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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?

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  • $\begingroup$ "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. $\endgroup$
    – The Photon
    Jul 14, 2020 at 15:25
  • $\begingroup$ 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 $$ W=QV$$ $$G=-nFE$$ $\endgroup$
    – ChemEng
    Jul 14, 2020 at 15:25
  • $\begingroup$ when a cell is charging voltage is greater than emf of cell, so voltage is always less than the emf is not always true $\endgroup$
    – maverick
    Jul 14, 2020 at 15:56

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The answer is the 2nd law of thermodynamics. (I'm assuming you're referring to the emf and potential difference associated with a battery.)

Emf is the work done by the battery in moving the charge from one terminal to the other, per unit charge. The potential difference is the change in potential energy of the charge, per unit charge, in the same process. That is, potential difference is the negative of the work done by a conservative force, namely, the force from the electric field, per unit charge. So the emf pushes the charge up the potential energy hill, so to speak.

Energy conservation says the net work done per unit charge (emf) must equal the work done against conservative forces plus the work done against nonconservative forces. The second law of thermodynamics says for a spontaneous chemical process, the work done by nonconservative forces must be dissipative (i.e. convert some of the emf into heat). That means for a passive device like a battery, the potential difference must be less than or equal to the emf, with equality holding only in the ideal limit of no dissipation.

(I say "passive devices" because you can construct a device, such as an amplifier or current source, where the internal emf is less than the potential difference, but there has to be some work put into it from outside, otherwise you would have made yourself a perpetual motion machine! Such a device appears to have a negative internal resistance, but it doesn't really.)

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?

You've got it a bit backwards - because there's resistance, there is more work required to move the electrons, and thus they gain less PE because some of the work gets dissipated (or, to say it another way, more work is required for the same gain in PE).

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