Lecture 15 The Redox Sequence

Oxidation StateHalf-ReactionsBalanced Oxidation-Reduction reactionsPredicted Sequence of Redox ReactionsTracers for these reactions

read Emerson and Hedges Section 3.5 and Chapter 12

The organic carbon that reaches the sedimentsdrives sedimentary diagenesis. This is 2% of B (i.e., f = 0.02 See Broecker (1971) and Lecture 5)

Oxidation StatesElement Oxidation State SpeciesNitrogen N (+V) NO3

-

N (+III) NO2-

N (O) N2

N (-III) NH3, NH4+

Sulfur S (+VI) SO42-

S (+II) S2O32-

S (O) SS(-II) H2S, HS-, S2-

Iron Fe (+III) Fe3+

Fe (+II) Fe2+

Manganese Mn (+VI) MnO42-

Mn (+IV) MnO2 (s)Mn (+III) MnOOH (s)Mn (+II) Mn2+

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 / 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 ↔ CO2 + H2O

e- donor e- acceptor e-acceptor e-donor

Why is organic matter an electron donor?

photosynthesis

Z-scheme for photosynthetic electron transport

to Calvin cycleand carbohydrate formation

Ferredoxin

energy from sun converted to C-C, energy rich, chemical bonds

photooxidationof water

ADP→ATP

ADP→ATP

Falkowski and Raven (2007)

Ene

rgy

Sca

le

e- from water

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-reactionsRedox reactions are written as half-reactions which are in the form of reductions

Ox + ne- = Red; Gr 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

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

ΔGr° = -2.3RTlogK

Redox Half Reactions

written as reductantsin terms of 1 e-

Balanced Redox ReactionsA 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.

Electron-Free Energy Diagram

Photosynthesis

Energy Scalee- acceptors

e- donors

Oxidation-Reduction reaction log K log KwAerobic Respiration1/4CH2O + 1/4O2 = 1/4H2O + 1/4CO2(g) 20.95 20.95

Denitrification1/4CH2O + 1/5NO3 + 1/5H+ = 1/4CO2(g) + 1/10N2(g) +7/20H2O

21.25 19.85Manganese Reduction1/4CH2O + 1/2MnO2(s) + H+ = 1/4CO2(g) + 1/2Mn2+ + 3/4H2O 21.0 17.0Iron Reduction1/4CH2O + Fe(OH)3(s) + 2H+ = 1/4CO2(g) + Fe2+ + 11/4 H2O

16.20 8.2Sulfate Reduction1/4CH2O + 1/8SO4

2- + 1/8H+ = 1/4CO2(g) + 1/8HS- + 1/4H2O 5.33

3.7Methane Fermentation1/4CH2O = 1/8CO2(g) + 1/8CH4 3.06 3.06

Tracers are circled

Free energy availableGr° = - 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)

Photosynthesis106CO2 + 16 NO3

- + HPO42- + 18H+ 122 H2O → “OM” + 138 O2

RespirationAerobic Respiration138 O2 + “OM” + 18 HCO3

- → 124 CO2 + 16 NO3- + HPO4

2- + 140 H2O

Denitrification94.4 NO3

- + “OM” → 13.6 CO2 + 92.4 HCO3- + 55.3 N2 + 84.8 H2O + HPO4

2-

Manganese Oxide Reduction236 MnO2 + “OM” + 364 CO2 + 104 H2O → 470 HCO3

- + 8N2 + 236 Mn2+ + HPO42-

Iron Oxide Reduction212 Fe2O3 + “OM” + 742 CO2 + 318 H2O → 848 HCO3

- + 16 NH3 + 424 Fe2+ + HPO42-

Sulfate Reduction53 SO4

2- + “OM” → 39 CO2 + 67 HCO3- + 16 NH4

+ + 53 HS- + 39 H2O + HPO42-

Methane Fermentation“OM” → 53 CO2 + 53 CH4 + 16 NH3 + HPO4

2- + 2H+

Indicator species are circled