Potential Change, Initial and Final States

Name: Greg
Status: AskAScientist
Grade: other
Location: TX

Question: Hi All, I am one of the volunteer scientists in this forum
and I was going to answer a question sent in (by Robine) concerning
Cu transformations to its ions when something jumped out at me from
my copy of a Standard Reductions Potentials table. I cannot figure
this one out. Cu(+) + e(-) = Cu(s) is +0.52V Cu(2+) + 2e(-) = Cu(s)
is +0.34V and Cu(2+) + e(-) = Cu(+) is +0.16V My question is, if we
take eq. 1, reverse it (giving it a -0.52V), and then add it to eq.
2, this would result in eq. 3 but with a potential of -0.18V! I
checked other sources and the potentials are as listed above. Am I
missing something? Isn't the potential change only a function of
initial and final states and therefore a kind of Hess's Law should
apply? (By the way, this seeming discrepancy also applies to the
three equations of Fe.)
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What you are missing is that potential is not a state function: free energy is.  In 
language less thermodynamics-specific, this is because energy, not potential, is 
conserved.  In other words, the ENERGY change going from Cu(2+) + 2e(-) to Cu(s) 
must be the same along any path taken to get there.  POTENTIAL change is not 
necessarily the same; potential will be subservient to energy conservation.

Potential is potential energy per unit charge.  In other words, when a charge moves 
through a potential, the energy expended (work done) on it is the potential 
multiplied by the charge.  Let's see how this checks out for the two paths to go 
from Cu(2+) + 2e(-) to Cu(s).

1.  Cu(2+) + 2 e(-) --> Cu(s);  V = 0.34V
  Since there are two electrons here, the energy expended is (2e)(0.34 V) = 0.68 eV 
  (electron-volts)

2. Cu(2+) + e(-) --> Cu(+);  V = 0.16 V; followed by Cu(+) + e(-) --> Cu(s); V = 
+0.52V
  One electron is added in the first step, and one in the second step.  For the 
  first step, the energy expended is 0.16 eV; for the second step, the energy 
  expended is 0.52 eV.  The total energy expended is 0.68 eV.

This works.  The energy expended on the two electrons is the same in both cases.  
Basically, the +0.34 V for the one-step, two-electron process is the AVERAGE 
potential change for the two electrons involved in the process.

Chemistry textbooks use the concept of "volt-equivalents" to explain how to combine 
half-cell reactions in this manner.  In a half-cell reaction in which n is the 
number of electrons transferred and V is the half-cell potential, nV is 
proportional to the free energy.  Thus, when adding two half-reactions together, 
to find the half-cell potential of the sum, you calculate nV for each of the two 
half-cell reactions, add THOSE together to find nV for the total half-reaction, 
and then divide that sum by n for the final half-reaction to find its V.

As a further example, let us examine the reduction of iron(III):

  Fe(3+) + e(-) --> Fe(2+);  V = 0.77 V
  Fe(2+) + 2 e(-) --> Fe;  V = -0.44 V

The sum of these two half-reactions is

  Fe(3+) + 3 e(-) --> Fe;  V = ?

For the reduction of Fe(3+) to Fe(2+), n = 1, so nV = 0.77 eV.  For the reduction 
of Fe(2+) to Fe, n = 2, so
nV = (2 e)(-0.44 V) = -0.88 eV.  Thus, for the reduction of Fe(3+) to Fe, nV = 
(0.77 - 0.88) eV = -0.11 eV.  Since n = 3 e, that means V = (-0.11 eV)/(3 e) = 
-0.037 V.  This is confirmed by the tables. 

(Imagine my chagrin after working through this to discover that it is the very same 
example worked through in Wikipedia's "Standard Electrode Potential" article at 

http://en.wikipedia.org/wiki/Standard_electrode_potential.)

Richard Barrans Jr., Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming
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