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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


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]


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


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


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


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.



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


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


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?


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.


K = (Fe thermodynamically 2+) / (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


pE – pH Diagrams thermodynamically

Based on rearrangement of

log K = log [(Ared)p / (Aox)m] + npH + pE

pE = log K - pH - log [(Ared)p / (Aox)m]


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


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


General procedure for constructing pE – pH diagrams phases and H

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)


6. phases and H

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


6.2 = ½ log[(SeO phases and H42-) / (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-)


Flooded Soils phases and H

Eh = 0.059pE


9. phases and H

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


Fe(OH) phases and H3 + 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


mA phases and Hox + 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


For which phases and H

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*


Substituting, log K = log K* + pE phases and H

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


Do problems 7 and 11. phases and H


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