Conditions at Equilibriums

[Marisol Meneses 2C]

At equilibrium, why is E=Enot not true? My TA mentioned that they're not equal at equilibrium, but why?

[Joshua Chandran 2D]

This is because Enaught represents the cell potential when the half-reactions of the redox reaction are in their standard states, which will always be a fixed value. However, at equilibrium, E will be 0 as there is no work that can be done by the galvanic cell when it's at equilibrium, so E will not be equal to Enaught at equilibrium.

[Caitlin Pan 1J]

E is the cell potential not at standard states, which means that it depends on the equation E = E°-RT/nF (lnQ). This value can change based on the concentration of the products/reactants.
E° is the standard cell potential, which does not change for a given reaction. This is dependent on the half reactions and the E cell for each.
Therefore if the situation is not at standard states, then E does not equal E°.

[Caleb Nguyen 1B]

At equilibrium, we know that regular E = 0.

However, the only time E naught = 0 is:

In a concentration cell, since E naught of the cathode and E naught of the anode are the same
or:
If the reaction is at equilibrium perfectly at standard conditions (1 M, 1 atm, and 298K) which is very rare.

[505673236]

E is the cell potential that is not at standard state. The standard state does not change and we can calculate the cell potential not at a standard state using the nernst equation allowing us to calculate cell potential as a function of concentration. When a redox reaction is at equilibrium (ΔG = 0) → nernst equation reduces because Q(constant for any point in reaction)=K(equilibrium constant) and there is no net transfer of electrons (Ecell = 0). The standard E cell does not change though and this is why they do not equal each other.

[405718358]

E doesn't equal Enot because E isn't a standard state, while Enot is the standard cell potential, since we know that a standard state doesn't change for a given reaction. So E is dependent on the E = E°-RT/nF (lnQ) equation.