lhh2003
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When we measure electrode potentials, we do so in comparison to a reference electrode because you can't measure the potential difference across an ion solution and metal/non-metal electrode directly.

Say we want to do this for Mg2+/Mg , how would this work ? Are we quantifying the influx of electrons received from each electrode and saying how much more positively/negatively charged the Mg cathode is in comparison to the platinum cathode involved in the H2/H+ equilibrium?

If this is the case, then how can you set up cells without a H2/H+ equilibrium occurring , for example when we set up 2 half cells for Cu and Mg? How would the voltmeter be able to compare the charge received from each in terms of H2/H+ when there is no Hydrogen molecules/ions present ?
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lhh2003
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(Original post by lhh2003)
When we measure electrode potentials, we do so in comparison to a reference electrode because you can't measure the potential difference across an ion solution and metal/non-metal electrode directly.

Say we want to do this for Mg2+/Mg , how would this work ? Are we quantifying the influx of electrons received from each electrode and saying how much more positively/negatively charged the Mg cathode is in comparison to the platinum cathode involved in the H2/H+ equilibrium?

If this is the case, then how can you set up cells without a H2/H+ equilibrium occurring , for example when we set up 2 half cells for Cu and Mg? How would the voltmeter be able to compare the charge received from each in terms of H2/H+ when there is no Hydrogen molecules/ions present ?
The Standard Hydrogen Electrode is the standard used.

So, all half-cells are measured with respect to this SHE, this gives the value of the electrode potential.

When two half-cells are compared directly there is no hydrogen present to set up an equilibrium with the hydrogen ions in thew solution.
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(Original post by charco)
The Standard Hydrogen Electrode is the standard used.

So, all half-cells are measured with respect to this SHE, this gives the value of the electrode potential.

When two half-cells are compared directly there is no hydrogen present to set up an equilibrium with the hydrogen ions in thew solution.
So you are comparing the amount of electrons on the SHE to the amount on the other electrode which is a quantitative indicator of the equilibrium position ?
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(Original post by lhh2003)
So you are comparing the amount of electrons on the SHE to the amount on the other electrode which is a quantitative indicator of the equilibrium position ?
No, it is not the amount of electrons, it is the "push" or "pull" that the electrons are given. This is also called the electromotive force, EMF, or potential difference, PD. It is measured in volts.

In one half-cell the electrons are pulled more than in the other. To measure the potential difference a high resistance voltmeter is used that prevents any movement of charge (electrons) and measures the relative difference in pull (potential) of each half-cell.

If the cell is allowed to pass current, it flows from the half-cell with the most negative potential to the half-cell with the more positive potential

Name:  ZnCu_cell.png
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In this example the zinc half-cell has a reduction potential of -0.76V and the copper half-cell has a potential of +0.34V.

When they are connected together the potential difference (literally the difference in potential between the two cells) is +1.10V.
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(Original post by charco)
No, it is not the amount of electrons, it is the "push" or "pull" that the electrons are given. This is also called the electromotive force, EMF, or potential difference, PD. It is measured in volts.

In one half-cell the electrons are pulled more than in the other. To measure the potential difference a high resistance voltmeter is used that prevents any movement of charge (electrons) and measures the relative difference in pull (potential) of each half-cell.

If the cell is allowed to pass current, it flows from the half-cell with the most negative potential to the half-cell with the more positive potential

Name:  ZnCu_cell.png
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In this example the zinc half-cell has a reduction potential of -0.76V and the copper half-cell has a potential of +0.34V.

When they are connected together the potential difference (literally the difference in potential between the two cells) is +1.10V.
Hmm. If I have a negative standard electrode potential, this would mean that my electrode would reduce the SHE. In other words, the more negative the Electrode potential, the greater the EMF in that half cell pushing the electrons through the wire.

So, what does a positive overall electrode potential of a cell represent qualitatively ? I know it always ends up being positive when working out the cell's overall Electrode potential, but don't understand the significance of this.

Thank you.
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(Original post by lhh2003)
Hmm. If I have a negative standard electrode potential, this would mean that my electrode would reduce the SHE. In other words, the more negative the Electrode potential, the greater the EMF in that half cell pushing the electrons through the wire.

So, what does a positive overall electrode potential of a cell represent qualitatively ? I know it always ends up being positive when working out the cell's overall Electrode potential, but don't understand the significance of this.

Thank you.
A positive electrode potential means that it gets reduced by the SHE. In other words the copper(II) ions are a better oxidising agent than the hydrogen ions.

If you connect a Cu||Cu2+(aq) half-cell to a SHE, then the value of the EMF is +0.34V

The SHE provides electrons that could reduce the copper ions to copper if the current were allowed to flow. There is a "pull" from the copper half-cell relatively stronger than the pull from the SHE

The overall cell reaction (if the current is allowed to flow) would be:

H2(g) + Cu2+(aq) --> 2H+(aq) + Cu(s)
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(Original post by charco)
A positive electrode potential means that it gets reduced by the SHE. In other words the copper(II) ions are a better oxidising agent than the hydrogen ions.

If you connect a Cu||Cu2+(aq) half-cell to a SHE, then the value of the EMF is +0.34V

The SHE provides electrons that could reduce the copper ions to copper if the current were allowed to flow. There is a "pull" from the copper half-cell relatively stronger than the pull from the SHE

The overall cell reaction (if the current is allowed to flow) would be:

H2(g) + Cu2+(aq) --> 2H+(aq) + Cu(s)
I understand that a positive standard electrode potential means the electrode will be reduced by an SHE, but if I use the formula
E cell = E right hand side - E left hand side, what is the significance of E cell always being positive ?
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(Original post by lhh2003)
I understand that a positive standard electrode potential means the electrode will be reduced by an SHE, but if I use the formula
E cell = E right hand side - E left hand side, what is the significance of E cell always being positive ?
What you are really doing is proposing a potential reaction.

If the result is negative it means that your proposed reaction is non-spontaneous and therefore cannot happen (under standard conditions)

If you take the Cu(s)||Cu2+(aq)||Zn2+(aq)||Zn(s) cell setup (no convention adopted).

You can determine the cell reaction as:

Cu2+(aq) + Zn(s) --> Cu(s) + Zn2+(aq)

using E(cell) = E(red) - E(ox) = +0.34 --0.76 = +1.10V

The result is positive, therefore the cell reaction is thermodynamically spontaneous (under standard conditions)

But if you determine the cell potential for:

Cu(s) + Zn2+(aq) --> Cu2+(aq) + Zn(s)

using E(cell) = E(red) - E(ox) = -0.76 - 0.34 = -1.10V

The result is negative, which tells you that the cell reaction is thermodynamically non-spontaneous, i.e. it cannot happen (under standard conditions)
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(Original post by charco)
What you are really doing is proposing a potential reaction.

If the result is negative it means that your proposed reaction is non-spontaneous and therefore cannot happen (under standard conditions)

If you take the Cu(s)||Cu2+(aq)||Zn2+(aq)||Zn(s) cell setup (no convention adopted).

You can determine the cell reaction as:

Cu2+(aq) + Zn(s) --> Cu(s) + Zn2+(aq)

using E(cell) = E(red) - E(ox) = +0.34 --0.76 = +1.10V

The result is positive, therefore the cell reaction is thermodynamically spontaneous (under standard conditions)

But if you determine the cell potential for:

Cu(s) + Zn2+(aq) --> Cu2+(aq) + Zn(s)

using E(cell) = E(red) - E(ox) = -0.76 - 0.34 = -1.10V

The result is negative, which tells you that the cell reaction is thermodynamically non-spontaneous, i.e. it cannot happen (under standard conditions)
I see. Thanks !
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lhh2003
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(Original post by charco)
What you are really doing is proposing a potential reaction.

If the result is negative it means that your proposed reaction is non-spontaneous and therefore cannot happen (under standard conditions)

If you take the Cu(s)||Cu2+(aq)||Zn2+(aq)||Zn(s) cell setup (no convention adopted).

You can determine the cell reaction as:

Cu2+(aq) + Zn(s) --> Cu(s) + Zn2+(aq)

using E(cell) = E(red) - E(ox) = +0.34 --0.76 = +1.10V

The result is positive, therefore the cell reaction is thermodynamically spontaneous (under standard conditions)

But if you determine the cell potential for:

Cu(s) + Zn2+(aq) --> Cu2+(aq) + Zn(s)

using E(cell) = E(red) - E(ox) = -0.76 - 0.34 = -1.10V

The result is negative, which tells you that the cell reaction is thermodynamically non-spontaneous, i.e. it cannot happen (under standard conditions)
Obviously an equilibrium will be met in a half cell before I connect the half cell with another half cell. One of the standard conditions is that al solutions have a 1M concentration, but as the concentrations change and I want to measure the concentration as soon as I connect the 2 half cells, does this mean the equilibrium mixtures formed in each half cell before I connect them must have ion solutions of 1 M ?
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(Original post by lhh2003)
Obviously an equilibrium will be met in a half cell before I connect the half cell with another half cell. One of the standard conditions is that al solutions have a 1M concentration, but as the concentrations change and I want to measure the concentration as soon as I connect the 2 half cells, does this mean the equilibrium mixtures formed in each half cell before I connect them must have ion solutions of 1 M ?
A high-resistance voltmeter is used in the circuit to prevent passage of charge between the two half-cells that would make the conditions non-standard.
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(Original post by charco)
A high-resistance voltmeter is used in the circuit to prevent passage of charge between the two half-cells that would make the conditions non-standard.
But the conditions will still become non-standard anyway just by the concentrations changing as the reaction progresses when the equilibrium is disrupted by connecting the 2 half cells.
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(Original post by lhh2003)
But the conditions will still become non-standard anyway just by the concentrations changing as the reaction progresses when the equilibrium is disrupted by connecting the 2 half cells.
You did not read/understand my previous post:

If no charge is passed around the external circuit there can be no reactions taking place in the half-cells. This is why the voltage is called the "potential" difference. It is a form of potential energy that is only released by allowing the external circuit to pass charge.
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(Original post by charco)
You did not read/understand my previous post:

If no charge is passed around the external circuit there can be no reactions taking place in the half-cells. This is why the voltage is called the "potential" difference. It is a form of potential energy that is only released by allowing the external circuit to pass charge.
So, I have a half cell. I let the half cell reach an equilibrium. When I connect the 2 half cells to measure the standard electrode potential, no electrons are actually transferred. In other words, the equilibrium positions do not shift because we are connecting the half cells. This contradicts the mark schemes that I have read, fore they say that connecting the half cells alters the [ions] which I infer can only be caused by changing [electrons].
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(Original post by charco)
You did not read/understand my previous post:

If no charge is passed around the external circuit there can be no reactions taking place in the half-cells. This is why the voltage is called the "potential" difference. It is a form of potential energy that is only released by allowing the external circuit to pass charge.
I think he's saying that if you dip a piece of e.g. Cu into a 1M solution of Cu2+, then some of the Cu will come out of solution, which will change [Cu2+] and this will happen even without an external circuit.

The chance in [Cu2+] would be probably so small that it could be ignored, but not it the volume of solution were also small, so I guess we just use a large enough volume (whatever that means).
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(Original post by Pigster)
I think he's saying that if you dip a piece of e.g. Cu into a 1M solution of Cu2+, then some of the Cu will come out of solution, which will change [Cu2+] and this will happen even without an external circuit.

The chance in [Cu2+] would be probably so small that it could be ignored, but not it the volume of solution were also small, so I guess we just use a large enough volume (whatever that means).
Please tell me this is the hardest part of the course conceptually...
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(Original post by lhh2003)
So, I have a half cell. I let the half cell reach an equilibrium. When I connect the 2 half cells to measure the standard electrode potential, no electrons are actually transferred. In other words, the equilibrium positions do not shift because we are connecting the half cells. This contradicts the mark schemes that I have read, fore they say that connecting the half cells alters the [ions] which I infer can only be caused by changing [electrons].
It is possible that the mark-scheme is giving a simplistic description of the processes occurring.

If you dip a piece of copper into a copper(II) ion aqueous solution there will be a potential difference between the metal and the ions, but there can be no movement of ions per se.

Were an ion to be deposited on the metal, the piece of metal would then develop a positive charge and the solution a relative negative charge. This simply cannot happen to two phases in contact. Positive charges attract negative charges!

There is however, the potential for this to happen if there is a means for the charge to be dissipated, i.e. flow around an external circuit.

When the external circuit is connected via a high-resistance voltmeter the electrons on the negative electrode will push the voltmeter and the positive half-cell will pull the positive side of the voltmeter (think of it like a spring). Still, no reactions can occur until the two solutions are connected by a salt-bridge.
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(Original post by lhh2003)
Please tell me this is the hardest part of the course conceptually...
Once you get the idea, it's not so bad...

Remember that concepts are often delivered in a way that give simplistic models and tools that allow students to develop their understanding.

It is good that you are trying to see beyond the simple explanations, as it will lead to greater clarity of thought. The fundamental ideas of chemistry are logical and simple, however, they are sometimes obscured by unnecessary "rules of thumb".
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