# Why is the EMF more than the voltage?

Why is the EMF of a battery always more than the voltage when current flows through the battery? Most answers say that because there is internal resistance in the battery there is a voltage drop, but isn't it true that when electrons are moved from low potential to higher potential within the battery they gain energy because work is done moving them between these two points. Thus wouldn't it follow that if there was resistance, more work would need to be done to move the electrons and thus they would gain more potential energy and increasing the voltage so that it's more than emf? I know that this is not the case so could someone please explain where I went wrong?

• "The" EMF isn't always larger than "the" voltage. In some cases, EMF is defined when voltage (electrostatic potential difference) isn't even a defined quantity. Sometimes we call an emf "the voltage" of some system. ... Oh wait, you're asking about the emf of an electrochemical cell or battery. You should say that right away to set the context of your question. Jul 14, 2020 at 15:25
• There are energy losses in the battery that cause the overpotential. The 3 major causes being activation overpotential (sluggish reaction kinetics), ohmic losses and mass transport/ concentration losses (slow mass transfer). We can also have parasitic side reactions and internal short circuit as losses among the others less considered. When charging these losses lead to a large voltage when discharging they lead to a lower voltage than the OCV (open circuit voltage). remember $$W=QV$$ $$G=-nFE$$ Jul 14, 2020 at 15:25
• when a cell is charging voltage is greater than emf of cell, so voltage is always less than the emf is not always true Jul 14, 2020 at 15:56