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 expor is buried. 98% of export is respired in the water column. Consumes O 2.

Oxidation States ElementOxidation StateSpecies NitrogenN (+V)NO 3 - N (+III)NO 2 - N (O)N 2 N (-III)NH 3, NH 4 + SulfurS (+VI)SO 4 2- S (+II)S 2 O 3 2- S (O)S  S(-II)H 2 S, HS -, S 2- IronFe (+III)Fe 3+ Fe (+II)Fe 2+ ManganeseMn (+VI)MnO 4 2- Mn (+IV)MnO 2 (s) Mn (+III)Mn 3+, MnOOH (s) Mn (+II)Mn 2+ 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 = -II Assign H = +I Charge on species Calculate Oxidation State V IV III II I O -I -II -III oxidized reduced more e -

Oxidation / Reduction Reactions One Reactant: is oxidized – it loses electrons = the e - donor (a reductant) is reduced – it gains electrons = the e - acceptor (an oxidant) Electrons are conserved! Example: CH 2 O + O 2 ↔ CO 2 + H 2 O e - donor e - acceptor e - acceptor e - donor (e - rich) Compare with acid / base reactions: HCl + H 2 O  Cl - + H 3 O + H + donor H + acceptor H + acceptor H + donor Reactions always coupled. No free e -. No free H +.

The driver of ocean redox chemistry is organic matter. Why is organic matter an electron donor? photosynthesis

Z-scheme for photosynthetic electron transport to Calvin cycle and carbohydrate formation Ferredoxin energy from sun converted to C-C, energy rich, chemical bonds inside chloroplasts of plants. Two photons absorbed. photooxidation of water H 2 O to O 2 ADP→ATP Falkowski and Raven (1997/2007) Energy Scale 2e - from water H 2 O O = -II O 2 O = 0

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 CO NADPH + 12 H ATP → C 6 H 12 O H 2 O + 12 NADP ADP + 18 P

Redox half-reactions Redox reactions are written as half-reactions which are in the form of reductions Ox + ne - = Red;  G  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 ΔG r ° = -2.3RTlogK

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

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 O 2 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 O 2 (g) + H + + e - = ½ H 2 OpE  = log K = (Rxn 18) 1/4 CO 2 (g) + H + + e - = 1/24 C 6 H 12 O 6 + 1/4 H 2 OpE  = We reverse reaction 18 (now it's log K = +0.20) and add it to reaction 1 to get: 1/4 O 2 (g) + 1/24 C 6 H 12 O 6 = 1/4 CO 2 (g) + 1/4 H 2 O log K = = Don ’ t like fractions: x 24 to get 6 O 2 ( g) + C 6 H 12 O 6 = 6 CO 2 (g) + 6 H 2 Olog K = x 24 =

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 O 2 → NO 3 → MnO 2 → Fe 2 O 3 → SO 4 2- (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 CO 2. Bacteria use the electron acceptors in the order of decreasing energy availability.

Electron-Free Energy Diagram Photosynthesis Energy Scale e - acceptors e - donors The spacing reflects how much energy is available.

Oxidation-Reduction reaction log Klog Kw Aerobic Respiration 1/4CH 2 O + 1/4O 2 = 1/4H 2 O + 1/4CO 2 (g) Denitrification 1/4CH 2 O + 1/5NO 3 + 1/5H + = 1/4CO 2 (g) + 1/10N 2 (g) +7/20H 2 O Manganese Reduction 1/4CH 2 O + 1/2MnO 2 (s) + H + = 1/4CO 2 (g) + 1/2Mn /4H 2 O Iron Reduction 1/4CH 2 O + Fe(OH) 3 (s) + 2H + = 1/4CO 2 (g) + Fe /4 H 2 O Sulfate Reduction 1/4CH 2 O + 1/8SO /8H + = 1/4CO 2 (g) + 1/8HS - + 1/4H 2 O Methane Fermentation 1/4CH 2 O = 1/8CO 2 (g) + 1/8CH Tracers are circled Free energy available  G r ° = RT logK = logK R = J deg -1 mol -1 T = °K = °C Increasing energy available

Organic Matter Degradation (using Redfield stoichiometry) “ OM ” = (CH 2 O) 106 (NH 3 ) 16 (H 3 PO 4 ) Photosynthesis 106CO NO HPO H H 2 O → “OM” O 2 Respiration Aerobic Respiration 138 O 2 + “OM” + 18 HCO 3 - → 124 CO NO HPO H 2 O Denitrification 94.4 NO “ OM ” → 13.6 CO HCO N H 2 O + HPO 4 2- Manganese Oxide Reduction 236 MnO 2 + “ OM ” CO H 2 O → 470 HCO N Mn 2+ + HPO 4 2- Iron Oxide Reduction 212 Fe 2 O 3 + “ OM ” CO H 2 O → 848 HCO NH Fe 2+ + HPO 4 2- Sulfate Reduction 53 SO “ OM ” → 39 CO HCO NH HS H 2 O + HPO 4 2- Methane Fermentation “ OM ” → 53 CO CH NH 3 + HPO H + Indicator species are circled