Falkowski et al, 2008

1. Falkowski et al, 2008 The Microbial Engines That Drive Earth’s Biogeochemical Cycles

2. Redox Chemistry in the Sea – the major driver of biogeochemical cycles • Chemical reactions that involve transfer of electrons are called RedOx reactions (for reduction-oxidation). • Redox active chemicals will spontaneously transfer electrons in order to achieve thermodynamic equilibrium (lowest free energy state). Equilibrium chemistry applies. Fe3+ + e-  Fe2+

3. Oxidation = loss of electrons Reduction = gain of electrons A chemical that loses elections undergoes oxidation and it oxidation number (valence state) increases. Conversely, a chemical that gains electrons undergoes reduction and its oxidation number (valence state) decreases. Oxidation # Oxidation # Reduction Oxidation

4. Oxidation states of elements Redox reactions are chemical reactions which involve the transfer of electrons, or more formally, a change in the oxidation state (or number) of the reactant which occurs as a result of e- transfer. Fe3+ + e- Fe2+ Fe(III)ox Fe(II)red

5. Some elements are Redox Active in the environment and some are not.Refer to handout table for key redox-active elements and their important redox states • Examples of elements without appreciable redox chemistry in the environment include: • Chlorine - nearly all Cl as a reduced form in Cl- in seawater • Major cations in seawater (Na+, K+, Mg2+ and Ca2+) (these elements are already oxidized relative to their native metallic forms)

6. Key redox-active elements and their valence states and forms in nature.

7. Rules for assigning oxidation states • The simplest things to remember are that: • 1) Any element in its native state will have an oxidation number of zero. i.e. O2, N2, S, Fe etc. (see the handout table of key elements and their important oxidation states and forms). • 2) In most other cases the element oxygen (O) is assigned the oxidation state of -2 and H = +1.There are exceptions to this, but these are not very common. Hydrogen peroxide, H2O2 is perhaps the most important of the exceptions, in which the O has an oxidation number of -1. • 3) The sum of the oxidation numbers in a molecule must equal the charge on the molecule. • Note: the oxidation state of elements is often represented by roman numerals (i.e. -II or +IV). This is to distinguish the oxidation state from the ionic charge on the molecule. For the purposes of calculation, you can use arabic numbers.

8. Thermodynamic equilibrium principles apply to the movement of electrons. When chemicals have electronic configurations which are out of equilibrium, relative to another chemical, they will spontaneously react together, transferring electrons to attain equilibrium – or the lowest possible state of free energy. Zn(s) + Cu2+ + SO42- <=> Zn2+ + Cu(s) + SO42- What has happened here?

9. Two redox active chemicals, at non-equilibrium concentrations, will have an electrical potential between them (i.e. a potential to transfer e-). The electrical potential (E) of the system is called Ecell which is the sum of all half reactions (oxidation and reduction). Ecell is the electrical potential between chemicals – in Volts The free energy change is related to the cell potential Where n= # moles of electrons transferred and F = 23.062 kcal/(V*mol electrons transferred) (Faraday constant) ΔG = -nFEcell

10. We can write the reaction for oxidation of zinc as follows: Zn(s) <=> Zn2+(aq) + 2e-(example of a half-reaction) where zinc metal (solid) loses 2 electrons to become the zinc ion. Because electrons cannot exist in a free state, this reaction would not occur if there was nothing to accept the electrons. In order for something to become oxidized, something else must become reduced - hence Red-Ox chemistry

11. Half cell potentials cannot be measured directly because they must occur in pairs. A standard half reaction is used is for comparison to all other half reactions. The standard half reaction used is: 2 H+ + 2 e- <=> H2 (the reduction of H+ to hydrogen) Standard Hydrogen Electrode The half cell potential of this hydrogen reduction is set to zero, by convention. It is said to have an Eho of zero. The superscript o designates standard conditions (1 molal concentrations, 1 atm pressure etc). Molal activity

12. Standard Hydrogen Electrode For the reaction: Zn2+(aq) + H2(g) <=> Zn(s) + 2 H+(aq) Under standard conditions Ecell = -0.76 Volts therefore the Eho of the half reaction: Zn2+(aq) + 2e-<=> Zn(s) is -0.76 V. ( Eho = Ecell relative to the hydrogen half cell potential). If the reaction is written in reverse, as an oxidation, the Eho is +0.76 V. (Eho values are usually tabulated for the reductions)

13. Consider again the chemical reaction between zinc metal (Zn(s)) and a solution of copper sulfate: Zn(s) + Cu2+ + SO42- <=> Zn2+ + Cu(s) + SO42- Comparing two half cell reactions we have the Cu2+ + 2 e- = Cu(s) Eho = +0.34 Zn2+ + 2 e- = Zn(s) Eho = -0.76 The reaction will occur spontaneously, if there is a difference in Eh between species From Table 7.1 in Libes The half reaction with the highest reduction potential will undergo reduction and the other half reaction will be an oxidation and will supply the electron(s). The potential of the oxidation reaction is reversed to +0.76V and the cell potential is therefore: Ecell = 0.34 + 0.76 = 1.10 Volts

14. Another way of expressing the electrical potential between two chemicals is by treating the electrons as reactants and calculating the “activity” of the electrons. This is often represented by pe-(-Log{e-}). Fe3++ e-  Fe2+ Here you can see that the value of {e-} will be inversely related to Keq Take (–Log) to get pe- which is directly related to Log Keq pe- is also directly related to Eh peo = EhoF/2.3RT

15. The peo column in Table 7.1 from Libes is messed up. The numbers in the column are “upside down” (they should be in reverse order; numbers at the bottom should be at the top). For a 1 electron reduction, Log K = peo. For a 2 electron reduction peo = ½ Log K, etc. From Libes Chap 7 (2nded) Correct peo Reduced form 31 5.7 23 3.7 21.8 2.7 20.8 2.4 17.1 0 -4.3 13.5 -4.75 13 8.8 A high peo means low electron activity (because peo= -Log{e-}), so rxns with high peo have a tendency to accept rather than donate e-. Those same rxns have high Log K indicating the equilibrium favors reduced form. -7.45 -13 -39.7 -46

16. Relating Red-Ox chemistry to thermodynamic equilibrium The free energy change in a chemical system governs whether the reaction is spontaneous and how strong the tendency is to proceed. For any reaction: G = Gproducts -Greactants The free energy change for the overall reaction is equal to the free energy of the products minus that of the reactants. The free energy change predicts the tendency of the reaction to proceed. If free energy of products is less than reactants, then rxn is spontaneous

17. For a redox reaction, the general free energy equation applies: G = Go + RT ln [(C)c(D)d]/(A)a(B)b] At equilibrium, ΔG = 0 and [(C)c(D)d]/(A)a(B)b] = Keq 0= Go + RT ln Keq Go = - RT ln Keq Which is the same as: Go = -2.303RT log Keq Where Eocell = Eh01 – Eho2 and Eho1 is the reduction reaction ΔGo = -nFEocell Go = 2.303 nRT (pe-o2 – pe-o1) Where peo-1 is the reduction reaction and peo-2 is the oxidation reaction Go = ƩGfoproducts – ƩGforeactants

18. Thermodynamics and redox chemistry The reaction with the greatest tendency to proceed spontaneously will be the one with the largest equilibrium constant or most negative ΔG value. For redox reactions, this is achieved by pairing the oxidizing agent with the largest Eh (or peo) to the reducing agent with the smallest Eh(or peo). In seawater these chemicals are most often O2 and organic matter (reduced carbon)

19. The superscript –naught (o) means standard conditions (1 molal activities, temperature of 0 oC, 100 kPa of pressure) The subscript – (water) represents special case of standard conditions but with pH at 7.0 and temperature of 25 oC (typical of natural water rxns).

20. pe and Eh are directly related to one another, and both to ΔG. For reactants with: Eh 1 Eh 1 ΔG small Eh 2 pεo =Log K pεo =Log K ΔG large Eh 2 Comparison of pe and Eh scales at 25 oC

21. Given a world with 21% O2 in the atmosphere, at thermodynamic equilibrium we would expect nearly all: • C to be found as CO2 • N as NO3- • S as SO42- • Fe as FeOOH • Mn as MnO2 • But in fact we find significant amounts of organic matter (reduced carbon -with -C-C- bonds) and NH4+, N2, and CH4 • Some reduced forms of S, Fe and Mn also exist (i.e. R-SH, Fe2+, Mn2+), especially in biological systems Thus, the world is in disequilibrium!

22. The large amount of “unstable” reduced compounds in nature results mainly from photosynthesis, which takes advantage of light energy to drive otherwise thermodynamically unfavorable reactions.(geochemical energy does so in certain places) Photosynthesis CO2 + H2O <=> CH2O + O2 The DGo for this reaction as written is +29.9 kcal/mol, so it is not spontaneous. Energy has to be put in to drive this reaction. The energy can come from the sun, or from chemical oxidation of other matter.

23. Organic matter does not react spontaneously with O2 under most circumstances (at least not on short time scales) because of kinetic factorsrelated to energy of activation. If you left a peanut butter sandwich in the air long enough there is the chance that it will explode in a puff of smoke - but not likely (activation energy too great). But what if you provide a spark?

24. Respiration with O2 is a perfect balance for photosynthesis: Photosynthesis CO2 + H2O <=> CH2O + O2 Respiration So, Why is there oxygen in the air? Preservation of organic carbon allows excess O2 to accumulate But, oxidation of all the organic matter in the current biosphere would lower atmospheric oxygen by only 1%. Therefore, a large amount of “reducing equivalents” must be buried. Most is as organic carbon in sediments, CH4 hydrates, and peat, but some is in the form of reduced sulfur (i.e. FeS2 -pyrite).

25. Life does not cease when oxygen disappears! Anaerobic respiration proceeds in the absence of oxygen and uses alternative electron acceptors. - The electron acceptor that yields the most energy when coupled with the oxidation of organic matter will be used in preference to all others. This generates a sequence in electron accepting processes that may be revealed in time or space as e- acceptors are depleted in turn (i.e. vertical profiles in sediments). The sequence in terms of energy yield is: O2 > NO3- > MnO2 > FeOOH > SO42- > CO2(see handout table)

26. Sequence of e-acceptors for the coupled oxidation of organic matter (OM is e- donor)

27. e-acceptors O2 - aerobic NO3- - denitrification MnO2 - Mn oxide FeO(OH) - Fe oxide SO42- - Sulfate reduction CO2 - Methanogenesis Organic matter Vertical segregation of electron accepting processes in sediments and water columns Interface The source of electron acceptors is typically from above. The thermodynamically most favorable electron acceptors become depleted at depth as they are used to oxidize organic matter. After depletion of one acceptor the next most favorable one is used, generating the vertical sequence at left. Depth

28. The concentration or supply rate of a given e- acceptor determines its importance for overall carbon oxidation in a given system • The concentration of O2 is relatively low, but it is diffusible • The concentration of nitrate is generally very low, but it can be produced in sediments provided oxygen is sufficient to sustain nitrification • The concentration of sulfate is very high and it can diffuse • The concentration of FeOOH is high, but it is a solid and so it cannot diffuse. The same can be said for MnO2. In the case of both Fe and Mn, the oxidized forms are insoluble whereas the reduced forms (Fe2+ and Mn2+) are highly soluble and diffusible.

29. The use of various electron acceptors in sediments results in sharp gradients of e- acceptors and reduced end-products SWI O2 NH4+ NO3- Mn2+ Fe2+ HS- SO42- CH4 Relative concentration of electron acceptors or reduced end-products in sediment pore waters Iron and manganese oxides are insoluble, thus they will be in the solid phase. The reduced Fe(II) and Mn(II) are soluble, hence they appear in the pore water. Both Mn(II) and Fe(II) form insoluble sulfides so they decrease with depth in the sulfide zone. Ammonium (NH4+) comes mainly from organic matter degradation – not from NO3- reduction!

30. Another view: Libes Chap 12 Example of hypothetical pore water distributions of selected redox species in a coastal sediment O2 would occur only in the very surface layer Note arbitrary depth scale – the depths over which these gradients occur will vary greatly, depending on several factors, including supply of labile organic matter, temperature (affects rates of respiration), sediment porosity, and availability of the different electron acceptors.

31. Oxidation of organic carbon (CH2O) with different electron acceptors (stoichiometry for 1 electron transfer)

32. Aerobes release NH4+, which is then oxidized by nitrifying prokaryotes Complete oxidation of “Redfield” ideal organic matter (C:N:P = 106:16:1) with different electron acceptors. Note that oxidation of 16 mole of ammonia level N to 16 moles of nitrate requires 32 moles of O2 Different mineral forms of MnO2 Different mineral forms of Fe2O3

33. e-acceptor Reduced product Electron accepting Process The Eoh’s of these redox couples is directly related to the Go values for coupling organic matter oxidation to the electron acceptors on the left. Aerobic respiration Denitrification Manganese reduction Nitrate reduction to ammonia Fe(II) Iron reduction Sulfate reduction CO2 reduction (Methanogenesis) Proton reduction Acetogenesis Modified from Canfield, Thamdrup & Kristensen. Aquatic Geomicrobiology, 2005

34. Microbially-mediated Red-Ox reactions All these reactions are for oxidation of organic matter coupled to reduction of listed electron acceptor All these reactions are for oxidation of the listed substrate with OXYGEN as the electron acceptor Chemo-autotrophic processes Does not occur  Should be H2O oxidation which produces O2 in photosynthesis Photo driven

35. Depth distributions of redox active chemical species in a portion of the water column of the Black Sea near the oxic-anoxic interface. The σt (density) values represent depth since density increases downward. From Konovalov et al, 2005

36. Aerobic respiration Use of molecular oxygen as electron acceptor in oxidation of organic matter

37. The steeper gradients in shallower sediments is due to more labile organic matter present in those compared to deep sea sediments From Canfield, Thamdrup & Kristensen. Aquatic Geomicrobiology, 2005 For sediments at these water column depths

38. Nitrate reduction (Denitrification) • Next most energetically favorable e- acceptor after O2. • Denitrification reduces nitrate (NO3-) to nitrogen gas (N2), with net production of small amounts of N2O • Denitrification removes biologically-available nitrogen from the ecosystem • Dentrification occurs in the marine water column in strong oxygen minimum zones, and possibly microzones • Denitrification in estuarine sediments can remove 50% or more of N inputs to estuaries • Oceanic (global) denitrification may control ocean primary production over long time scales i.e. glacial/interglacial

39. Metal oxide reduction (FeOOH and MnO2) • Oxidized metals Fe(III) and Mn(IV) are highly insoluble at pH of seawater and in the presence of O2 they form insoluble oxides (e.g. FeOOH and MnO2) • Metal oxides can be used as e- acceptors by bacteria, but these oxides also are chemically labile • Some of the most primitive of life forms among the Bacteria and Archaea are metal reducers, suggesting a role for metal reduction in early evolution. • Reduced end-products (Fe2+ and Mn2+) are highly soluble under anoxic conditions and therefore diffusible. They are subject to oxidation either chemically or biologically – especially when they reach zones where O2 is around. • Reduction/oxidation cycles of the most abundant metals (typically Fe and Mn) greatly influence the chemistry of other trace metals Iron oxide Manganese oxide

40. Dissimilatory Sulfate Reduction • Redfield organic matter • (CH2O)106 (NH3)16 (H3PO4)1 + 53 SO42- • --> 106 CO2 + 16 NH3 + 53 S2- + H3PO4 + 106 H2O • note that the sulfide and ammonia stay in reduced forms under anoxic conditions. • 2 moles of carbon are oxidized per mole of sulfate reduced. • No free intermediates of oxidation state between +6 and -2 are known to be released during sulfate reduction. • In contrast, many intermediates (So, SO3-, S2O3- etc) are released during oxidation of sulfide to sulfate.

41. Sulfate reduction is one of the most important biogeochemical processes responsible for oxidation of organic matter (due to high [SO42-) concentration in seawater; ~ 28 mM. • Responsible for ~50% of carbon oxidation in coastal marine sediments • Generates highly reactive sulfide (HS-) and contributes to alkalinity • Sulfide reacts with important metals especially Fe, forming insoluble metal sulfides, thereby greatly affecting metal chemistry • Dissimilatory sulfate reduction dominates the natural sulfur cycle in terms of mass flux. However, most of this is within aquatic systems - exchange of sulfur with atmosphere is primarily via organic sulfur (i.e. DMS, COS etc.)

42. Methanogenesis Biogenesis of methane occurs by two main pathways: CO2 + 4 H2  CH4 + 2 H2O Autotrophic methanogenesis H2 is produced during fermentation and other anaerobic processes CH3COOH  CH4 + CO2 Acetate fermentation (acetoclastic methanogenesis) But can also have Methylotrophic Methanogesis where methylated compounds such as methanol, methylamines and dimethylsulfide (DMS) are converted to CH4 and CO2.

43. Segregation of methane accumulation from zone of sulfate reduction. When sulfate is depleted, CH4 is produced (if enough organic matter is present to allow methanogenesis) CO2 is the electron acceptor for CH4 formation but it is generally not limiting in most anoxic waters) Chapter 13, Bianchi.

44. Various abiological chemical reactions involving electron acceptors can take place such as 2FeOOH + 3 H2S  2FeS + So + 4 H2O where iron oxides are reduced by H2S chemically. Similarly, sulfide can react with MnO2 or O2 and become oxidized. In fact, most of the sulfide generated in anoxic environments is reoxidized, either abiotically or biologically.

45. The electron Bully My pe-is so low that I guess I have no choice My Eh is higher than yours, so give me those electrons! e- e- e- e- e- e- e- e- Oxygen (O2) Hydrogen sulfide (H2S) Oxidation of H2S with O2 occurs spontaneously (abiotically) but is relatively slow, which allows microbes to enzymatically do the same reaction and harness the free energy released

46. e-acceptors Vertical segregation of electron accepting processes in sediments and water columns O2 - aerobic NO3- - denitrification MnO2 - Mn oxide FeO(OH) - Fe oxide SO42- - Sulfate reduction CO2 - Methanogenesis Organic matter Interface Less energy is gained from the organic matter as less favorable electron acceptors are used Depth Where did the energy go?

47. H2O N2/NH4 Mn(II) Fe(II) S2- CH4 H2 Free Energy Content (when coupled to O2 reduction) Low High Completing the biogeochemical cycles - oxidation of reduced end-products The reduced end products of respiration reactions (H2O, N2/NH4, Mn(II), Fe(II), S2-, CH4 and H2) contain “free energy” in amounts inverse to that of the yield from the electron accepting (organic carbon respiration) processes that produced them.

48. Just as respiration generates oxidized carbon and reduced inorganic chemicals as end-products (H2S, Fe2+, Mn2+, NH4+ etc), Chemoautotrophy completes the biogeochemical cycles and utilizes the energy in reduced chemicals for the fixation of inorganic carbon (and hence production of reduced carbon biomass for growth of microorganisms). Examples include: Sulfide oxidation H2S + 2O2 SO42- + 2H+ Ammonia Oxidation NH4+ + O2 NO2- NO2-+ O2  NO3- Methane oxidation CH4 + 4O2 CO2 + 2 H2O Iron and manganese oxidation (etc) Chemoautotrophic processes Can be coupled with CO2 fixation into biomass

49. Oxidation of reduced chemicals by molecular oxygen Organic matter ΔGow = -29.95 Reactions written for 1 electron transfers. The free energy change would be multiplied by # of electrons per mole of substrate oxidized

50. Anaerobic oxidation of ammonia Anammox – a recently discovered reaction in the nitrogen cycle. (it is a form of denitrification) 16NH4+ + 16 NO2- 16 N2 + 32 H2O • The nitrite comes from the denitrification pathway • Discovered only in the mid 1990’s! • Carried out by a unique group of bacteria within the Planctomyces • Major role in the ocean N cycle identified only in 2003! May account for 15-30% of N2 production • Occurs in sediments and anoxic water columns (e.g. Black Sea)