Are the potentials of the electrolytes equal in Galvanic cell?

Question

My book defines Electrode Potential for a Galvanic Cell as follows:

A potential difference that develops between the electrode and the electrolyte is called Electrode Potential

$$E= V_{electrode} - V_{electrolyte}\tag1$$

It further defines cell potential as:

The potential difference between the two electrodes of a galvanic cell is called Cell Potential.

Mathematically, I interpret it as $$E_{Cell} = V_{electrode,anode} - V_{electrode,cathode}\tag2$$

However my book further mentions that

The cell potential is the difference between the electrode potential of the anode and cathode.

I interpret this as: $$E_{Cell} = E_{anode} - E_{cathode}$$ From (1) $$E_{cell}=(V_{electrode,anode} - V_{electrolyte,anode} )- (V_{electrode,cathode} - V_{electrolyte,cathode})$$ To keep the above consistent with (2), $$V_{electrolyte,anode} = V_{electrolyte,cathode}\tag4$$

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I find (4) illogical as there is no basis for the two electrolytes to be at same potential.

What is going on here?

Answer 1

Yes, the potentials of electrolytes are equal. If they are of different type and connected by a salt-bridge then they are at same potential, because we assume that the salt bridge offers no resistance to electrons, it is essentially short-circuiting the two electrolytes. If the electrolytes are the same then no salt bridge is needed, and they are obviously at same potential.

Answer 2

The voltage drop about the electrolyte is one of the major energy losses. You do have to subtract the voltage drop across the electrolyte from the theoretical open circuit voltage to get the cell voltage. One of the major losses in the electrolyte is because the current is due to ions flowing as a liquid not electrons in a conductive material (bigger power drops p=IV where I is the flow of ions). In the end the voltage is measured between 2 points whether you go external circuit or internal through the battery.