oxidation - reduction reactions

1. OXIDATION - REDUCTION REACTIONS Assigned reading: Sparks Chapter 8, pp 245 - 255 and Lindsay pp.23-30 Additional reading: : McBride 1.2f, 7.1a, b and e, 7.2; Sposito Chap. 6 and Essington Chap. 9.

3. The scale of potential were developed assuming: H+ + e- = 1/2 H2 Eo = 0.00V The ?Go for this reaction is zero.

4. Electrode potential Measurements Electrode Potentials Standard Electrode Potential (EHo) Example: Reduction of Fe3+ Fe3+ + e- = Fe2+ Eo = 0.771 V oxidize reduced

5. Defined vs. Standard Hydrogen Electrode H+ + e- = 1/2 H2 Eo = 0.000 Note: This also defines the ? Gf = 0 for H+ The overall reaction is: Fe3+ + 1/2H2 = Fe2+ + H+

6. Basic Cell for Measurement of Electrode Potentials

7. Half-cell Electrode Potentials (EH) Calculate the potential at other than standard conditions. Nernst equation: Note the negative sign

8. Also

9. At equilibrium with respect to SHE , EH = 0 and ?G = 0 and: and

10. Rearrange and

11. Then the relationship between ?Go and Eo is: ?Go = -nFEo also ?G = -nFEH

12. Half-Cell Reactions of Common Electron Acceptors in Soils Oxidized + mH+ + n electrons = reduced + H2O Nernst equation At 25 0C

13. In Volts

14. In millivolts

15. Eo values for reactions in soils (McBride, Table 7.1)

16. EH vs pH (Mcbride Figure 7.1)

17. Sparks Fig 8.1

18. Boundary reactions for electron acceptors that limit EH in aqueous systems Upper boundary redox in water 1/4 O2 + H+ + e- = 1/2 H2O Eo =1.23v Oxygen is an electron acceptor Lower boundary H+ + e- = 1/2 H2 Eo = 0.00v Hydrogen is an electron acceptor

19. Half Cell Reaction involving Organic Electron Donors e. g. Glucose Organic electron donors supply the energy to microbes in soils and sediments. The reaction below is a half-reaction that illustrates this process. In the soil the reaction goes from right to left. 1/4 CO2 + H+ + e- = 1/24 C6H12O6 + 1/4 H2O EHo = -0.014 Carbon compounds are abundant electron donors in soils

20. Reaction in airSucrose donates ectrons to oxygen 1/4 O2 + H+ + e- = 1/2 H2O Eo =1.23V 1/24 C6H12O6 + 1/4 H2O = 1/4 CO2 + H+ + e- + 0.014 V 1/24 C6H12O6 + 1/4 O2 = 1/4 CO2 + 1/4 H2O 1.24 V.

21. "Electron Activity" (pe) We can imagine that electrons have a measurable activity and use -log (e) = pe, instead of EH. For a half cell reaction. oxidized + mH+ + ne = reduced

22. "Electron Activity" (pe) (cont.) at 25�C Let pe = -log(e-) then

23. "Electron Activity" (pe) (cont.) log K is sometimes called pe0 ***Remember*** Electron activity is a convenient fiction, not a physical reality.

24. We also included protons in the Nernst equation

25. Comparison of the Nernst and pe equations

26. Equate the two equations

27. At 25 C

28. pe can easily be calculated from EH

29. Combining half cell reactions to describe a complete reaction. Question: What is the equilibrium constant for the oxidation of Fe2+ by O2? log K Fe2+ = e- + Fe3+ -13.04 H+ + 1/4O2 + e- = 1/2H2O 20.78 --------------------------------------------------------------- H+ + 1/4O2 + Fe2+ = 1/2H2O + Fe3+ 7.74 This equation says that O2 can readily oxidize Fe2+ at acid pH values

30. Combining half cell reactions (cont.) log K = pH -1/4log Po2 + log(Fe3+)/(Fe2+)]

31. Combining half cell reactions (cont.) Find the pH at which (Fe3+) = (Fe2+) at Po2 = 0.21 atm

32. Answer 7.7 = pH + .17 pH = 7.57

33. Combining half cell reactions with solubility reactions If (Fe3+) is controlled by the solubility of Fe(OH)3 in a soil then under reducing (flooded soil) conditions Fe2+ can be calculated as a function of pH and EH. Fe(OH)3 + 3H+ = Fe3+ + 3H2O log K = 2.70 (Lindsay) Predict Fe2+ as a function of pe in soil

34. Combining half cell reactions with solubility reactions(cont.) log K Fe3+ + e- = Fe2+ 13.04 Fe(OH)3 + 3H+ = Fe3+ + 3H2O 2.70 Soil ---------------------------------------------------------------------- Fe(OH)3 + e- + 3H+ = Fe2+ + 3H2O 15.74 EH = 0.93 V Similar to the value of pe in Sparks Table 1 McBride Table 1 is different (is for freshly ppt Fe(OH)3)

35. Combining half cell reactions with solubility reactions(cont.)

36. Combining half cell reactions with solubility reactions(cont.) If measured pH = 7.0 and (Fe2+) = 1.0 x 10-5 M, what is pe?

37. Answer pe = 15.74 + 5 - 3(7) pe = 15.74 - 16.0 pe = -0.26 and EH = -0.014

38. Combining half cell reactions with solubility reactions(cont.) Plot pe (or EH) vs. pH Example: Fe(OH)3 + 3H+ + e- = Fe2+ + 3H2O pe = log K - 3pH - log(Fe2+) at log (Fe2+) = -5 pe = 20.74 - 3pH

39. McBride Fig. 7.1

40. pe - pH plot for Fe

41. Consequences of flooding a soil 1. Oxygen is depleted. 2. Obligate and facultative anerobes utilize other electron acceptors. In order they are , nitrate, MnO2 , Fe(OH)3 and other oxides of FeIII, and sulfate. At very low EH methane is produced.

42. Compare reactions using conditional log K (log cK7.0 or Log Kw) at pH = 7.0 Conditional constants for pH dependent reactions can be recalculated for pH 7.0 [log(H+) = -7]. This allows for ranking of redox reactions at near neutral pH values. For reaction (1) log K = pe + pH - 1/4 log PO2

43. Reactions in Flooded SoilsSee Sparks Table 8.7 Note: Sparks assumes soluble FeII and MnII = 10-4M (Lindsay uses 10-5 Electron acceptors Electron donors pe at pH 7 logK logcK7.0 (1) 1/4 O2(g) + H+(aq) + e- = 1/2 H2O 20.8 13.6 (2) 1/4 NO3- + 5/4 H+ + e- = 1/8 N2O(g) + 5/8 H2O 18.9 9.6

44. (3) 1/2 MnO2 + 2H+ + e = 1/2 Mn2+ + H2O 20.7 8.8 (4) Fe(OH)3 + 3H+ + e- = Fe2+ + 3H2O 15.8 -1.2 (5) 1/8 SO42- + 5/4 H+ + e- = 1/8 H2S(aq) + 1/2 H2O 5.2 -3.5

45. (6) 1/8 CO2(g) + e- + H+ = 1/8 CH4(g) + 1/2 H2O 2.9 -4.1 (7) 1/4 CO2(g) + e- + H+ = 1/24 C6H12O6 + 1/4H2O -0.2 -7.2 (8) H+ + e- = 1/2 H2 0.0 -7.0

46. Oxygen consumption in by organic carbon, when O2 is the electron acceptor Oxidation of organic carbon in soils generates energy for microorganisms. The most energy is gained by using electron acceptors with highest log K (E0) . Example, sucrose Combine reaction (1) with reaction (7) 1/24 C6H12O6 + 1/4 O2(g) = 1/4 CO2(g) + 1/4 H2O log K = 21.0 oxidation ---> <------ photosynthesis

47. Can calculate ?Go fromlog K or E0 Energy units kJ/mole

48. Using nitrate as an electron acceptor Combine with reaction 7 and 2 1/24 C6H12O6(aq) + 1/4 NO3- + 1/4 H+ = 1/8 N2O(g) + 3/8 H2O log K = 19.1 Can calculate Gibbs free energy ?G = - 109.2 kJ/mole

49. When all electron acceptors are depleted then fermentation Combine with reaction 7 and 6 1/24 C6H12O6(aq) = 1/8 CH4(g) + 1/8 CO2(g) log K =3.1 eg. In methane digester or soil flooded for a long time. Low energy yield.

50. Review of consequences of flooding a soil With flooding O2 is used up because O2 diffusion in water is very slow. When O2 is depleted NO3- becomes the electron acceptor which yields the most energy, This followed followed by MnO2, Fe(OH)3, and then SO42-. Kinetic considerations can be important. Thus, Fe(OH)3 reduction can begin before all of the MnO2 is depleted. Because of pH dependence, the exact order of these reactions can vary slightly with pH,

51. McBride Figure 7.1

52. Chemical changes after flooding of a soil (Sposito chapter 6)

53. Changes in N chemistry with flooding (Sposito Chapter 6)

54. Changes in N chemistry with flooding (Sposito Chapter 6)

55. Determination of EH Measurements Use a bright Pt electrode and a reference electrode. Essington suggests the use of a Calomel (Hg2Cl2/Hg) electrode. In fact, most people use an AgCl/Ag electrode.

56. Reference electrode log K AgCls + e- = Ags + Cl- +3.75 Inside the electrode is an Ag plated wire with AgCl and a fixed KCl concentration. The KCl is generally in the range of 3 M to saturated. Potential is about 0.2 v. (Compare to standard)

57. Use reference electrode Measure E vs. the above electrode then add 0.2 V to get EH In practice standardize meter and reference electrode vs. a solution like quihydrone: quinone/hydroquinone = 1.0 (at pH 4.01) EH = 0.461v

58. Reference standardQuinhydrone reaction + 2H+ + 2e- ? + H2O quinone hydroquinone

59. What does a Pt electrode measure in soils? Measures EH only of electrode reactive reactions (electrons can be transferred to or from electrode). Fe3+ �> Fe2+ is electrode reactive. 1/4 O2 + H+ + e- = 1/2 H2O is not Can�t measure O2 in soils or waters using Pt electrodes Sucrose �>H2O and CO2 is not electrode reactive.

60. Electrode data allows for general statements about redox status McBride (Fig.7-6)

61. Range of measured pe and pH in soils

62. Reactions in flooded soils vs. measured EH (Fig. 7.5)

63. Compare to EH - pH diagrams McBride Fig. 7.1

64. pH Changes Following Flooding In low pH soils the pH increases to near neutrality because H+ is consumed during the reduction of most electron acceptors. Most notably: 3H+ + Fe(OH)3 + e- = Fe2+ + 3H2O In calcareous soils pH decreases because of the precipitation of carbonates at the high Pco2. Ca2+ + H2O + CO2 = CaCO3 + 2H+

65. pH Changes Following Flooding (cont.) Also Mn(II) and Fe(II) carbonates precipitate Siderite, FeCO3 Rhodochrosite, MnCO3, These minerals are important in controlling the pH in many flooded soils. These minerals both have a calcite type structure with very similar unit cell sizes. They form a solid solution series.

66. Precipitation of Siderite log K Fe2+ + CO32- = FeCO3 10.8 H2O + CO2 = H++ HCO3- - 7.81 HCO3- = H+ + CO32- -10.33 ------------------------------------------------------- Fe2+ + H2O + CO2 = 2H+ + FeCO3 - 7.34

67. FeCO3 (cont.) From earlier slide Fe(OH)3 + e- + 3H+ = Fe2+ + 3H2O +15.74 soil Fe2+ + H2O + CO2 = 2H+ + FeCO3 - 7.34 -------------------------------------------------------------------- Fe(OH)3 + e- + CO2 + H+ = FeCO3 + 2H2O 8.40

68. Example, FeCO3 (cont.) log K = pH + pe - log Pco2 Thus at equilibrium pe is a function of pH and Pco2, only. Set the PC02 then calc. pH At pH = 7.0 and Pco2 = 0.10 pe = log K - pH + log Pco2 pe = 8.40 - 7.0 - 1.0 pe = 0.4 EH = (0.059)( 0.4) = - 0.024 V

69. pe - pH plot for Fe

70. Sulfide precipitation in reduced soils Sulfate is reduced to S2- and can precipitate with Fe2+, Mn2+, Zn2+, Cd2+ etc. FeS is the predominant form of sulfide. Sulfide formation can reduce the bioavailablity of heavy metals.

71. Formation of sulfidic coastal swamps

72. Formation of acid sulfate soils by draining sulfidic swamps

73. Formation of surface oxic layer in flooded soils Slow O2 diffusion can result in an oxic surface. The thickness of the layer is result of the O2 diffusion rate and O2 consumption rate. In low OM mineral soils this layer may be > 1 cm In high OM soil it may be only a few mm thick.

74. Oxic surface in a rice paddy soil

75. N transformations in the surface of a rice paddy soil

76. Phosphate is generally released by flooding Flooding releases P by: 1. Raising pH of acid soils. 2. Lowering the pH of alkaline soils. 3. Reduction Fe(III) oxides that sorb P.

77. Reduction Fe(III) phosphate (strengite) Fe3+ + e- = Fe2+ 13.04 FePO4�2H2O + 2H+ = Fe3+ + H2PO4- - 6.85 --------------------------------------------------------------- FePO4�2H2O + e- + 2H+ = Fe2+ + H2PO43- 6.19

78. Redox reaction of strengite (cont.) pe = 6.2 - log(Fe2+) - log(H2PO4) - 2pH Let pH = 7.0, (Fe2+) = 10-5 M, and (H2PO43-) = 10-5 M pe = 6.2 + 5 + 5 -14 = 2.2 EH = 0.129v

79. Redox reaction of strengite (cont.) Comparison with previous computations shows strengite is reduced at a similar pe as Fe(OH)3 Reduction of iron oxides releases adsorbed P. Drainage and restoration of oxic conditions can tie up P. Following draining of rice soils, phosphate can be tied up causing P deficiency for rotation crops.

80. Short Summary The log K (E0) of half cell reactions can be used to calculate the log K of oxidation reduction reactions. EH can be converted to pe. The pe values do not represent real activities of electrons. Pt electrodes can be use to measure EH in flooded soils. The measured EH can only be approximately predicted from measured concentrations and half cell reactions.

81. In flooded soils consumption of O2 and other electron acceptors by oxidation of reduced C (organic C) lowers pe (EH) Flooding increases the pH of acid soils and decreases the pH of calcareous soils. At very low EH sulfides can precipitate. A thin layer on the surface of flooded soils can be oxic. Flooding causes the release of bound P.