Help to understand the passage about reversibility in galvanic cell

[determined daisy]

I am confused about some lines in my textbook about galvanic cell .These 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
Please help.
Regards.

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[EParker37]

I am confused about some lines in my textbook about galvanic cell .These 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
Please help.
Regards.

Think of Le Chatlier's principle, 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.

The gibbs stuff is just referring to the fact that you have to have zero current in order to find the maximum amount of useful work that the system can provide, IE: have a cell potential of 1.1 and an external voltage of 1.1. Then you use the deltaG = -nFE equation to find the -deltaG.

 

[determined daisy]

How does a salt bridge complete circuit?As no electrons flow within salt bridge.Current is a movement of charge. Electrons are not the only charged objects.Ions can also cause flow of current. Ions are responsible for completing the circuit via salt bridge,right?But to complete the circuit ions/electrons should move from cathode to anode ,how that occurs?

 

[EParker37]

How does a salt bridge complete circuit?As no electrons flow within salt bridge.Current is a movement of charge. Electrons are not the only charged objects.Ions can also cause flow of current. Ions are responsible for completing the circuit via salt bridge,right?But to complete the circuit ions/electrons should move from cathode to anode ,how that occurs?

The salt bridge allows the Ions to move freely from one half cell to the other. If we didn't have the salt bridge, then we would accumulate negative charge and positive charge in the respective half cell and the reaction would eventually cease. The ions actually travel from one half cell to the other half cell based on charge, IE: the cation will head towards the cathode half cell and the anion will move towards the anode half cell. The electrons move through the wire connecting the two electrodes.

Edit: Cawolf pointed out my mistake with the charge movement.

 

[Cawolf]

IE: the cation will head towards the anode half cell and the anion will move towards the cathode half cell.

It is actually the opposite of this.

 

[EParker37]

It is actually the opposite of this.

Whoops, yes, that is right, it has been a hectic morning at work with traumas rolling in and I am half asleep haha.