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A-level chemistry salt bridge questions (how do they work?)

In a Galvanic cell, how exactly does the salt bridge work? I've been told that it completes the circuit by allowing ions to flow between the half cells and balance out the charges as the system has to be electrically neutral for a voltage to exist.

1.) Why does it have to be electrically neutral for a voltage to exist?
2.) From my GCSE physics knowledge (I don't do A-level physics) current is the flow of electrons. If electrons are not flowing between the salt bridge, how is current maintained?
3.) Does the salt bridge not dry out after a while when ions are transferred to each half cell? Is it therefore replaced? Would ions from the solution in one half cell not transfer to the other by diffusion?

I would honestly appreciate any help with any of these questions as I've spent about 2 hours trying to figure it out myself with minimal success
Original post by aque1408
In a Galvanic cell, how exactly does the salt bridge work? I've been told that it completes the circuit by allowing ions to flow between the half cells and balance out the charges as the system has to be electrically neutral for a voltage to exist.

1.) Why does it have to be electrically neutral for a voltage to exist?
2.) From my GCSE physics knowledge (I don't do A-level physics) current is the flow of electrons. If electrons are not flowing between the salt bridge, how is current maintained?
3.) Does the salt bridge not dry out after a while when ions are transferred to each half cell? Is it therefore replaced? Would ions from the solution in one half cell not transfer to the other by diffusion?

I would honestly appreciate any help with any of these questions as I've spent about 2 hours trying to figure it out myself with minimal success

1.) Why does it have to be electrically neutral for a voltage to exist?

If a charge were to develp in the half-cell it would oppose the forces pulling the electrons. For example when an electron moves from the half-cell around the external circuit the half-cell would develop a positive charge. This would prevent the loss of a further electron by electrostatic attraction. (simplistic, but it works)

2.) From my GCSE physics knowledge (I don't do A-level physics) current is the flow of electrons. If electrons are not flowing between the salt bridge, how is current maintained?

Current is actually a flow of charge, not necessarily electrons. The movement of ions constitutes a current.

3.) Does the salt bridge not dry out after a while when ions are transferred to each half cell? Is it therefore replaced? Would ions from the solution in one half cell not transfer to the other by diffusion?


The ions are not "wet", they are solute particles in the solvent. The solvent remains in the salt bridge and is unaffected by the flow of ions. Ions are prevented from moving in the opposite direction by the same forces that cause them to move in the first place, electrostatic attraction.
Reply 2
Original post by charco
1.) Why does it have to be electrically neutral for a voltage to exist?

If a charge were to develp in the half-cell it would oppose the forces pulling the electrons. For example when an electron moves from the half-cell around the external circuit the half-cell would develop a positive charge. This would prevent the loss of a further electron by electrostatic attraction. (simplistic, but it works)

2.) From my GCSE physics knowledge (I don't do A-level physics) current is the flow of electrons. If electrons are not flowing between the salt bridge, how is current maintained?

Current is actually a flow of charge, not necessarily electrons. The movement of ions constitutes a current.

3.) Does the salt bridge not dry out after a while when ions are transferred to each half cell? Is it therefore replaced? Would ions from the solution in one half cell not transfer to the other by diffusion?


The ions are not "wet", they are solute particles in the solvent. The solvent remains in the salt bridge and is unaffected by the flow of ions. Ions are prevented from moving in the opposite direction by the same forces that cause them to move in the first place, electrostatic attraction.

Thank you, this is extremely helpful!

Just one final question with number 3.), is it the ions in the salt bridge that move into the two solvents in the half cells (e.g. if the salt bridge was NaCl, are Cl- ions moving into the half cell that is being oxidised?). If so, will the solvent in the salt bridge not eventually 'run out' (possibly better phrase than 'dry out') as the Na+ and Cl- ions are being used up to balance the charge in each half cell? Or to be more specific, does it typically 'run out' before each electrode is used up?
Original post by aque1408
Thank you, this is extremely helpful!

Just one final question with number 3.), is it the ions in the salt bridge that move into the two solvents in the half cells (e.g. if the salt bridge was NaCl, are Cl- ions moving into the half cell that is being oxidised?). If so, will the solvent in the salt bridge not eventually 'run out' (possibly better phrase than 'dry out') as the Na+ and Cl- ions are being used up to balance the charge in each half cell? Or to be more specific, does it typically 'run out' before each electrode is used up?

You wouldn't use a chloride for the salt bridge as the chloride ions could interfere with the half-cells.
Potassium nitrate is the solute of choice.

The half-cells would have to "work" for a long time to exhaust the supply of ions from the salt bridge. In general, the cell is not allowed to pass current as this would change the characteristics of the half-cells and make voltage measurement irrelevant.
Reply 4
Original post by charco
You wouldn't use a chloride for the salt bridge as the chloride ions could interfere with the half-cells.
Potassium nitrate is the solute of choice.

The half-cells would have to "work" for a long time to exhaust the supply of ions from the salt bridge. In general, the cell is not allowed to pass current as this would change the characteristics of the half-cells and make voltage measurement irrelevant.

Ah ok, that makes a lot more sense now

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