Galvanic cell - open circuit voltage and EMF

Summary
understanding the open circuit voltage (emf) reached across galvanic cell electrodes
Hi,

having not a deep knowledge of electrochemistry I've some doubts about processes involved in a galvanic cell. Take for instance a Zn/Cu Daniell cell for which E0cell is 1,10V. That means emf for it is 1,10V.

Starting to read from how battery works I had a first understanding of how controlled Zn/Cu Galvanic/Voltaic process works.
Consider a Daniell cell with no external conductive path across electrodes: under that condition electrochemical processes involved at anode (Zinc) and cathode (Copper) are not going on. Connect then an "ideal" voltmeter across electrodes (a device with an infinite input impedance).

Do you think it would show a non-zero value (basically the cell emf value 1,10V) ? In other words, taking in account the fact chemical processes are not running when external conductive path does not exist, could be there nevertheless a charge accumulation on the electrodes themselves (electrons on anode and positive ions on cathode) ?
 
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BvU

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Yes it would. The emf is actually present between the electrodes.
 
Yes it would. The emf is actually present between the electrodes.
ok I believe too...but let's consider what actually happens when you "setup" a Daniell cell. Start from the 2 isolated half-cells with either no external electric connection or salt bridge: no oxidation or reduction process is going on.

Then add the salt bridge across half-cells: why suddenly redox reactions should start to take place accumulating charges on anode (electrons) and cathode electrodes (copper positive ions) ?
 
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Borek

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It is not like there is no redox process before closing the circuit. The reaction won't go far, just far enough to charge the electrodes a little bit. This charge is what 1. stops the reaction from going further, 2. produces the voltage difference.
 

BvU

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Start with the electrodes above the liquid and lower them in position.

If there is no bridge,
The Cu electrode will accumulate Cu until the potential of the electrode wrt the liquid is + 0.34 V -- then nothing happens anymore.
At the same time, the Zn electrode will loose Zn ions until the electrode is at - 0.718 V wrt the liquid -- then nothing happens any more: there is equilibrium in the chemical reaction.
 

BvU

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I interpreted this as: yes salt bridge, no bulb
correct

Start with the electrodes above the liquid and lower them in position.

If there is no bridge,
The Cu electrode will accumulate Cu until the potential of the electrode wrt the liquid is + 0.34 V -- then nothing happens anymore.
What actually is going on is that cathode metal electrode "donate" electrons to be employed for the reduction of Cu cations that are inside the half-cell solution ?

What happens when salt bridge is inserted between half-cells without any external electric connection as before ?
 
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gleem

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It is not like there is no redox process before closing the circuit. The reaction won't go far, just far enough to charge the electrodes a little bit. This charge is what 1. stops the reaction from going further, 2. produces the voltage difference.
If the charge is the source of the potential difference when you change the physical construct of the electrode (make them longer and wider) to accommodate more charge wouldn't that then change the potential difference between the electrodes?
 

Borek

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If the charge is the source of the potential difference when you change the physical construct of the electrode (make them longer and wider) to accommodate more charge wouldn't that then change the potential difference between the electrodes?
No.

Note: in the static system (static as in "no changing magnetic/electrical fields") you can't have potential difference without charge separation, that's Physics 101. But the potential difference is not just a matter of charge sizes, it also depends on the system geometry and other factors.
 

gleem

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But the potential difference is not just a matter of charge sizes, it also depends on the system geometry and other factors.
Isn't that I was aiming at when I said:

If the charge is the source of the potential difference when you change the physical construct of the electrode (make them longer and wider) to accommodate more charge wouldn't that then change the potential difference between the electrodes?
?
 

Borek

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Yes, but they don't have to all work in the same direction. Higher pointlike charge would make the potential difference larger, but the charge is not pointlike.

And actually you are trying to put the cart before horse, as it is potential difference at which the system is at equilibrium that defines by how much the given electrode will be charged, not the other way around.
 
Yes, but they don't have to all work in the same direction. Higher pointlike charge would make the potential difference larger, but the charge is not pointlike.
Not sure to understand, could you elaborate a bit please ?
 

BvU

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What happens when salt bridge is inserted between half-cells without any external electric connection as before ?
Good question ! @Borek ?
My estimate is that some ions will move but that stops almost immediately once the charge balance is restored
wikipedia said:
nitrate anions in the salt bridge move into the zinc half-cell in order to balance the increase in Zn2+ ions. At the same time, potassium ions from the salt bridge move into the copper half-cell in order to replace the Cu2+ ions being precipitated
 

gleem

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...... as it is potential difference at which the system is at equilibrium that defines by how much the given electrode will be charged, not the other way around.
Yes I fully agree.

To me your previous statement

This charge is what 1. stops the reaction from going further, 2. produces the voltage difference.
suggested that the charges defined the potential and to you my statement

If the charge is the source of the potential difference when you change the physical construct of the electrode (make them longer and wider) to accommodate more charge wouldn't that then change the potential difference between the electrodes?
apparently also suggested the same yet we both understand that the potential difference is intrinsic to the cations and regulates the charges produced.
 

256bits

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Start with the electrodes above the liquid and lower them in position.

If there is no bridge,
The Cu electrode will accumulate Cu until the potential of the electrode wrt the liquid is + 0.34 V -- then nothing happens anymore.
At the same time, the Zn electrode will loose Zn ions until the electrode is at - 0.718 V wrt the liquid -- then nothing happens any more: there is equilibrium in the chemical reaction.
If that happens, then,
when the solid zinc and copper electrodes are inserted, from where do the Cu ions acquire the electrons to deposit upon the Cu(s) electrode, and to where do the electrons go when Zn(s) is deposited into solution as positive zinc ions?
 
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Not quite what was asked, but I remember using a 'standard' voltaic cell at Uni. The 'gotcha' was that it only gave the correct voltage, temperature corrected, when no current was being drawn. Else polarisation set in...
The way around was to repeatedly 'balance' it using a Wheatstone bridge...
 

Borek

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when the solid zinc and copper electrodes are inserted, from where do the Cu ions acquire the electrons to deposit upon the Cu(s) electrode
Directly from the electrode, charging it in the process.
 

Borek

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What happens when salt bridge is inserted between half-cells without any external electric connection as before?
There will be some shift of the charges - positive charges will be attracted to the negative electrode till they neutralize it (screen the electric field between the charged electrodes).

Again, this is hardly chemistry, this is just how charges behave.
 
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256bits

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There will be some shift of the charges - positive charges will be attracted to the negative electrode till they neutralize it (screen the electric field between the charged electrodes).

Again, this is hardly chemistry, this is just how charges behave.
That is where I was eventually going with this.
Sorry for not being more explicit .

A different chain of events thought experiment as that proposed earlier.
If we take the two solutions, same molarity, and first add the salt bridge no movement of ions will result at that point.
Then we add the electrodes. The electrodes will become charged - and +.
There will be a congregation of ZN+2 ions around the Zn(s) electrode, and also a few more in solution making the solution to have a positive charge, relatively speaking.
At the copper electrode. there will be less Cu+2 ions in solution, making it slightly negative.
It seems that there should be a slight movement of excess Zn+2 ions to the right through the salt bridge and a movement of sulfate ions to the left so as to remove this imbalance.

Question is, if we add the electrodes to the solution firstly, before adding the salt bridge, does the Zn(s) stay neutral or does a Zn ion enter into solution.
 

Borek

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Question is, if we add the electrodes to the solution firstly, before adding the salt bridge, does the Zn(s) stay neutral or does a Zn ion enter into solution.
Note that just the presence of water means a redox system is present, so some of the metal must always react.
 

256bits

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Thanks for the answer.
 
There will be some shift of the charges - positive charges will be attracted to the negative electrode till they neutralize it (screen the electric field between the charged electrodes).
Just to be sure I got the point: positive charges you're referring to are basically the cations (positive ions) belonging to the salt bridge that will shift towards the negative electrode (Zn(s)) neutralizing some electrons on it (analogus process about anions that will shift from salt bridge towards anode) ? In that case the electric field between the charged electrodes will suffer a small decrease as the net emf of the cell itself ?
 
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Borek

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You have ions (cations, anions) everywhere, not only in the salt bridge. Ions neutralizing the electric field will be just on the electrode surface (google double layer) and there will be no net electric field in the solution.

That will change when you close the circuit, but that's a bit another story.
 
You have ions (cations, anions) everywhere, not only in the salt bridge. Ions neutralizing the electric field will be just on the electrode surface (google double layer) and there will be no net electric field in the solution.

That will change when you close the circuit, but that's a bit another story.
Thus, from my understanding assuming the open circuit condition, the cell electrodes just behave as, let me say, the plates of a small capacitor charged up until the generated electric field between them is able to stop the ongoing redox process (oxidation on anode and reduction on cathode): the electric potential difference between them equals (let me say by definition) the emf of the cell itself.

Make sense ? thanks.
 

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