# Help to understand reversibility in galvanic cell

## Homework Statement

:-I am confused about some lines in my textbook about galvanic cell[/B]

## Homework Equations

:-The lines are as follows............
The cell is connected to an external source of potential that opposes and exactly balance the cell potential . If the external potential is reduced infinitesimally ,the cell reaction occurs in it's spontaneous direction.An opposite infinitesimal change in external potential will cause the reaction to proceed in the reverse direction.The cell reaction is then said to occur under reversible conditions.Thus,the cell can be made to behave in a reversible manner by balancing their potentials by an external potential so that no current flows.
The reversible electrical work done in a galvanic cell by cell reaction is equal to decrease in gibb's energy.Hence,
Electrical work= - deltaG
[/B]

## The Attempt at a Solution

:-Let us take if I have a single displacement, reversible reaction, AB + C <--> AC + B. Then I increase the concentration of B on the right, what happens? The AC molecule splits, and forms more AB to re-balance the equation. This is the same thing happening in the galvanic cell. Lets assume the cell has a potential of 1.1V and I apply a current of exactly 1.1V to the circuit, the electrodes will no longer undergo a redox reaction, as there is no potential gradient that would cause the electrons to move from one electrode to the other. But if I drop the external voltage to 1V, then the cell will begin undergoing a redox reaction, albeit at a slower rate to bring the voltage back to 1.1. If I apply a greater voltage of say 1.2V, then electrons are going to go the opposite direction, and oxidize the cathode and reduce the anode.
Actually someone has explained me in this manner,i want to verify this.Is this correct?
Regards.[/B]

mfb
Mentor
Sounds fine.
Note that you don't have to care about the internal chemical details if you want to analze circuits.

I apply a current of exactly 1.1V to the circuit,
How one can apply current of 1.1 v as volt is unit of potential.

Lets assume the cell has a potential of 1.1V and I apply a current of exactly 1.1V to the circuit, the electrodes will no longer undergo a redox reaction, as there is no potential gradient that would cause the electrons to move from one electrode to the other.
Are we charging galvanic cell here,so voltage of 1.1 is applied from external source.To charge it, a power/voltage source is connected positive terminal to positive terminal of the battery, and negative terminal to negative terminal of the battery (parallel connection),right?

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But if I drop the external voltage to 1V, then the cell will begin undergoing a redox reaction, albeit at a slower rate to bring the voltage back to 1.1.
What does this mean?

Bystander
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How one can apply current of 1.1 v as volt is unit of potential.
It's a typo.
positive terminal to positive ---- negative terminal to negative
Correct.

• gracy
Bystander
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What does this mean?
Higher voltage goes one way, lower, the other.

Why when there is parallel connection we subtract the two potentials?

Higher voltage goes one way, lower, the other.
Sorry,still didn't understand.

Bystander
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This gives you the magnitude of the emf driving the reaction. Zero, it goes nowhere. Positive it goes one way, negative the other. Whichever you choose to define as forward or reverse as far as calling positive or negative forward or reverse (European or American conventions).

• gracy
I understood the whole thing which i had written in my attempt to a solution ,but how this explains reversibility in galvanic cell?

Bystander
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When the textbook is discussing "reversibility," it's attempting to state that whatever electrical work is done on (or by) the cell in either direction can be completely recovered only when the cell is run reversibly, and that the cell can be run reversibly only by making infinitesimal changes in the potential it is working against, or that is working on it. What it means is that any real use of the cell, work done charging it, or work done by it during discharge is NOT reversible.

• gracy
making infinitesimal changes in the potential
Increasing or decreasing the potentials by 1 volt,will it be considered as infinitesimal change?

One more thing about reversibility is
In my textbook it is written that for the reaction to be reversible the system should be in mechanical equilibrium with it's surroundings.(the external pressure should be equal to pressure of the gas ,this balance of pressure is called as mechanical equilibrium)
I don't understand if there is no pressure difference between system and surrounding,how work will be done by/on the system?

Bystander
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1 volt,
Not infinitesimal. This is the same "infinitesimal" as is used in your math courses.
how work will be done by/on the system?
There is mechanical work, and there is also electrical work, "Eit."

. But if I drop the external voltage to 1V, then the cell will begin undergoing a redox reaction, albeit at a slower rate to bring the voltage back to 1.1. If I apply a greater voltage of say 1.2V, then electrons are going to go the opposite direction, and oxidize the cathode and reduce the anode.
Can I draw a conclusion from this that emf resist changes as we have learnt in physics class.

There is mechanical work, and there is also electrical work, "Eit."
How this answers my post 14.

Bystander
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conclusion from this that emf resist changes
Yes.
Electrical work is voltage times current times time.

Electrical work is voltage times current times time.
But in my textbook this mechanical equilibrium is given in context of piston cylinder .How electricity can come in this type of problems.

Bystander
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How electricity can come in this type of problems.
Textbook writers are lazy --- they like to stick to the historical development of thermodynamics, pV work in steam engines, and very often (almost always) fail to generalize the "work term" in the first law to include ALL work, chemical, mechanical as pV, mechanical as surface tension times area, electrical, magnetic, .....

• gracy
Electrical work is voltage times current times time.
How this explains mechanical equilibrium?
I mean in piston cylinder type ,when the external pressure is equal to pressure of the gas ,this balance of pressure is called as mechanical equilibrium.
How can we define mechanical equilibrium when we take electrical system?

Bystander
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define mechanical equilibrium when we take electrical system?
"Equilibrium" in its most general sense means that nothing is changing over time; no mechanical work is being done, no electrical, no chemical, or any other. This means that whatever the work term is for the system, the pressure, or electrical potential, or chemical potential is constant AND equal to that of the exterior or surroundings.

Equilibrium" in its most general sense means that nothing is changing over time; no mechanical work is being done, no electrical, no chemical, or any other.
I am quite surprised.
Textbook writers are lazy
And I think they are wrong at times.Because according to my textbook this mechanical equilibrium is one of the conditions for maximum work.

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Bystander
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equilibrium is one of the condition for maximum work.
The balance of forces is necessary, and this is where the "infinitesimal" difference comes into play; at balance, nothing changes (equilibrium), while "infinitesimal" imbalance allows change, it's just infinitesimally slow. Maximum work takes literally forever, but can be approached to within any specified limit. Back to the bicycle pump: you're late for class and have a flat tire, and you pump as fast as you can; if you count the pump strokes to inflate the tire, you get some number; if you later decide to see how efficiently you can inflate the tire, you pump as slowly as you can stand to waste time, getting maximum efficiency from the pump, and you'll reach the same inflation pressure in fewer pump strokes --- that's if there isn't a slow leak in the tire, the pump, or the valve stem connecting the pump to the tire --- so don't waste your time with this "thought experiment."

• gracy
Borek
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And I think they are wrong at times.Because according to my textbook this mechanical equilibrium is one of the conditions for maximum work.

Nope, that's quite correct (even if sounds nonsensical). Trick is, you can get more work from the system if you do the work more slowly. In the limit maximum work is done when it takes infinitely long time to do it, which technically means at any point in time system is in equilibrium.

Problem is our intuition stops to work correctly in such cases. Don't worry, you are not the first person having problems.

• gracy
epenguin
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It is. If it wasn't charged there would be no potential difference between cathode and anode.

Charge is typically pretty small, but it definitely is there.

I apologise for coming back late on this thread, it is because I have myself some comprehension problems with the processes, but I think it is safe to say what follows:

If we are talking about the terminals I don't think we should encourage the student to think there is a charge there or that it matters. If they ask what is the charge there shouldn't the answer be it depends on the capacitance between them? Which I think is pretty small. More important it is insignificant. If it were a hundred times more than what it is, and more importantly if it were zero the potential would be just the same. For actual thermodynamic measurement, I.e. when nothing is happening, the battery is brought to balance against a standard battery with its output potential adjusted with a potential divider e.g. a sliding contact, and there is no current and no charge at the junctions between batteries.

It is surely not charge that causes current or potential of a battery, it is a chemical disequilibrium.
The reactants are so disposed that the electron for reduction can only arrive through the metal conductor and for oxidation be yielded through one.

Some ballpark figures seem of interest. (Checks and comments welcome).
1 Coulomb is about 6.2 X 1018electrons.

The capacitance between battery terminals is surely orders of magnitude below that of a designed capacitor. Usual capacitors e.g. in radio circuits used to be of the order of a few microfarads (though we can do much better now). A 1μF capacitor across a 1 V battery would hold only 1 μCoulomb. Which ≈ 6 X 1012 /6 X 1023 ≈10-11 moles of electrons. The electric charge of the battery I just said is orders of magnitude less even than this.
By contrast i read that a typical battery delivers in a lifetime of the order of 5000 Coulomb.
Which I calculate to be about 0.05 moles of electrons. Equivalent to about 1.5 g of Zn. (Seems about right ballpark).
A 1 A incandescent torch bulb (confusingly called flashlight by certain people ) would use up any charge on the battery in orders of magnitude less than a μs.

Inside the battery there is some charge separation with a charge on the metal and a cloud of opposite charge in the medium near it, but I thought this charge separation was only over at most a few molecular diameters. It does not seem to me relevant to the elementary fundamentals. The Nernst equation is independent of such details.

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Borek
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It is surely not charge that causes current or potential of a battery, it is a chemical disequilibrium.

By the very definition of the potential - if the potential difference exists, it means there is an electric field between the electrodes. No field, no potential difference. And the only way the field can exist is when the charges are there. As I said they are pretty small, but they just have to be there (unless I was taught wrong physics).

The difference between a capacitor and battery is that the charge in the battery is recreated by the chemical reaction, while in the capacitor once it is gone - it is gone.

epenguin
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I think I was taught not so well - the important thing was to be able to do the standard calculations. Recently looked at a textbook and online edu sites and am not much wiser - they still are about the same things. I think it may be generally badly taught, surely many people find difficulties. Whatever the charges actually are, if they were exactly zero you would still get the flows you get due to the chemical disequilibrium it seems to me. Also if charges caused the potential if you joined two negatives, say, they would repel each other slightly but you force them together, having twice the charge at the junction wouldn't you have twice the potential there? - which at least I'm sure I was not taught at school or anywhere else.

I know the difference between a battery and a capacitor, I mentioned them not so much as part of the argument but as useful illustrations and because these typical numbers and orders of magnitude are useful to have an appreciation of and actually they do not get mentioned much in texts, only the formulae.

We need a physicist, is this the right place? Borek
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Whatever the charges actually are, if they were exactly zero you would still get the flows you get due to the chemical disequilibrium it seems to me.

Looks to me like that's the part you get wrong - no separation of charges would mean no current flowing.

To reiterate - lack of chemical equilibrium expresses itself by a minute charges on the electrodes, these charges are responsible for the observed potential difference.

Also if charges caused the potential if you joined two negatives, say, they would repel each other slightly but you force them together, having twice the charge at the junction wouldn't you have twice the potential there?

If you were capable of keeping everything else identical - yes. Actually that's what happens in the case of capacitor - charge and potential are directly proportional to each other(note that in the case of capacitor "everything else" is kept unchanged and combined into one parameter, capacity).

Imagine attaching a capacitor to the battery. It will get charged. Charge can be easily calculated knowing battery voltage and capacitor capacity. Now, let's separate capacitor plates and reshape them so that they become just battery contacts. Charge doesn't disappear - it is still there. Sure, it has changed as the capacity changed, but technically it is still exactly the same circuit, and there are still charges on both ends.