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
Suboxic
Anoxic
~ pE > 7
~ 2 < pE < 7
~ pE < 2
at pH = 7
at pH = 7
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.
Redox Reactions
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-) = 10-47.6
What 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 = (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
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]
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
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.
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[(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-)
Flooded Soils
Eh = 0.059pE
+
2H 2 O  O 2 + 4H + 4e
o
Eh = E + (R T/n F)ln([oxidized]/[reduced])
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 = -2pH – log (PCO2) – log (Fe2+)
-7.5 = -2pH + 2.0 + 7.0
pH = 8.25
log K = -7.5
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
12
pE
6
F e (O H ) 3
0
-6
Fe
2+
FeC O 3
-1 2
4
6
8
pH
10
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
mAox + nH+ + e
= H+ + e
= pAred + qH2O
½ H2 + mAox + nH+
= H+ + pAred + qH2O
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*
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
Do problems 7 and 11.