# Intuitive explanation for EMF of cell being zero at equilibrium?

I am not able to understand why $$E_{cell=0}$$.

I went through this answer of the site Electrode Potential at Equilibrium

It says

This eventually comes to a stop when the equilibrium is reached. That is the rate of negative charge build-up is equal to the positive charge. Like you say at this point there is no more drive for the reaction to go one way or the other.

But if rate of negative charge build up is equal to positive charge but we would still have charge separation so why should potential difference be zero?

Like you say at this point there is no more drive for the reaction to go one way or the other.

I know that the Galvanic cells are drove through the Gibbs' energy of redox reaction and I agree that at equilibrium there is no change in Gibbs' energy but we still have charge separation between cathode and anode so why should the potential difference between them be zero ?

I ask my query as a new question here because answerer to the linked question has been inactive for long time.

• When a system reaches equilibrium it reaches the minimal free energy and no energy can be spontaneously withdrawn from it. As mechanical analogy, you can draw energy from a ball rolling downhill. When it is at mechanical equilibrium at the bottom of the pit or valley, you cannot draw more energy from it (by mechanical means). May 1 at 13:22
• @Poutnik As I mentioned in question , I know why EMF should be zero due to energy considerations , simply because we don't have and Gibb's energy to do a non P-V work but my query is that we do have charge separation and thus a potential difference May 1 at 13:25
• No, we do not have, if equilibrium is reached. Note that some systems may need practically infinite time to reach equilibrium. May 1 at 13:34
• @Poutnik Can you explain how we do not have charge separation at equilibrium ? May 1 at 13:37
• Rather, can you explain how we can have it? Having potential electrode difference and being at equilibrium are mutually excluding states ( I mean cell equilibrium, not electrode/half cell equilibrium) May 1 at 13:39

Let's discuss the Daniell cell. In the beginning, the cell is made of a zinc plate in a $$\ce{ZnSO4}$$ solution (maybe $$1$$ M), and a copper plate in a $$\ce{CuSO4}$$ solution (may be $$1$$ M). Zinc is the anode and copper the cathode.
If the cell delivers some current, the amount of zinc metal decreases, the concentration of $$\ce{Zn^{2+}}$$ increases. On the other electrode, the amount of metallic copper increases, the concentration of $$\ce{Cu^{+2}}$$ decreases, and the corresponding anions $$\ce{SO4^{2-}}$$ are crossing the bridge between the two compartments. So that the cathodic region (around Cu electrode) is slowly loosing all its ions. Its potential decreases from $$+0.34$$ V to less and less. Simultaneously, the potential of the cathode increases from $$-0.76$$ V to a less negative potential. The importance of these changes is described by Nernst's law.
When the reaction is finished, both electrodes have the same redox potential, which should be around $$-0.2$$ V. No potential difference. No current any more. The cell is made of a heavily corroded zinc anode in a concentrated $$\ce{ZnSO4}$$ solution (maybe $$2$$ M), and a thick cathode made of much copper, in a rather diluted $$\ce{CuSO4}$$ solution. $$\pu{E_{cell}}$$ is zero
• @Lalit. The potential of an electrode (anode or cathode) is described by Nernst's law : $\pu{E = E° + \frac{0.059 V}{n}·ln\frac{[Ox]}{[Red]}}$. In the case of the cathode, made of copper in a $\ce{Cu^{2+}}$ solution, $\pu{[Ox] = [Cu^{2+}]}$. So the potential $\pu{E}$ of the cell increases when the concentration $\ce{Cu^{2+}}$ increases May 2 at 11:41