Geochemistry
DM Sherman, University of Bristol
2010/2011
Oxidation-Reduction
Equilibria
Geochemistry, DM Sherman
University of Bristol
Oxidation States of Atoms and Ions
The oxidation state of an atom is defined with the following
convention:
• The oxidation state of an atom in an elemental form is 0.
In O2, O is in the 0 oxidation state. • When bonded to something else, oxygen is in oxidation
state -2 and hydrogen is in oxidation state of +1 (except for
peroxide and superoxide).
In CO32-, O is in -2 state, C is in +4 state.
• The oxidation state of a single-atom ion is the charge on
the ion.
For Fe2+, Fe is in +2 oxidation state.
Page 1
Geochemistry
DM Sherman, University of Bristol
2010/2011
Oxidation-Reduction Reactions
Consider the reaction
2Fe2+ + MnO2 + 4H2O = 2Fe(OH)3 + 2H+ + Mn2+
• Fe2+ is being oxidized to Fe3+ (as Fe(OH)3)
• Mn4+ (as MnO2) is being reduced to Mn2+
We can express the overall reaction as two halfreactions:
2Fe2+ + 6H2O = 2Fe(OH)3 + 6H+ + 2eMnO2 + 4H+ + 2e- = Mn2+ + 2H2O
_________________________________
2Fe2+ + MnO2 + 4H2O = 2Fe(OH)3 + 2H+ + Mn2+
K for half-reactions
For each half-reaction,
A + e" = B
We can define an equilibrium constant
!
K=
[B]
[A][e " ]
Where [e-] is the activity of electrons. (This does not
mean that bare electrons are floating around in
solution!!)
!
Page 2
Geochemistry
DM Sherman, University of Bristol
2010/2011
Important half reactions
Reaction
Cu+2
+ e- =
pK
Cu+
-2.7
Fe3+ + e- = Fe2+
-13.0
1/2MnO2 + 2 H+ + e- = 1/2Mn+2 + H2O
-20.7
(1/8)CO3-2 + (5/4) H+ + e= (1/8)CH4 +(3/8) H2O
-5.1
(1/8)SO4-2 + (9/8) H+ + e= (1/8)HS- + (1/2) H2O
-4.25
(1/5)NO3- + (6/5) H+ + e= (1/10)N2 + (3/5) H2O
-21.1
The pK ladder
Electrons can fall
from species on the
left side to species on
the right. The energy
released can do work
(e.g., biology).
Page 3
Geochemistry
DM Sherman, University of Bristol
2010/2011
The pe concept..
For convenience, take -log (= p) of the K expression
to get
pK = pB " pA " pe
By analogy with pH, the pe (= -log[e-]) can be used
!
to characterize
the redox state of a system.
Knowing the pe, we can predict the oxidation state
of the species in the system.
Speciation and pe
Example: A soil has pH = 6.0 and pe= 5.0 (Eh = 0.3 V).
Calculate speciation of Cr given pK = -60.6 for the reaction
HCrO4- + 7H+ + 3e- = Cr3+ + 4H2O.
Solution:
K=
[Cr 3+ ][H2O]
[HCrO4 ][H + ]7 [e " ]3
pK = (p[Cr] - p[HCrO4]) - 7pH - 3pe
!Since pH = 6.0 and pe 5.0
"
p[Cr ] " p[HCrO4 ] = "3.6 or
[Cr 3+ ]
"
[HCrO4 ]
= 103.6
!
Page 4
Geochemistry
DM Sherman, University of Bristol
2010/2011
Calculating pe from Concentrations
Example: Given that pK = -16.5 for the half-reaction
Fe(OH)3(s) + 3H+ + e- = Fe2+ + H2O
calculate the pe of groundwater in which [Fe2+] = 10-4 M,
pH = 7 and the solution is saturated in Fe(OH)3.
Solution:
K=
[Fe 2+ ]
[H + ]3 [e " ]
pK = p[Fe] - 3pH - pe
!
-16.5 = 4 - 21 - pe Hence, pe = -0.5.
Predicting Stability of Species
Would (NO3)- be stable in this groundwater? pK = -104.6
for the reaction
NO3- + 6H+ + 5e- = 1/2N2(g) + 3H2O
K=
(PN2 (g) )1/2
-
[NO3 ][H+ ] 6[e " ]5
(Note: P = partial pressure)
pK = -(1/2)log(PN2) -p[NO3-] - 6pH - 5pe
!
Since, PN2 = 0.8, pe = -0.5 and pH = 7, we get
-104.6 = -(1/2)(-0.097) -p[NO3-] - 42.0 +2.5
p[NO3-] = 65.1
Page 5
Geochemistry
DM Sherman, University of Bristol
2010/2011
Example: Speciation of Fe
The half-reaction for the Fe redox couple
Fe3+ + e- = Fe2+
[Fe ] = 10
K=
[Fe ][e ]
2+
13.0
3+
"
so that
!
pK = pFe2+ - pFe3+ - pe
Example: Speciation of Fe
-13.0 = pFe2+ - pFe3+ - pe
When pe << pK, then pFe2+ = pFetot
so that pFe3+ = 13.0 - pe + pFetot
When pe >> pK, then pFe3+ = pFetot
so that pFe2+ = -13.0 + pe + pFetot
Page 6
Geochemistry
DM Sherman, University of Bristol
2010/2011
Example: Speciation of Fe
Range of pe of Aqueous Solutions
By convention, ΔG0 = 0.0 for the reaction
H+(aq) + e- = 1/2H2(g)
K=
(pH 2 )1/ 2
[H + ][e " ]
=1
The most reducing condition that is possible at the Earth’s
surface will have pH2 = 1 bar. Hence, pH + pe = 0 or
!
pH = -pe
Page 7
Geochemistry
DM Sherman, University of Bristol
2010/2011
Range of pe of Aqueous Solutions (cont.)
The most oxidizing condition under which an aqueous
solution can exist is buffered by the half-reaction
1/2H2O = e- +1/4O2 + H+
K=
(pO 2 )1/ 4 [H +][e"]
1/ 2
[H2O]
= 10 "20.75
Under the most oxidizing condition, pO2 =1. Since
[H2O] = 1, we have pH + pe = 20.75 or
!
pe = 20.75-pH
pe-pH Environments
Page 8
Geochemistry
DM Sherman, University of Bristol
2010/2011
The Mn2+-MnO2 couple
pK = -41.6 for the half-reaction
MnO2 + 4H+ + 2e= Mn2+ + 2H2O
Hence,
seawater
pMn - 4pH - 2pe = -41.6
or
pe = 20.8 + pMn/2 - 2pH
pe-pH Diagrams: The Fe-H2O system
A
B
Fe3+ + e- = Fe2+
pe = -pK
Fe(OH)3 + 3H+ = Fe3+ + 3H2O
pH = (-pK + pFe3+)/3
Fe(OH)3 + 3H+ + e-
C
*[Fe]total = 10-3 m
= Fe2+ + 3H2O
pe = -pK- 3pH + pFe2+
Page 9
Geochemistry
DM Sherman, University of Bristol
2010/2011
Cu-H2O System
Acidic solutions are necessary to
mobilize Cu for supergene enrichment.
Summary
• Be able to decompose an oxidation-reduction reaction
into half-reactions.
• Plot pe-pH diagrams..
• Know pe-pH environments in nature.
• Calculate pe from concentrations.
• Prediction of stable oxidation states.
Page 10