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Re: emf and batteries

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?

Bob at PC

*********** REPLY SEPARATOR ***********

On 3/23/2004 at 10:02 AM Edmiston, Mike wrote:

The emf for a reaction represents
the maximum work that can be obtained
=66rom that reaction. In order to obtain
the maximum work, the cell must operate
reversibly. Operating reversibly to
get the maximum work typically means
slowly, and certainly precludes
operating in a fashion that liberates

As the current (from a battery)
increases, the electrochemistry is
certainly moving from reversible
to irreversible, and heat is clearly
a factor. Therefore we cannot be
getting the maximum work. If we are
not getting the maximum work then
we must measure a potential difference
that is less than the emf.

This explanation never uses the
word "resistance."