# Standard hydrogen electrode doubts

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#1

Suppose I want the reduction potential of Fe i.e Fe2+lFe so I connect it to the SHE as shown below.
I just don't grasp how this can even work.

Which electrode is getting reduced? Which oxidised?
From where to where do the electrons flow?
The electrons flow through the platinum wire right?
Which ions flow through the salt bridge?

Edit:
I now know that electrons move from the anode to the cathode and normally the left electrode is the anode so electrons would prob move from left to right through the platinum wire. Correct me if I'm wrong please
But I still don't know which ions would move through the salt bridge and in which direction.

Last edited by Presto; 1 year ago
0
1 year ago
#2
(Original post by Presto)

Suppose I want the reduction potential of Fe i.e Fe2+lFe so I connect it to the SHE as shown below.
I just don't grasp how this can even work.

Which electrode is getting reduced? Which oxidised?
From where to where do the electrons flow?
The electrons flow through the platinum wire right?
Which ions flow through the salt bridge?

Edit:
I now know that electrons move from the anode to the cathode and normally the left electrode is the anode so electrons would prob move from left to right through the platinum wire. Correct me if I'm wrong please
But I still don't know which ions would move through the salt bridge and in which direction.

It's no different than any other electrochemical (voltaic) cell. The SHE is assigned a value of zero potential, so that any potential registered on the (high resistance) voltmeter is due to the other half-cell.

In this case the value registered on the voltmeter is -0.41V

This means that the iron half-cell has a negative potential when compared to the SHE. In other words it tends to release electrons, which are trying to go around the external circuit to the SHE, at which reduction would occur, 2H+ + 2e ==> H2

The iron/iron(II) equilibrium tends to the left hand side:

Fe2+ + 2e <==> Fe

In reality, there is no actual flow of electrons as the resistance of the voltmeter is too high, it just measures the "push" of the electrons, which we call the electromotive force, or the voltage. If current were allowed to flow, the characteristics of the half-cells would change as they react and the system would become non-standard.

The half-cells are assigned names, anode and cathode, assuming that the cell is actually working, producing a current. The anode is the half-cell at which oxidation occurs, in this case the iron/iron(II) half cell, which is the negative electrode (electron producer). Electrons always flow around the external circuit from negative to positive (when the cell is allowed to "work").

The ions in the salt bridge flow in the same (circuitous, in this case anticlockwise) direction as the electrons, i.e. negative ions flow towards the iron(II) solution and positive ions towards the SHE solution. This compensates for possible (actually impossible) charge build up, which would prevent the cell from operating.
1
#3
(Original post by charco)
It's no different than any other electrochemical (voltaic) cell. The SHE is assigned a value of zero potential, so that any potential registered on the (high resistance) voltmeter is due to the other half-cell.

In this case the value registered on the voltmeter is -0.41V

This means that the iron half-cell has a negative potential when compared to the SHE. In other words it tends to release electrons, which are trying to go around the external circuit to the SHE, at which reduction would occur, 2H+ + 2e ==> H2

The iron/iron(II) equilibrium tends to the left hand side:

Fe2+ + 2e <==> Fe

In reality, there is no actual flow of electrons as the resistance of the voltmeter is too high, it just measures the "push" of the electrons, which we call the electromotive force, or the voltage. If current were allowed to flow, the characteristics of the half-cells would change as they react and the system would become non-standard.

The half-cells are assigned names, anode and cathode, assuming that the cell is actually working, producing a current. The anode is the half-cell at which oxidation occurs, in this case the iron/iron(II) half cell, which is the negative electrode (electron producer). Electrons always flow around the external circuit from negative to positive (when the cell is allowed to "work").

The ions in the salt bridge flow in the same (circuitous, in this case anticlockwise) direction as the electrons, i.e. negative ions flow towards the iron(II) solution and positive ions towards the SHE solution. This compensates for possible (actually impossible) charge build up, which would prevent the cell from operating.
Thank you soo much for your help 0
1 year ago
#4
(Original post by charco)
It's no different than any other electrochemical (voltaic) cell. The SHE is assigned a value of zero potential, so that any potential registered on the (high resistance) voltmeter is due to the other half-cell.

In this case the value registered on the voltmeter is -0.41V

This means that the iron half-cell has a negative potential when compared to the SHE. In other words it tends to release electrons, which are trying to go around the external circuit to the SHE, at which reduction would occur, 2H+ + 2e ==> H2

The iron/iron(II) equilibrium tends to the left hand side:

Fe2+ + 2e <==> Fe

In reality, there is no actual flow of electrons as the resistance of the voltmeter is too high, it just measures the "push" of the electrons, which we call the electromotive force, or the voltage. If current were allowed to flow, the characteristics of the half-cells would change as they react and the system would become non-standard.

The half-cells are assigned names, anode and cathode, assuming that the cell is actually working, producing a current. The anode is the half-cell at which oxidation occurs, in this case the iron/iron(II) half cell, which is the negative electrode (electron producer). Electrons always flow around the external circuit from negative to positive (when the cell is allowed to "work").

The ions in the salt bridge flow in the same (circuitous, in this case anticlockwise) direction as the electrons, i.e. negative ions flow towards the iron(II) solution and positive ions towards the SHE solution. This compensates for possible (actually impossible) charge build up, which would prevent the cell from operating.
Is there a reason hydrogen was chosen to make the standard reference electrode ?
0
1 year ago
#5
(Original post by Leah.J)
Is there a reason hydrogen was chosen to make the standard reference electrode ?
It made sense considering that most systems are aqueous. In reality, the SHE is not used experimentally for measuring Eº values, rather a calomel electrode, or another easily used stable reference electrode system.
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