Internal Resistance of the Battery - Ohmic Resistance

Question

When current starts flowing in a circuit, voltage of the battery drops from EMF value. There are three sources of this however here I am only interested in ohmic resistance of the battery.

EMF of the battery is created because of electrochemical processes on both electrodes which create potential difference on electrode/electrolyte interface due to charge separation.

When talking about ohmic resistance of the battery I am not sure how does ohmic resistance of battery components affect charge separation on the interface. What does ohmic resistance of electrolyte or electrodes have to do with equilibrium of electrochemical reactions on the electrodes?

Answer 1

Nothing. The Ohmic resistance does not change the electrochemical reactions on the electrodes, it changes the electric field between the electrodes. By Ohm’s law, if there is a current through a resistor there is also a voltage across the resistor and the two are related by $V=IR$.

Suppose that the circuit is grounded at the conductor attached to the negative terminal of the battery and that the half cell at the negative side is 0.5 V and the half cell at the positive side is 1 V.

Now, in an open circuit condition there is no current so there is no Ohmic voltage drop. So the negative terminal is at 0 V, the electrolyte is at 0.5 V, and the positive terminal is at 1.5 V.

Suppose the circuit is closed and there is enough current to produce a 0.1 V Ohmic voltage drop (and neglect the other electrochemical effects at the electrodes). Now, the negative terminal is at 0 V and the electrolyte immediately adjacent is at 0.5 V. However, because of Ohm’s law the electrolyte is not all at 0.5 V, but instead the electrolyte adjacent to the positive terminal is at 0.4 V, and thus the positive terminal is at 1.4 V.

So Ohmic resistance gives a voltage drop without changing the electrochemistry at the electrodes, but instead by changing the E field between the electrodes.