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Electrochemical Phenomena Eh and pE Approaches Redox Reactions pE-pH Diagrams Flooded Soils

Electrochemical Phenomena Eh and pE Approaches Redox Reactions pE-pH Diagrams Flooded Soils. Eh and pE Approaches aA + bB = cC + dD + ne K = (C) c (D) d / (A) a (B) b where (X) is the activity of X ΔG = ΔG o + RT ln [(C) c (D) d / (A) a (B) b ]

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Electrochemical Phenomena Eh and pE Approaches Redox Reactions pE-pH Diagrams Flooded Soils

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  1. Electrochemical Phenomena Eh and pE Approaches Redox Reactions pE-pH Diagrams Flooded Soils

  2. Eh and pE Approaches aA + bB = cC + dD + ne K = (C)c (D)d / (A)a (B)b where (X) is the activity of X ΔG = ΔGo + RT ln [(C)c (D)d / (A)a (B)b] If this reaction is an oxidation – reduction reaction, ΔG = nEhF and ΔGo = nEhoF where n is the number of e-s E is the potential of the reaction F is the Faraday constant  Eh = Eho + (RT / nF) ln [(C)c (D)d / (A)a (B)b]

  3. Alternatively, can write an equilibrium expression for the reduction half-reaction and take logs cC + dD + ne = aA + bB log K = log [(A)a (B)b / (C)c (D)d (e)n] = log [(A)a (B)b / (C)c (D)d] + npE where pE = - log (e) Large values of pE favor electron-poor (oxidized) chemical species Small values of pE favor electron-rich (reduced) chemical species

  4. Note that these are generally of the form mAox + nH+ + e = pAred + qH2O From which can be written K = (Ared)p (H2O)q / (Aox)m (H+)n (e) or log K = log [(Ared)p / (Aox)m] + npH + pE Eh = Eho + (RT / F) ln [(Aox)m (H+)n / (Ared)p (H2O)q] Eh = Eho + 0.059 log [(Aox)m / (Ared)p] – 0.059n pH

  5. Oxic ~ pE > 7 at pH = 7 Suboxic ~ 2 < pE < 7 at pH = 7 Anoxic ~ pE < 2 at pH = 7 Microbial limits and observed soil limits to pE – pH domain

  6. Above pE = 5 to pE = 11, O2 is consumed by aerobic respiration At ~ pE = 8, NO3- is reduced oxic to anoxic At ~ pE = 7, Mn4+ is reduced oxic to anoxic At ~ pE = 5, Fe3+ is reduced suboxic to anoxic At ~ pE = 0, SO42- is reduced anoxic Microbial activity forces depletion of O2, NO3- and so forth in this sequence and enrichment in es. Easy to see reduced activity of O2 reducers, NO3- reducers as concentration of the e acceptor decreases with decreasing pE. However, the activity of anaerobic organisms is also reduced by high values of pE.

  7. Redox Reactions

  8. Reduction half-reactions are coupled with oxidation half-reactions as with 1/24 C6H12O6 + 1/4 H2O = 1/4 CO2 + H+ + e 1/8 NO3- + 5/4 H+ + e = 1/8 NH4+ + 3/8 H2O

  9. Example use of Table 6.2 mAox + nH+ + e = pAred + qH2O From which can be written K = (Ared)p (H2O)q / (Aox)m (H+)n (e) or log K = log [(Ared)p / (Aox)m] + npH + pE • Can calculate ratio of a redox pair, pH or pE, given any two of these variables

  10. 1/8 SO42- + 9/8 H+ + e- = 1/8 SH- +1/2 H2O K = (SH-)1/8 (H2O)1/2 / (SO42-)1/8 (H+)9/8 (e-) log K = 1/8 log [(SH-) / (SO42-)] + 9/8 pH + pE pE = log K – 9/8 pH + 1/8 log [(SO42-) / (SH-)] Given log K = 4.3, pH = 7 and (SO42-) = (SH-), what is pE? pE = 4.8 – (9/8)7 = -3.6 anoxic For pH = 7 and pE = 2, what is the activity of SH- if (SO42-) = 0.001? log (SH-) = 8 (log K – 9/8 pH - pE) + log (SO42-) = -47.6 (SH-) = 10-47.6What volume of water contains one SH- ion?

  11. Note that although reduction or oxidation may be thermodynamically favorable, equilibrium may not exist. Often these redox reactions are slow. However, the activity of microoganisms accelerates (catalyzes) these reactions so that equilibrium is closely approached. Given the following two reactions, show that Fe3+ (aq) and S2 - (aq) are unstable equilibrium species in soil solutions. Fe 3+ (aq) + e (aq) = Fe 2+ (aq) log K = 13.0 S2- (aq) + H+ (aq) = HS-(aq) log K = 13.9 This is problem 1.

  12. K = (Fe2+) / (Fe3+)(e) 13.0 = log [(Fe2+) / (Fe3+)] + pE pE = 13 + log [(Fe3+) / (Fe2+)] which for (Fe3+) > (Fe2+) is not seen K = (HS-) / (S2-)(H+) 13.92 = log [(HS-) / (S2-)] + pH pH = 13.92 + log [(S2-) / (HS-)] which for (S2-) > (HS-) is not seen

  13. pE – pH Diagrams Based on rearrangement of log K = log [(Ared)p / (Aox)m] + npH + pE pE = log K - pH - log [(Ared)p / (Aox)m]

  14. For example, let’s determine under what conditions of pE and pH that Mn2+(aq) is favored respect to MnO2(s) or MnCO3 and under what conditions one or the other of these minerals is favored. 1/2MnO2 + 2H+ + e = 1/2Mn2+ + H2O log K1 = 20.7 1/2MnO2 + 1/2CO2 + H+ + e = 1/2MnCO3 + 1/2H2O log K2 = 16.3 Combine the above to give MnCO3 – Mn2+ equation 1/2MnCO3 + H+ = 1/2Mn2+ + 1/2CO2 + 1/2H2O log K3 = 4.4

  15. From which one writes, assuming unit activity for solid phases and H2O, log K1 = 20.7 = 2pH + pE + 1/2log(Mn2+) log K2 = 16.3 = pH + pE – 1/2log(PCO2) log K3 = 4.4 = 1/2 log(Mn2+) + 1/2log(PCO2) + pH Set (Mn2+) = 10-6 and PCO2 = 10-2 pE = 23.2 - 2pH pE = 15.3 – pH pH = 8.4

  16. General procedure for constructing pE – pH diagrams Choose sets of redox pairs (reactions along with log K values) Assume unit activities for solid phases and water Assume set values for solution activities (other than H+ and e, of course)

  17. 6. ½ SeO42- + H+ + e = ½ SeO32- + ½ H2O log K = 14.5 ½ SeO32- + ½ H2O = ½ SeO42- + H+ + e ½ MnO2 + 2H+ + e = ½ Mn2+ + H2O log K = 20.7 ½ SeO32- + ½ MnO2 + H+ = ½ SeO42- + ½ Mn2++ ½ H2O log K = 6.2 6.2 = ½ log[(SeO42-) / (SeO32-)] + ½ log [(Mn2+) / (H+)2] 20.7 = ½ log (Mn2+) – log (H+) + pH + pE ½ log[(Mn2+) / (H+)2] = 20.7 – pH – pE 6.2 = ½ log[(SeO42-) / (SeO32-)] + 20.7 – pH - pE

  18. 6.2 = ½ log[(SeO42-) / (SeO32-)] + 20.7 – pH – pE ½ log[(SeO42-) / (SeO32-)] = -14.5 + pH + pE pH + pE = 14.5 for (SeO42-) = (SeO32-) and pH + pE > 14.5 for (SeO42-) > (SeO32-)

  19. Flooded Soils Eh = 0.059pE

  20. 9. Fe(OH)3 + 3H+ + e = Fe2+ + 3H2O log K = 16.4 log K = log(Fe2+) + 3pH + pE pE = log K – log(Fe2+) – 3pH pE = 16.4 + 7.0 – 3pH = 23.4 – 3pH Fe2+ + CO2 + H2O = FeCO3 + 2H+ log K = -7.5 log K = -2pH – log (PCO2) – log (Fe2+) -7.5 = -2pH + 2.0 + 7.0 pH = 8.25

  21. Fe(OH)3 + 3H+ + e = Fe2+ + 3H2O log K = 16.4 Fe2+ + CO2 + H2O = FeCO3 + 2H+ log K = -7.5 Fe(OH)3 + H+ + CO2 + e = FeCO3 + 2H2O log K = 8.9 log K = pH – log (PCO2) + pE pE = 8.9 – 2.0 – pH = 6.9 - pH

  22. mAox + nH+ + e = pAred + qH2O K = (Ared)p(H2O)q / (Aox)m(H+)n(e) log K = log [(Ared)p(H2O)q / (Aox)m(H+)n] + pE = log K* + pE Eh for half-reaction measured with respect to the standard H2 electrode Pt, H2 / H+ // Ared / Aox, Pt which gives the overall reaction ½ H2 = H+ + e mAox + nH+ + e = pAred + qH2O ½ H2 + mAox + nH+ = H+ + pAred + qH2O

  23. For which K* = {(Ared)p(H2O)q / (Aox)m(H+)n} {(H+) / (H2)1/2} And according to E = Eo - (RT / nF) ln K* E = Eoreduction + Eooxidation – (RT / nF) ln K* And since Eooxidation = 0 and (H+) = (H2) = 1 Eh = Eoreduction – (RT / F) ln {(Ared)p(H2O)q / (Aox)m(H+)n} Eh = Eoreduction – (RT / F) ln K* Eh = Eoreduction – (2.303RT / F) log K*

  24. Substituting, log K = log K* + pE Eh = Eoreduction – (2.303RT / F) {log K - pE} At equilibrium Eoreduction = -(2.303 RT / F) log K so, Eh = (2.303RT / F) pE which for standard conditions becomes Eh = 0.0592 pE volts Eh = 59.2 pE millivolts

  25. Do problems 7 and 11.

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