PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Lecture 16 The Redox ReactionsOxidation StateHalf-ReactionsBalanced Oxidation-Reduction reactionsPredicted Sequence of Redox ReactionsTracers for these reactionsDistributions in NatureThe organic carbon that reaches the sedimentsdrives sedimentary diagenesis. 2% of exportis buried.Oxidation StatesElement Oxidation State SpeciesNitrogen N (+V) NO3-N (+III) NO2-N (O) N2N (-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+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.How to determine:Assign 0 = -IIAssign H = +ICharge on speciesCalculate Oxidation StateVIVIIIIIIO-I-II-IIIoxidizedreducedOxidation / Reduction ReactionsOne 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 + H2Oe- donor e- acceptor e-acceptor e-donorWhy is organic matter an electron donor?photosynthesisZ-scheme for photosynthetic electron transportto Calvin cycleand carbohydrate formationFerredoxinenergy from sun converted to C-C, energy rich, chemical bondsinside chloroplasts of plants. Two photons absorbed.photooxidationof water H2O to O2ADP→ATPADP→ATPFalkowski and Raven (1997/2007)Energy Scalee- from waterThe ATP produced is the energy used to make glucose in the Calvin/Bensen CycleThe 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 PiRedox half-reactionsRedox reactions are written as half-reactions which are in the form of reductions Ox + ne- = Red; GrKwhere 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-)nWe 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) ]orEh = 2.3 RT pE / FΔGr° = -2.3RTlogKRedox Half Reactions written as reductantsin terms of 1 e-Awkward fractionsBalanced Redox ReactionsA balanced reaction has an electron passed from an electron donor to an electron acceptor. Thus:Ox1 + Red2 = Red1 + Ox2In 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 glucoseThe 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 = -0.20We 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 get6 O2(g) + C6H12O6 = 6 CO2(g) + 6 H2O log K = 20.95 x 24 = 502.80Ideal 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.Oxidation-Reduction reaction log K log KwAerobic Respiration1/4CH2O + 1/4O2 = 1/4H2O + 1/4CO2(g) 20.95 20.95Denitrification1/4CH2O + 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 + 1/8SO42- + 1/8H+ = 1/4CO2(g) + 1/8HS- + 1/4H2O 5.33 3.7Methane Fermentation1/4CH2O = 1/8CO2(g) + 1/8CH4 3.06 3.06Tracers are circledFree energy availableGr° = - 2.3 RT logK = -5.708 logK R = 8.314 J deg-1 mol-1 T = °K = 273 + °CIncreasing energy availableElectron-Free Energy DiagramPhotosynthesisEnergy Scalee- acceptorse- donorsOrganic Matter Degradation (using Redfield stoichiometry)“OM” = (CH2O)106(NH3)16(H3PO4)Photosynthesis106CO2 + 16 NO3- + HPO42- + 18H+ 122 H2O → “OM” + 138 O2RespirationAerobic Respiration138 O2 + “OM” + 18 HCO3- → 124 CO2 + 16 NO3- + HPO42- + 140 H2ODenitrification94.4 NO3- + “OM” → 13.6 CO2 + 92.4 HCO3- + 55.3 N2 + 84.8 H2O + HPO42-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 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 circledSampling stations overlaid on an image highlighting the area of the hydrocarbon intrusion in the deep Gulf of Mexico.J D Kessler et al. Science 2011;331:312-315The integrated dissolved oxygen anomaly was about3.0 x 1010 moles O2. If this O2 anomaly was due to CH4 oxidation, how much CH4 wouldit account for?¼ O2(g) + H+ + e- = ½ H2O log K = 20.751/8 CO2(g) + H+ + e- = 1/8 CH4 (g) + ¼ H2O log K =