Is standard electrode potential of a galvanic half-cell is zero at equilibrium under standard conditions?

Question

At standard conditions (at 1 atm pressure and unit activity (1 molal or 1 molar concentration) of all dissolved compounds), the electrode potential is equal to the standard electrode potential by the Nernst equation: $\Delta E = \Delta E^\circ.$

For the electrochemical reaction at equilibrium the electrode potential of the cell is zero: $\Delta E = 0.$

This implies that for any reaction at equilibrium under the standard conditions $\Delta E = \Delta E^\circ = 0,$ which is quite surprising. Is this correct?

Answer 1

There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential.

$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$.

Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.

For $\Delta E$ however, consider the Nernst equation, $$E = E^\circ - (RT/nF)\ln Q$$ Change in $E$ $$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$ therefore $$\Delta E = -(RT/nF)\Delta\ln Q$$

At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero).