Ana+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. what is the change in the ion's electric potential energy as it moves from inside to outside the cell?

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:

where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: , and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find: