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 = ΔG o + RT ln [(C) c (D) d / (A) a (B) b ] If this reaction is an oxidation – reduction reaction, ΔG = nEhF and ΔG o = nEh o F where n is the number of e - s E is the potential of the reaction F is the Faraday constant  Eh = Eh o + (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

5. Note that these are generally of the form mA ox + nH + + e = pA red + qH 2 O From which can be written K = (A red ) p (H 2 O) q / (A ox ) m (H + ) n (e) or log K = log [(A red ) p / (A ox ) m ] + npH + pE Eh = Eh o + (RT / F) ln [(A ox ) m (H + ) n / (A red ) p (H 2 O) q ] Eh = Eh o + 0.059 log [(A ox ) m / (A red ) p ] – 0.059n pH

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

7. Above pE = 5 to pE = 11, O 2 is consumed by aerobic respiration At ~ pE = 8, NO 3 - is reducedoxic to anoxic At ~ pE = 7, Mn 4+ is reducedoxic to anoxic At ~ pE = 5, Fe 3+ is reducedsuboxic to anoxic At ~ pE = 0, SO 4 2- is reducedanoxic Microbial activity forces depletion of O 2, NO 3 - and so forth in this sequence and enrichment in es. Easy to see reduced activity of O 2 reducers, NO 3 - 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.

8. Redox Reactions

9. Reduction half-reactions are coupled with oxidation half-reactions as with 1/24 C 6 H 12 O 6 + 1/4 H 2 O = 1/4 CO 2 + H + + e 1/8 NO 3 - + 5/4 H + + e = 1/8 NH 4 + + 3/8 H 2 O

10. Example use of Table 6.2 mA ox + nH + + e = pA red + qH 2 O From which can be written K = (A red ) p (H 2 O) q / (A ox ) m (H + ) n (e) or log K = log [(A red ) p / (A ox ) m ] + npH + pE  Can calculate ratio of a redox pair, pH or pE, given any two of these variables

11. 1/8 SO 4 2- + 9/8 H + + e - = 1/8 SH - +1/2 H 2 O K = (SH - ) 1/8 (H 2 O) 1/2 / (SO 4 2- ) 1/8 (H + ) 9/8 (e - ) log K = 1/8 log [(SH - ) / (SO 4 2- )] + 9/8 pH + pE pE = log K – 9/8 pH + 1/8 log [(SO 4 2- ) / (SH - )] Given log K = 4.3, pH = 7 and (SO 4 2- ) = (SH - ), what is pE? pE = 4.8 – (9/8)7 = -3.6anoxic For pH = 7 and pE = 2, what is the activity of SH - if (SO 4 2- ) = 0.001? log (SH - ) = 8 (log K – 9/8 pH - pE) + log (SO 4 2- ) = -47.6 (SH - ) = 10 -47.6 What volume of water contains one SH - ion?

12. 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 Fe 3+ (aq) and S 2 - (aq) are unstable equilibrium species in soil solutions. Fe 3+ (aq) + e (aq) = Fe 2+ (aq) log K = 13.0 S 2- (aq) + H + (aq) = HS - (aq) log K = 13.9 This is problem 1.

13. K = (Fe 2+ ) / (Fe 3+ )(e) 13.0 = log [(Fe 2+ ) / (Fe 3+ )] + pE pE = 13 + log [(Fe 3+ ) / (Fe 2+ )] which for (Fe 3+ ) > (Fe 2+ ) is not seen K = (HS - ) / (S 2- )(H + ) 13.92 = log [(HS - ) / (S 2- )] + pH pH = 13.92 + log [(S 2- ) / (HS - )] which for (S 2- ) > (HS - ) is not seen

14. pE – pH Diagrams Based on rearrangement of log K = log [(A red ) p / (A ox ) m ] + npH + pE pE = log K - pH - log [(A red ) p / (A ox ) m ]

15. For example, let’s determine under what conditions of pE and pH that Mn 2+ (aq) is favored respect to MnO 2 (s) or MnCO 3 and under what conditions one or the other of these minerals is favored. 1/2MnO 2 + 2H + + e = 1/2Mn 2+ + H 2 Olog K 1 = 20.7 1/2MnO 2 + 1/2CO 2 + H + + e = 1/2MnCO 3 + 1/2H 2 Olog K 2 = 16.3 Combine the above to give MnCO 3 – Mn 2+ equation 1/2MnCO 3 + H + = 1/2Mn 2+ + 1/2CO 2 + 1/2H 2 Olog K 3 = 4.4

16. From which one writes, assuming unit activity for solid phases and H 2 O, log K 1 = 20.7 = 2pH + pE + 1/2log(Mn 2+ ) log K 2 = 16.3 = pH + pE – 1/2log(P CO2 ) log K 3 = 4.4 = 1/2 log(Mn 2+ ) + 1/2log(P CO2 ) + pH Set (Mn 2+ ) = 10 -6 and P CO2 = 10 -2 pE = 23.2 - 2pH pE = 15.3 – pH pH = 8.4

18. 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)

19. 6. ½ SeO 4 2- + H + + e = ½ SeO 3 2- + ½ H 2 O log K = 14.5 ½ SeO 3 2- + ½ H 2 O = ½ SeO 4 2- + H + + e ½ MnO 2 + 2H + + e = ½ Mn 2+ + H 2 Olog K = 20.7 ½ SeO 3 2- + ½ MnO 2 + H + = ½ SeO 4 2- + ½ Mn 2+ + ½ H 2 O log K = 6.2 6.2 = ½ log[(SeO 4 2- ) / (SeO 3 2- )] + ½ log [(Mn 2+ ) / (H + ) 2 ] 20.7 = ½ log (Mn 2+ ) – log (H + ) + pH + pE ½ log[(Mn 2+ ) / (H + ) 2 ] = 20.7 – pH – pE 6.2 = ½ log[(SeO 4 2- ) / (SeO 3 2- )] + 20.7 – pH - pE

20. 6.2 = ½ log[(SeO 4 2- ) / (SeO 3 2- )] + 20.7 – pH – pE ½ log[(SeO 4 2- ) / (SeO 3 2- )] = -14.5 + pH + pE pH + pE = 14.5 for (SeO 4 2- ) = (SeO 3 2- ) and pH + pE > 14.5 for (SeO 4 2- ) > (SeO 3 2- )

21. Flooded Soils Eh = 0.059pE

23. 9. Fe(OH) 3 + 3H + + e = Fe 2+ + 3H 2 Olog K = 16.4 log K = log(Fe 2+ ) + 3pH + pE pE = log K – log(Fe 2+ ) – 3pH pE = 16.4 + 7.0 – 3pH = 23.4 – 3pH Fe 2+ + CO 2 + H 2 O = FeCO 3 + 2H + log K = -7.5 log K = -2pH – log (P CO2 ) – log (Fe 2+ ) -7.5 = -2pH + 2.0 + 7.0 pH = 8.25

24. Fe(OH) 3 + 3H + + e = Fe 2+ + 3H 2 Olog K = 16.4 Fe 2+ + CO 2 + H 2 O = FeCO 3 + 2H + log K = -7.5 Fe(OH) 3 + H + + CO 2 + e = FeCO 3 + 2H 2 Olog K = 8.9 log K = pH – log (P CO2 ) + pE pE = 8.9 – 2.0 – pH = 6.9 - pH

26. mA ox + nH + + e = pA red + qH 2 O K = (A red ) p (H 2 O) q / (A ox ) m (H + ) n (e) log K = log [(A red ) p (H 2 O) q / (A ox ) m (H + ) n ] + pE = log K* + pE Eh for half-reaction measured with respect to the standard H 2 electrode Pt, H 2 / H + // A red / A ox, Pt which gives the overall reaction ½ H 2 = H + + e mA ox + nH + + e= pA red + qH 2 O ½ H 2 + mA ox + nH + = H + + pA red + qH 2 O

27. For which K* = {(A red ) p (H 2 O) q / (A ox ) m (H + ) n } {(H + ) / (H 2 ) 1/2 } And according to E = E o - (RT / nF) ln K* E = E o reduction + E o oxidation – (RT / nF) ln K* And since E o oxidation = 0 and (H + ) = (H 2 ) = 1 Eh = E o reduction – (RT / F) ln {(A red ) p (H 2 O) q / (A ox ) m (H + ) n } Eh = E o reduction – (RT / F) ln K* Eh = E o reduction – (2.303RT / F) log K*

28. Substituting, log K = log K* + pE Eh = E o reduction – (2.303RT / F) {log K - pE} At equilibrium E o reduction = -(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 pEmillivolts

29. Do problems 7 and 11.