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?

  • $\begingroup$ Why would you expect the standard electrode potential of a reaction in equilibrium not to be 0? $\endgroup$
    – Jerry
    Apr 1, 2013 at 8:15
  • $\begingroup$ given that the standard electrode potential determines the tendency of the system to follow the forward reaction,does the value of $$ E^0 $$ change as the reaction proceeds towards equilibrium? $\endgroup$ Apr 1, 2013 at 8:46

1 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).

  • $\begingroup$ But then we do have a list of standard electrode potentials of different half cells with respect to the standard hydrogen electrode,right?how can $$ \Delta E^0 $$ be zero? $\endgroup$ Apr 1, 2013 at 18:40
  • $\begingroup$ That is quite different... $E^o$ itself is the standard electrode potential of the electrode with respect to the Standard Hydrogen Electrode. So the value of $E^o$ for a given redox couple would never change unless you changed the reference electrode (you find the electrode potential of the electrode with respect to some other electrode instead of the SHE). $\endgroup$
    – kaliaden
    Apr 2, 2013 at 3:06
  • $\begingroup$ I am confused.I have solved some problems,where I had to calculate $ \Delta E^0 $ for calculating emf of the cell,or $ \Delta G^0 $ where I simply put the Nernst equation at equilibrium.All questions consider the state of equilibrium,standard conditions,and given half cell connected to SHE.is it correct to get non zero $ \Delta E^0 $ ? $\endgroup$ Apr 2, 2013 at 11:07

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