor lead will be at 0.909 volts.
9) Across the resistor itself, the potential will vary from 0 to 909 mV. Depending on the type of resistor, this is likely to be a roughly linear gradient from one lead, through the resistive media, to the other lead.
10) There is also a progression of potential through the battery itself - dependent entirely on the current, the temperature and the battery structure and chemistry.

2) Can the behavior of curcuit divided also in several characteristic phases if we consider the following experiment:

We start with battery- wire- resistor circuit as in Question 1 and connect to the wire somewhere the "Earth" (considered again as before as idealized enormous electrical conductor). By experience there should nothing happen. Or does it happen like the case you discussed above that at the moment we connect the Earth to the wire, there happen an ultrashort discharge current flow, but then the circuit stabilizes again?

Yes. If the circuit holds a charge relative to Earth, there will be a momentary static discharge.
Notice that I measured all potentials relative to the negative battery terminal - so none of those numbers change.

 

[hutchphd]

10) There is also a progression of potential through the battery itself - dependent entirely on the current, the temperature and the battery structure and chemistry.

I like the rest but this must be very carefully qualified. The internals of a battery are very complicated and often the concept of electric potential therein is not useful. There is some wet (or gooey) and often very reactive chemistry that will conspire to produce, by design, a fixed potential difference across the conductive terminals. Unless you are designing batteries that is all one needs to know. Attempts to define "a progression of potentials" and fields are destined to be incomplete and therefore confusing.
Also just because we model a battery as having a series resistance does not mean that this is really the physics. This is the Thevenin equivalent...in some ways the Norton equivalent is more physical but either will produce the correct answers in a circuit...the Thevenin is usually more useful.