What is the voltage of an electrochemical cell at equilibrium?

Question

I am having a hard time understanding what happens at equilibrium in an electrochemical cell (a full cell with two electrodes). At equilibrium, according to Nernst equation, the cell potential is $E_{cell}=E^0_{cell}-\frac{RT}{zF}lnQ_r$. Is this cell potential equal to the voltage of the cell ? At equilibrium, the voltage is zero, so does it mean that $E_{cell}=0$ and that both half cells have the same potential ? I am confused about this, as I also understood that $E_{cell}$ at equilibrium is the electromotive force of the cell, which means that it isn't zero.

Thank you for your help !

Answer 1

There is a cell equilibrium and 2 independent half-cell equilibrii.

There are 2 half-cell equilibrium potentials and a cell open voltage as EMF as the difference of 2 half-cell equilibrium potentials, because the cell in the whole is not in equlibrium.

When a cell gets into equlibrium as the whole system, the cell voltage is zero.