Is the subtrahend in the Nernst equation related to ohmic drop?

Question

Let us consider the Nernst equation $$E= E^\circ -\frac{2.303RT}{nF} \log Q$$ which once at $\pu{298.15 K}$, is the equivalent of $$E= E^\circ - \frac{0.059}{n} \log Q .$$

Does the part of the Nernst equation $-\frac{0.059}{n} \log Q$ address Ohmic drop? In other words, is $-\frac{0.059}{n} \log Q$ related to solution (electrolyte) resistance of an electrochemical cell?

Answer 1

In short: No.

The Nernst equation describes the thermodynamic equilibrium state. By definition, there is no current flowing through the cell. Therefore, there is no ohmic drop in any term within the Nernst equation.

The Q term relates to the activities of the reactants and products at equilibrium.