Homework Statement:-I am confused about some lines in my textbook about galvanic cell[/B]
Homework Equations:-The lines 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[/B]
The Attempt at a Solution:-Let us take 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. Let's 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.
Actually someone has explained me in this manner,i want to verify this.Is this correct?