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Lecture 16 The Redox Reactions

Lecture 16 The Redox Reactions. Oxidation State Half-Reactions Balanced Oxidation-Reduction reactions Predicted Sequence of Redox Reactions Tracers for these reactions Distributions in Nature. The organic carbon that reaches the sediments drives sedimentary diagenesis . 2% of export

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Lecture 16 The Redox Reactions

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  1. Lecture 16 The Redox Reactions Oxidation State Half-Reactions Balanced Oxidation-Reduction reactions Predicted Sequence of Redox Reactions Tracers for these reactions Distributions in Nature

  2. The organic carbon that reaches the sediments drives sedimentary diagenesis. 2% of export is buried.

  3. 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 ElementOxidation StateSpecies Nitrogen N (+V) NO3- N (+III) NO2- N (O) N2 N (-III) NH3, NH4+ Sulfur S (+VI) SO42- S (+II) S2O32- S (O) S S(-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+ How to determine: Assign 0 = -II Assign H = +I Charge on species Calculate Oxidation State V IV III II I O -I -II -III reduced oxidized

  4. 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

  5. Why is organic matter an electron donor? photosynthesis

  6. Z-scheme for photosynthetic electron transport Falkowski and Raven (1997/2007) ADP→ATP Ferredoxin to Calvin cycle and carbohydrate formation Energy Scale ADP→ATP photooxidation of water H2O to O2 e- from water energy from sun converted to C-C, energy rich, chemical bonds inside chloroplasts of plants. Two photons absorbed.

  7. 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

  8. Redox half-reactions Redox reactions are written as half-reactions which are in the form of reductions Ox + ne- = Red; 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 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

  9. Redox Half Reactions written as reductants in terms of 1 e- Awkward fractions

  10. 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 caseRed2 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

  11. 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.

  12. Oxidation-Reduction reactionlog Klog 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 19.85 • Manganese Reduction • 1/4CH2O + 1/2MnO2(s) + H+ = 1/4CO2(g) + 1/2Mn2+ + 3/4H2O • 21.0 17.0 • Iron Reduction • 1/4CH2O + Fe(OH)3(s) + 2H+ = 1/4CO2(g) + Fe2+ + 11/4 H2O • 16.20 8.2 • Sulfate Reduction • 1/4CH2O + 1/8SO42- + 1/8H+ = 1/4CO2(g) + 1/8HS- + 1/4H2O • 5.33 3.7 • Methane Fermentation • 1/4CH2O = 1/8CO2(g) + 1/8CH4 3.06 3.06 Increasing energy available Free energy available DGr° = - 2.3 RT logK = -5.708 logK R = 8.314 J deg-1 mol-1 T = °K = 273 + °C Tracers are circled

  13. e- acceptors Energy Scale Electron-Free Energy Diagram Photosynthesis e- donors

  14. Organic Matter Degradation (using Redfield stoichiometry) “OM” = (CH2O)106(NH3)16(H3PO4) 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 + HPO42- Manganese Oxide Reduction 236 MnO2 + “OM” + 364 CO2 + 104 H2O → 470 HCO3- + 8N2 + 236 Mn2+ + HPO42- Iron Oxide Reduction 212 Fe2O3 + “OM” + 742 CO2 + 318 H2O → 848 HCO3- + 16 NH3 + 424 Fe2+ + HPO42- Sulfate Reduction 53 SO42- + “OM” → 39 CO2 + 67 HCO3- + 16 NH4+ + 53 HS- + 39 H2O + HPO42- Methane Fermentation “OM” → 53 CO2 + 53 CH4 + 16 NH3 + HPO42- + 2H+ Indicator species are circled

  15. Sampling stations overlaid on an image highlighting the area of the hydrocarbon intrusion in the deep Gulf of Mexico. The integrated dissolved oxygen anomaly was about 3.0 x 1010 moles O2. If this O2 anomaly was due to CH4 oxidation, how much CH4 would it account for? ¼ O2(g) + H+ + e- = ½ H2O log K = 20.75 1/8 CO2(g) + H+ + e- = 1/8 CH4 (g) + ¼ H2O log K = 2.87 Reverse 2nd reaction 1/8 CH4(g) + 1/4H2O = 1/8 CO2(g) + H+ + e- log K = -2.87 Add to first reaction 1/4O2(g) + 1/8 CH4(g) = 1/8 CO2(g) + ¼ H2O log K = 17.88 Multiply by x8 2O2(g) + CH4(g) = CO2(g) + 2 H2O log K = 143.04 So: 3.0 x 1010 mol O2 = 1.5 x 1010 mol CH4 (upper limit due to C2H6, C3H8) J D Kessler et al. Science 2011;331:312-315

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