Electrochemical Phenomena

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Electrochemical Phenomena Eh and pE
Approaches Redox Reactions pE-pH
Diagrams Flooded Soils
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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 ?G ?Go RT ln (C)c (D)d / (A)a
(B)b If this reaction is an oxidation
reduction reaction, ?G nEhF and ?Go
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
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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
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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
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Oxic pE gt 7 at pH 7 Suboxic 2 lt pE lt 7
at pH 7 Anoxic pE lt 2 at pH 7 Microbial
limits and observed soil limits to pE pH domain
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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
curtailed activity of O2 reducers, NO3- reducers
as concentration of the e acceptor decreases
with decreasing pE. However, the activity of
anaerobic organisms is also curtailed by high
values of pE.
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Redox Reactions
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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
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• 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

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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-) 10-47.6
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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
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K (Fe2) / (Fe3)(e) 13.0 log (Fe2) /
(Fe3) pE pE 13 log (Fe3) /
(Fe2) which for (Fe3) gt (Fe2) is not
seen K (HS-) / (S2-)(H) 13.92 log (HS-)
/ (S2-) pH pH 13.92 log (S2-) /
(HS-) which for (S2-) gt (HS-) is not seen
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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
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For example, lets 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 20.2 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
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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
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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)
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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
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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 gt 14.5 for (SeO42-) gt
(SeO32-)
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Flooded Soils
Eh 0.059pE
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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
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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
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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
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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
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Substituting, log K log K pE Eh
Eoreduction (2.303RT / F) log K - pE So that
since at equilibrium Eoreduction -(2.303 RT /
F) log K Eh (2.303RT / F) pE Which for
standard conditions becomes Eh 0.0592 pE
volts Eh 59.2 pE millivolts