lecture #48 redoxchemistry: log c vs pe diagrams · iron redox diagram analogous to log c vs ph...
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Lecture #48Redox Chemistry:
Log C vs pe Diagrams(Stumm & Morgan, Chapt.8 )
Benjamin; Chapter 9
David Reckhow CEE 680 #48 1
Updated: 29 April 2020 Print version
The Iron Mystery If:
The Fe+2/Fe+3 boundary is at a peo and pe(w) of 13.03 Oxygen saturated water should have a pe(w) of 13.6, but
Pankow says the effective pe(w) of surface water is more like 12.6
Will reduced iron spontaneously oxidize to ferric in surface waters?1. Yes2. No
David Reckhow CEE 680 #48 2
Then why is it so hard to keep reduced iron (ferrous) from oxidizing to the ferric form?
Redox and pH effects Often oxidation of metals results in a more hydrolyzed
species Acidity of oxidized species is higher, resulting in release of
protons Speciation changes and affects the overall reaction
A good example is the oxidation of ferrous iron to ferric Fe+3 + e- ↔ Fe+2
peo = 13.03 This is very close to the theoretical peo defined by saturated O2
in water (13.6), or the effective peo (e.g., 12.6) But this is deceptive, because Fe+3 isn’t the dominant species at
neutral pH
David Reckhow CEE 680 #48 3
Ferrous Hydroxides: α diagram
David Reckhow CEE 680 #48 4
pH0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
α
1e-10
1e-9
1e-8
1e-7
1e-6
1e-5
1e-4
1e-3
1e-2
1e-1
1e+0
α0
α1 α2
α3Fe+2
FeOH+
Fe(OH)2o
Fe(OH)3-
pH 7.0 Fe+2 dominates at neutral pH
Ferric Hydroxides: α diagram
David Reckhow CEE 680 #48 5
pH0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
α
1e-10
1e-9
1e-8
1e-7
1e-6
1e-5
1e-4
1e-3
1e-2
1e-1
1e+0
1e+1 α0 α1α2 α3
Fe+3
FeOH+2
Fe(OH)2-
Fe(OH)3o
α4
Fe(OH)4-
pH 7.0 Fe+3 is a minor
species at neutral pH
Iron redox diagram
Analogous to log C vs pH diagram
David Reckhow CEE 680 #47 6
Stumm & Morgan, 1996; Fig. 8.1, pg. 435Similar to: Benjamin, 2002Fig 9-3, pg.486
][][}{ 3
2
+
+− =
FeFeeK
𝐾𝐾 = 1013.03
HOCl and H2S example Neutral pH (~7.0)
0.1 mM HOCl, 1 mM Cl-
1 mM H2S and SO4-2
David Reckhow CEE 680 #48 8
Hypochlorite
5 x 10-4M ClT Where:
ClT = [HOCl] + [OCl-] + [Cl-]
David Reckhow CEE 680 #48 10
OHCleHHOCl 221
21
21
21 +↔++ −−+
1.255.05.0
5.0
10}{}{}{
}{ +−+
−
==eHHOCl
ClK
Kn
peo log1=
pHHOClCl
HHOClCl
Oxd
npp o
21
21
5.05.0
5.0
}{}{log1.25
}{}{}{log1.25
][][Relog1
−
−=
−=
−=
−
+
−
εε
Determining Equilibrium Concentrations Graphical solution analogous to acid/base
problems Create LogC vs pe diagram Determine location on graph using electron balance
Analogous to proton balance in acid/base problems
Example: HOCl and NaHS Reduced species: Cl- which is 2e- poor Oxidized species: SO4
-2 which is 8e- rich
David Reckhow CEE 680 #48 11
Electron Balance Oxidation Reduction
e- Balance:
David Reckhow CEE 680 #48 13
8[SO4-2] = 2[Cl-]
HS-HOCl
OHCleHHOCl 22 +↔++ −−+
OHHSeHSO 22
4 489 +↔++ −−+−
Constants
Reference reaction
Where {e-}=1, if all chemical species activities are also unity
David Reckhow CEE 680 #47 14
)(221 gHeH ↔+ −+
0.1}}{{
)}({ 5.02 == −+ eH
gHK
pE bounds for water I Oxygen and Hydrogen half cell reactions
David Reckhow CEE 680 #48 15
Stumm & Morgan, 1996; Fig. 8.2, pg. 437
εppHP
HeH
HeHeH
HK
H
g
g
g
22log
}log{5.0}log{}log{
}{}}{{
0.1}}{{
}{
2
)(2
5.0)(2
5.0)(2
−−=
=+
=
==
−+
−+
−+
Electron Balance for HOCl & HS-
Oxidation Reduction
e- Balance:
David Reckhow CEE 680 #48 17
4[O2(aq)] + 8[SO4-2] = 2[Cl-] + 2[H2(aq)]
HS-HOCl
OHCleHHOCl 22 +↔++ −−+
OHHSeHSO 22
4 489 +↔++ −−+−
H2O
OHeHO 22 244 ↔++ −+
2244 HeH ↔+ −+
To next lecture
David Reckhow CEE 680 #48 23
DAR
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