Half-cells and reduction potentials
Recall that we could set up tables of fGo and fHo for ions in solution by referencing to
a standard, say fGo for H+ equals zero by definition. For example, consider the Na+(aq)
ion. From the tables of thermodynamics properties we find,
o
 f GNa
 261.91kJ/mol. (1)
+
( aq )
What is the reaction?
Na(s)  Na+(aq) + e
(2)
This is a “half-reaction, a hypothetical reaction, written as an oxidation. By international
convention, electrochemists have agreed to write half-reactions as reductions. Rewrite
our half-reaction as a reduction,
Na+(aq) + e  Na(s).
(3)
Then the Gibbs free energy change for the reduction reaction is just the negative of the
Gibbs free energy of formation of the positive ion,
o
 rG o   f GNa
. (4)

( aq )
We can calculate an emf associated with this half reaction from,
E 
o

o
 rGNa

( aq )
nF

o
 f GNa

( aq )
1F
J
(5a,b)
mol  2.715 v.
Coul
96,485
mol
261.91  103
This E o is called the “half-cell” potential for the Na+(aq)|Na half cell.
o
Recall that  f GNa
was not measured in an absolute sense, but was measured relative

( aq )
to an arbitrary standard which we (by convention) pick to be pick to be the formation of
the H+(aq) ion. That is, we measure the fH o of all other ions relative to H o of the
reaction,
1/2 H2(g)  H+(aq) + e,
for which (by convention) we set,
 f GHo + ( aq )  0. (6)
Then
o
ENa
is relative to
+
( aq )
EHo + ( aq )  0 . (7)
This is easy to show. Write the half-reaction for the reduction of hydrogen ion,
H+(aq) + e  ½ H2(g), (8)
for which
 rG o   f GHo + ( aq )  0 . (9)
Then,
Eo 
0
 0. (10)
1F
There are tables of half-cell potentials or you can make your own from tables of
fGoion(aq). (Modern half-cell tables will always be written for half-reactions written as
reductions. Tables from older American books have the half-reactions written as
oxidations and the tables are oxidation potentials.)
Conventions and Usage
Given a cell, A,B|C,D,…| …..|Z, write the reaction for the Right-hand-side as a
Reduction, and left-hand-side as an oxidation.
Right = Reduction --- (call value from table EoR.)
Left = Oxidation --- (call the table value, which will be a reduction potential, EoL.)
Then, for the complete reaction,
Eo = EoR  EoL. (11)
I emphasize that both values are table REDUCTION potentials. You do not change the
sign of either one to compensate for anything. All the signs have been taken care of in
the above equation. Then,
 rG o  nFE o . (12)
If Eo > 0 then Go < 0, reaction proceeds spontaneously as written,
If Eo < 0 then Go > 0, reaction does not proceed spontaneously as written.
Example:
Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s) (13)
R Cu2+ + 2 e  Cu
L Zn
 Zn2+ + 2 e
------------------------------------Cu2+ +Zn  Cu  Zn2+ (13)
n=2
o
ECu
 0.337v
o
EZn
 0.763v
From tables, both as reductions. (14a, b)
E o  ERo  EL0  0.337  (0.763)  1.10v. (15)
Then
G o  2F 1.10  212.3 kJ. (16)
Recall, that if Eo > 0 then Go < 0 so the reaction is spontaneous.
Let the reaction run free, which side is positive?
Notice that the half-reaction,
Zn  Zn2+ + 2 e, (17)
is running forward.
Zn is "giving off" electrons, Zn is negative, therefore Cu is positive.
We can get the equilibrium constant from Eo:
Recall,
G o   RT ln K  nFE o
nFE o
ln K 
,
RT
K e
nFE o
RT
(18a, b, c)
.
Other Thermodynamic Functions From Electrochemical Cell Data
We have seen that we can get rGo for a reaction from the emf of its electrochemical cell,
can we get other thermodynamic functions? Yes, we can get other thermodynamic
functions, but it requires additional data.
Recall,
 G 

  S ,

T

p
or
(19a, b)
 G 

  S .
 T  p
Then,
 G o 
 (nFE o ) 
 E o 
S o   




n
F




 . (20)

T

T

T

p

p

p
So we can get So if we know Eo as a function of T. Then we can also get Ho from,
Ho = Go + TSo. (21)
Sometimes we want to do this graphically. Plot Eo vs T.
S o
, but this is not likely to give a straight line. Ho varies much more
nF
slowly with T than does So. Try the Gibbs-Helmoltz equation,
The slope is
 G o
 T

  1
 T


o
  H . (22)

p
Plug
G o  nFE o (23)
into Equation 22 to get,
 Eo

nF  T
  1
 T


o
  H .

p
(24)
Then we plot Eo/T vs 1/T. We should get a line much closer to a straight line, with slope
H o
equal to 
.
nF
We can still get So from
S o 
H o  G o
. (25)
T