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Lecture 15 The Redox Sequence
Oxidation State
Half-Reactions
Balanced Oxidation-Reduction reactions
Predicted Sequence of Redox Reactions
Tracers for these reactions
read Emerson and Hedges Section 3.5 and Chapter 12
The organic carbon that reaches the sediments
drives sedimentary diagenesis.
This is 2% of B (i.e., f = 0.02 See Broecker (1971) and Lecture 5)
Many elements in the periodic table can exist in more than one oxidation state.
Oxidation states are indicated by Roman numerals (e.g. (+I), (-II), etc).
The oxidation state represents the "electron content" of an element which
can be expressed as the excess or deficiency of electrons relative to the
elemental state.
Oxidation States
Element
Nitrogen
Sulfur
Iron
Manganese
Oxidation State
N (+V)
N (+III)
N (O)
N (-III)
S (+VI)
S (+II)
S (O)
S(-II)
Fe (+III)
Fe (+II)
Mn (+VI)
Mn (+IV)
Mn (+III)
Mn (+II)
Species
NO3NO2N2
NH3, NH4+
SO42S2O32S
H2S, HS-, S2Fe3+
Fe2+
MnO42MnO2 (s)
MnOOH (s)
Mn2+
Oxidation / Reduction Reactions
One Reactant:
is oxidized – it loses electrons = the e- doner (a reductant)
is reduced – it gains electrons = the e- acceptor (an oxidant)
Example:
CH2O
+
O2
e- donor e- acceptor
↔
CO2
+
e-acceptor
H2O
e-donor
Why is organic matter an electron donor?
photosynthesis
Z-scheme for photosynthetic electron transport
Falkowski and Raven (2007)
ADP→ATP
Energy Scale
Ferredoxin
ADP→ATP
to Calvin cycle
and carbohydrate
formation
photooxidation
of water
e- from water
energy from sun converted to C-C, energy rich, chemical bonds
The ATP produced is the energy used to make glucose in the Calvin/Bensen Cycle
The sum of reactions in the Calvin cycle is the following:
6 CO2 + 12 NADPH + 12 H+ + 18 ATP → C6H12O6 + 6 H2O + 12 NADP+ + 18 ADP + 18 Pi
Redox half-reactions
Redox reactions are written as half-reactions which are in
Ox + ne- = Red;
the form of reductions
DGr
K
where the more oxidized form of an element is on the left and the reduced form is on the right. n
is the number of electrons transferred.
We can write an equilibrium constant for this reaction as we can any other reaction. Formally the
concentrations should be expressed as activities. Thus:
K = (Red) / (Ox)(e-)n
ΔGr° = -2.3RTlogK
We can also rearrange the equation to determine the activity of the electron for any redox couple:
(e-) = [ (Red) / K (Ox) ] 1/n
Electron activities are usually expressed on either the pE or Eh scales as shown below.
pE = - log (e-) = 1/n [logK - log (Red)/(Ox) ]
or
Eh = 2.3 RT pE / F
Redox Half Reactions
written as reductants
in terms of 1 e-
Balanced Redox Reactions
A balanced reaction has an electron passed from an electron donor to an electron acceptor.
Thus:
Ox1 + Red2 = Red1 + Ox2
In this case Red2 is the electron donor, passing electrons to Ox1 which is the electron acceptor.
Thus Red2 is oxidized to Ox2 and Ox1 is reduced to Red1.
The equilibrium constant for an oxidation-reduction reaction can be determined by combining
the constants from Table 1 as follows for O2 with glucose
The two half reactions (written as reductions in terms of one electron) with their appropriate
values of log K, are:
(Rxn 1) 1/4 O2(g) + H+ + e- = ½ H2O
pE = log K = 20.75
(Rxn 18) 1/4 CO2(g) + H+ + e- = 1/24 C6H12O6 + 1/4 H2O
pE =
-0.20
We reverse reaction 18 (now it's log K = +0.20) and add it to reaction 1 to get:
1/4 O2(g) + 1/24 C6H12O6 = 1/4 CO2(g) + 1/4 H2O
log K = 20.75 + 0.20 = 20.95
Don’t like fractions: x 24 to get
6 O2(g) + C6H12O6 = 6 CO2(g) + 6 H2O
log K = 20.95 x 24 = 502.80
Ideal Redox Sequence
There is an ideal sequence of redox reactions driven by e- rich organic matter that is
based on the energy available for the microbes that mediate the reactions.
In this sequence organic matter is combusted in order by
O2 → NO3 → MnO2 → Fe2O3 → SO42- (decreasing energy yield).
Most of these reactions have slow kinetics if not mediated by bacteria.
Bacteria mediate most of these reactions and get the energy for their life processes.
Because the energy of the sun is trapped in the C-C bonds, bacteria are indirectly
using sunlight when they combust natural organic matter to CO2. Bacteria use the
electron acceptors in the order of decreasing energy availability.
e- acceptors
Electron-Free Energy Diagram
Photosynthesis
e- donors
Energy Scale
Oxidation-Reduction reaction
log K
log Kw
Aerobic Respiration
1/4CH2O + 1/4O2 = 1/4H2O + 1/4CO2(g)
20.95
20.95
Denitrification
1/4CH2O + 1/5NO3 + 1/5H+ = 1/4CO2(g) + 1/10N2(g) +7/20H2O
21.25
Manganese Reduction
1/4CH2O + 1/2MnO2(s) + H+ = 1/4CO2(g) + 1/2Mn2+ + 3/4H2O
21.0
Iron Reduction
1/4CH2O + Fe(OH)3(s) + 2H+ = 1/4CO2(g) + Fe2+ + 11/4 H2O
16.20
Sulfate Reduction
1/4CH2O + 1/8SO42- + 1/8H+ = 1/4CO2(g) + 1/8HS- + 1/4H2O
5.33
Methane Fermentation
1/4CH2O = 1/8CO2(g) + 1/8CH4
3.06
Tracers are circled
19.85
17.0
8.2
3.7
3.06
Free energy available
DGr° = - 2.3 RT logK = -5.708 logK
R = 8.314 J deg-1 mol-1
T = °K = 273 + °C
Organic Matter Degradation (using Redfield stoichiometry)
“OM” = (CH2O)106(NH3)16(H3PO4)
Indicator species are circled
Photosynthesis
106CO2 + 16 NO3- + HPO42- + 18H+ 122 H2O → “OM” + 138 O2
Respiration
Aerobic Respiration
138 O2 + “OM” + 18 HCO3- → 124 CO2 + 16 NO3- + HPO42- + 140 H2O
Denitrification
94.4 NO3- + “OM” → 13.6 CO2 + 92.4 HCO3- + 55.3 N2 + 84.8 H2O + HPO42Manganese Oxide Reduction
236 MnO2 + “OM” + 364 CO2 + 104 H2O → 470 HCO3- + 8N2 + 236 Mn2+ + HPO42Iron Oxide Reduction
212 Fe2O3 + “OM” + 742 CO2 + 318 H2O → 848 HCO3- + 16 NH3 + 424 Fe2+ + HPO42Sulfate Reduction
53 SO42- + “OM” → 39 CO2 + 67 HCO3- + 16 NH4+ + 53 HS- + 39 H2O + HPO42Methane Fermentation
“OM” → 53 CO2 + 53 CH4 + 16 NH3 + HPO42- + 2H+