Redox andRedox and
pE pE –– pH DiagramspH Diagrams
OCN 623 – Chemical Oceanography
Reading: Libes, Chapter 7
Equations are
written as
REDUCTIONS!!
This lecture will go briefly over the principles involved in
deriving and using these equations. These principles
are all based on electrochemical cells.
Redox Equations Worth KnowingRedox Equations Worth Knowing
nRT
GpE
KpE
EhRT
FpE
reduc
oxidpEpE
reduc
oxid
nEE
nFEG
n
nb
a
b
a
3.2
log
3.2
log
log0592.0
1
1
∆−=•
=•
=•
+=•
+=•
−=∆•
ο
ο
ο
Electrochemical CellsElectrochemical Cells
Consider the following simple electrochemical cell
operating at 25°C, used to measure the electrochemical
potential of this reaction: Cu2+(aq) → Cu0(s)
ions
We arbitrarily assign a potential of 0 to
the reaction in the left cell:
2H+(aq) + 2e- → H2(g) E°= 0.000 V
Then the potential for the reaction in
the right cell is:
Cu2+(aq) + 2e- → Cu0(s) E°= 0.337 V
(always write as a reduction)
The standard potentials for all redox reactions are similarly
determined against the standard hydrogen electrode:
= E° = E°H
An electrochemical cell is capable of doing work – by
driving electrons across a potential difference.
This can be measured as a change in free energy:
∆Gr° = -nFE°
where
n = number of moles of electrons (equivalents) involved in
the reaction
F = Faraday constant = 23.1 kcal V-1 equiv-1
E° = the cell potential (V) at standard state
For the general case: ∆Gr = -nFE
Nernst EquationNernst Equation
We know from a previous class:
y
x
reactants
products RTG∆G
}{
}{ln+°∆=
Substituting ∆G = -nFE, we get the Nernst Equation:
Or:
yreactants
xproducts
nF
RTEE
}{
}{ln −°=
xproducts
yreactants
nF
RTEE
}{
}{ln +°=
reduced species
oxidized species
At 298°K:
xproducts
yreactants
nEE
}{
}{log
0592.0 +°=
Important points:
• Geochemists usually use the symbol EH instead of E
(indicating the hydrogen scale is being used)
• The Nernst Equation relates the EH of a cell to the
standard EH and to the activities of reactants and
products under given conditions
• When at standard state (all activities = 1), EH = E°
• We can use EH as an indicator of the state of natural
waters:
+−+ ↔+ 23 VeV
What species of V dominates in seawater with
measured Eh = 0.729 V?
1. From Table 7.1: Ehº= -0.26
2. Plug into equation:
+=+
+
2
3
log1
059.0
V
VEhEh ο
16
2
3
2
3
2
3
2
3
106
log8.16
log059.0989.0
log059.026.0729.0
x=
=
=
+−=
+
+
+
+
+
+
+
+
V
V
V
V
V
V
V
V
∴ V3+ dominates
Example #1: Use Eh values to Example #1: Use Eh values to
Calculate Vanadium SpeciationCalculate Vanadium Speciation
{ }{ } { }
{ }{ }{ }
{ }{ }{ }b
a
b
a
n
na
b
reduc
oxidK
en
reduc
oxidK
e
eoxid
reducK
loglog1
log
:logtheTake
1
+=
=
=
−
−
−
Definition of pEDefinition of pE
a(oxidized-species) + ne- → b(reduced-species)
From equilibrium theory:
{ } { }
{ }{ }
{ }{ }b
a
b
a
reduc
oxid
npEpE
KpE
reduc
oxidK
npE
npEee
n
n
log1
log:but
loglog1
bydivideandlog1
log:but
1
1
+=∴
=
+=∴
≡−= −−
ο
ο
Example #2: Calculation of pEExample #2: Calculation of pE
Assume:
• pE in a given environment is controlled by this reaction:
Fe3+ + e- → Fe2+
(∴n = 1)
• {Fe3+} = 10-5
• {Fe2+} = 10-3
( )
{ } 11
3
5
10
0.1110
10log0.13
0.13log
log1
−−
−
−
=
=
+=∴
==
+=
e
pE
n
KpE
reduc
oxid
npEpE
ο
ο
Table 7.1
Example #3: Effect of Atmospheric OExample #3: Effect of Atmospheric O22
Assume: Natural water at pH 7.5 in equilibrium with the atmosphere
Conclusion: This environment has lower electron activity than
Example #2, and is thus more oxidizing
{ }
( ) ( ) ( )
( )
{ } { }{ }
{ }
{ } 08.13
5.7
2
1
2
4
1
2
22
2
1008.13
10log
Libes7.3,Table75.20
21.021.0
1
2
1
4
1
2
−−
−++
−+
=∴=
=+=
=
↔++
=∴=
epE
H
OH
HOpEpE
pE
OHeHO
OatmP
n
laqg
O
ο
ο
1
Redox Reactions Have Characteristic pRedox Reactions Have Characteristic pE ValuesE Values
HERT
FpE
3.2=
nRT
GpE
3.2
∆−=
Garrels & Christ (1965)
¼ O2 + H+ + e- ← ½ H2O
H2O + e- → ½ H2 + OH-
Thus, if an environment is characterized by a Thus, if an environment is characterized by a
certain redox reaction, it has a characteristic pEcertain redox reaction, it has a characteristic pE
pEpE--pH DiagramspH Diagrams
pE-pH stability field diagrams show in a comprehensive
way how protons (pH) and electrons (pE)
simultaneously shift equilibria of reactions under various
conditions
These diagrams also indicate which species predominate
under any given condition of pE and pH
Two equations are used to produce the diagrams:
{ }{ }r
o
reduc
oxid
npEpE log
1+°=
Kn
pE log1
=°
Oxidizing limit of diagrams: ¼ O2 + H+ + e- → ½ H2O
{ } { }{ }
pHpE
npEpE
−=
+°=+
75.20
OH
HOlog
1
21
2
41
2
Reducing limit of diagrams: H2O + e- → ½ H2 + OH-
OH- + H+ → H2O
H+ + e- → ½ H2
{ }{ }
pHpE
npEpE
−=
+°=+
21
2H
Hlog
1
pE° = +20.75 (Table 7.3)
n = 1
pH = -log{H+}
Set limit: {O2} = 1
pE° = 0.0 (Table 7.3)
n = 1
pH = -log{H+}
Set limit: {H2} = 1
Oxidizing limit of diagrams:
O2 + H+ + e- → ½ H2O
pHpE −= 75.20
Reducing limit of diagrams:
H+ + e- → ½ H2
pHpE −=
• Phase-boundary lines on a
pE-pH diagram indicate stability
field boundaries – defined as
lines where activities of both
adjacent dominant species are
equal.
• Lines are defined by reactions
between adjacent dominant
species
• Reactions must have known
log K or pE° values.
Construction of pEConstruction of pE--pH DiagramspH Diagrams
{ }{ }{ }
{ }{ }
{ } { } 0.2,
log0.2log
2.0logKHSOHSO6.
4
2
4
4
2
4
4
2
4
2
44
==
−=−=
=
−=+↔
−−
−
−
−
+−
+−−
pHHSOSOWhen
pHHSO
SOK
HSO
HSOK
Acid-base reactions with no pE dependency
{ }{ }{ }{ }
{ }{ }
{ } { } pHpEHSSOWhen
pHHS
SOpE
Kn
pE
OHHS
HSOpEpE
8
9
8
34,
8
9log
8
1
8
34
8
34log
1
log8
1
34.0KlogO4HHS8e9HSO5.
2
4
2
4
4
2
92
4
2
2
4
−==
−+=
==
+=
=+↔++
−−
−
−
−
+−
−−+−
ο
ο
=1
Redox reactions of dissolved species
Redox reactions of dissolved and solid species
( )
{ } { }( ){ }{ }
{ }
{ } pHpESOWhen
pHSOpE
OHS
HSOpE
s
6
870.5,10
6
8log
6
1
6
2.36
log6
1
6
2.36
36.2KlogO4HS6e8HSO1.
22
4
2
4
4
2
82
4
2s
2
4
−==
−+=
+=
=+↔++
−−
−
+−
−+−
=1
Activity of dissolved
species must be given
SI
Homework Homework –– Due Thurs, Jan. 24, 2013Due Thurs, Jan. 24, 2013
1) Write a balanced equation for the nitrification of NH4+ to
NO3- (oxidation by O2).
2) Calculate the Eh at 25°C for this reaction in offshore
Hawaiian seawater at 100 m depth. Assume [NH4+] = 1 µM,
[NO3- ] = 150 nM, and the partial pressure of O2 = 0.21 atm;
assume the activity coefficients for NH4+ and NO3
- are 0.7 .
(Hint: calculate ∆G° and then Eh.)
3) In which direction would the reaction proceed in an
environment with an Eh of -200 mV? Why?