Determining emf of a cell using a potentiometer

When we are to determine the emf of a cell, we find the null point and balance length using the cell whose emf is to be found out.
But there is some internal resistance of the cell.
Why isn’t any potential of the cell (whose emf is to be found) dropped at its internal resistance?
I may reframe my question as: why is the potential drop in the balance length of potentiometer wire exactly the same as that of the emf of the cell concerned?
(I think by Kirchhoff’s Law, when there is no current in the circuit, there is no potential drop in the loop containing the concerned cell, but I need a qualitative clarification.)

Additionally, what happens when the jockey is shifted towards the positive terminal of the cell(s). Does any potential get dropped at the internal resistance?

I have yet another thing bothering me. In the driving circuit with the cell with known emf, does the current flow all the time, or after null point is found, there is no current in any part of the setup?