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Re: emf and batteries



Bob LaMontagne said...
One of the nice features of the resistance
model is that you can explain why the
terminal potential difference is greater
than the EMF when you run a current backwards
through the battery. Does this
reversible-irreversible approach easily
do the same?

Yes. Assuming the reaction is reversible,
so that you are running the same reaction
(backwards) when you run current backwards,
then we would say...

The work you are doing (actually rate of work)
when you run the current backwards is
V*I. If you have a very small current such that
you are doing a thermodynamically reversible
process, then V*I equals emf*I.

However, for larger currents you have not
satisfied the conditions of a thermodynamic
reversible process, so your work goes other
places in addition to running the chemical
reaction backwards. In this case your
work (power) is

V*I equals emf*I + losses (heat)

This means that V must be greater than emf.

Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton College
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu