Nernst Equation
Consider the half reaction:
NO3- + 10H+ + 8e-  NH4+ + 3H2O(l)
We can calculate the Eh if the activities of H+, NO3-,
and NH4+ are known. The general Nernst equation
is
2.303RT
0
Eh  E 
n
log Q
The Nernst equation for this reaction at 25°C is
 aNH 
0
.
0592
4
Eh  E 0 
log 
 aNO  a10
8
 3 H




Let’s assume that the concentrations of NO3- and
NH4+ have been measured to be 10-5 M and
310-7 M, respectively, and pH = 5. What are the
Eh and pe of this water?
First, we must make use of the relationship
 G
E 
n
0
o
r
For the reaction of interest
rG° = 3(-237.1) + (-79.4) - (-110.8)
= -679.9 kJ mol-1
679.9
E 
 0.88 volts
(8)( 96.42)
0
The Nernst equation now becomes
0.0592  aNH 4
Eh  0.88 
log
10

8
aNO  aH 
 3




substituting the known concentrations
(neglecting activity coefficients)
0.0592  3  107 
Eh  0.88 
log

0
.
521
volts
10
 105 105  
8


and
pe  16.9 Eh  16.9(0.521)  8.81
Biology’s view  upside down?
Reaction directions for 2 different redox couples brought together??
More negative potential  reductant // More positive potential  oxidant
Example – O2/H2O vs. Fe3+/Fe2+  O2 oxidizes Fe2+ is spontaneous!
Stability Limits of Water
• H2O  2 H+ + ½ O2(g) + 2eUsing the Nernst Equation:
0.0592
1
Eh  E 
log 12 2
n
pO2 aH 
0
• Must assign 1 value to plot in x-y space (PO2)
• Then define a line in pH – Eh space
UPPER STABILITY LIMIT OF
WATER (Eh-pH)
To determine the upper limit on an Eh-pH
diagram, we start with the same reaction
1/2O2(g) + 2e- + 2H+  H2O
but now we employ the Nernst eq.
0.0592
1
Eh  E 
log 12 2
n
pO2 aH 
0
0.0592
1
Eh  E 
log 12 2
2
pO2 aH 
0
 G
 ( 237.1)
E 

 1.23 volts
n
(2)( 96.42)
0
0
r
Eh  1.23  0.0296 log pO22 aH2 
1
Eh  1.23  0.0148 log pO2  0.0592 pH
As for the pe-pH diagram, we assume that
pO2 = 1 atm. This results in
Eh  1.23  0.0592 pH
This yields a line with slope of -0.0592.
LOWER STABILITY LIMIT OF
WATER (Eh-pH)
Starting with
H+ + e-  1/2H2(g)
we write the Nernst equation
1
2
p
0
.
0592
H2
0
Eh  E 
log
1
aH 
We set pH2 = 1 atm. Also, Gr° = 0, so E0 =
0. Thus, we have
Eh  0.0592 pH
C2HO
Making stability diagrams
• For any reaction we wish to consider, we
can write a mass action equation for that
reaction
• We make 2-axis diagrams to represent
how several reactions change with respect
to 2 variables (the axes)
• Common examples: Eh-pH, PO2-pH, T-[x],
[x]-[y], [x]/[y]-[z], etc
Construction of these diagrams
• For selected reactions:
Fe2+ + 2 H2O  FeOOH + e- + 3 H+
3

a
0.0592  H 
0
Eh  E 
log
 a 2
1
 Fe




How would we describe this reaction on a 2-D
diagram? What would we need to define or
assume?
• How about:
• Fe3+ + 2 H2O  FeOOH(ferrihydrite) + 3 H+
Ksp=[H+]3/[Fe3+]
log K=3 pH – log[Fe3+]
How would one put this on an Eh-pH diagram,
could it go into any other type of diagram
(what other factors affect this equilibrium
description???)
Redox titrations
• Imagine an oxic water being reduced to
become an anoxic water
• We can change the Eh of a solution by
adding reductant or oxidant just like we can
change pH by adding an acid or base
• Just as pK determined which conjugate
acid-base pair would buffer pH, pe
determines what redox pair will buffer Eh
(and thus be reduced/oxidized themselves)
Redox titration II
100
--
H2S(aq)
SO4
90
4
--
Some species w/ SO (umolal)
• Let’s modify a bjerrum plot to reflect pe
changes
80
70
60
50
-4
-2
0
2
4
pe
6
8
10
12