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Reduction-Oxidation (Redox) Reactions
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Chemical reactions in a geologic setting are usually
either…
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Proton exchange reactions, i.e.
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Acid dissociation
Metal complexation
Acid weathering of rocks
Silicate alteration reactions
Or…
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Electron exchange reactions, i.e.
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Oxidation-reduction chemical reactions
Metabolism (Life)
Redox Reactions
• Reduction: electrons are gained by an electron acceptor
(oxidizing agent)
• Oxidation: electrons are lost from an electron donor
(reducing agent)
• Oxidation and reduction reactions always occur in
balanced, symmetric reaction couples where something
is oxidized and something else is reduced… no excess
electrons are allowed.
Half-Reactions
•
In any given redox couple, the oxidation step and the reduction step
can each by separated into “half reactions”
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•
Only in a half-reaction are excess electrons allowed
It is understood that any half-reaction, if it actually occurs, is coupled
to some other half-reaction going the opposite direction…
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•
EX:
Reduction half reaction:
Oxidation half reaction:
O2 + 4H+ + 4e- = 2H2O
CH2O + H2O = CO2 + 4H+ + 4e_______________________________
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Result:
CH2O + O2 = CO2 + H2O
Add one reaction to the other, and the electrons must cancel out
The electron donor (“CH2O”) gives up electrons…
And the electron acceptor (O2) gains those electrons.
Half Reactions
• Example: dissolution of iron metal in a strong acid
Fe0(s) = Fe3+ + 3e2H+ + 2e- = H2
•
Must first multiply each reaction by a common denominator to get
equal numbers of electrons on both sides
2Fe0(s) = 2Fe3+ + 6e6H+ + 6e- = 3H2
2Fe0(s) + 6H+ = 2Fe3+ + 3H2
or
Fe0(s) + 3H+ = Fe3+ + 1.5H2
Half Reactions
• Example: catalytic reduction of sulfate by plutonium(II)
SO42- + 10H+ + 8e- = H2S + 4H2O
Pu2+ = Pu6+ + 4e•
Must first multiply each reaction by a common denominator to get
equal numbers of electrons on both sides
SO42- + 10H+ + 8e- = H2S + 4H2O
2Pu2+ = 2Pu6+ + 8eSO42- + 2Pu2+ + 10H+ = H2S + 4H2O + 2Pu6+
Half Reactions
• Example: oxidation of trichloroethene (TCE) by ozone
4H2O + C2H3Cl3(aq) = 2CO2 + 3Cl- + 11H+ + 8eO3(aq) + 2H+ + 2e- = O2(aq) + H2O
•
Must first multiply each reaction by a common denominator to get equal
numbers of electrons on both sides
4H2O + C2H3Cl3(aq) = 2CO2 + 3Cl- + 11H+ + 8e4O3(aq) + 8H+ + 8e- = 4O2(aq) + 4H2O
C2H3Cl3(aq) + 4O3(aq) = 2CO2 + 3Cl- + 3H+ + 4O2(aq)
Balancing Redox Reactions
• Helps to break down overall reaction couple into two half
reactions… each half-reaction can be handled individually
• Use correct formal valence numbers for elements
• Valence numbers you should know like your own name:
• Element in pure form: 0
• Hydrogen: +1 (except in metal hydrides… can ignore in earth
systems)
• Oxygen: -2 (except in peroxides: H2O2, O is -1)
• All halogens: -1
• All alkalis: +1
• All alkaline earths: +2
Balancing Redox Reactions
• Other valences you should try to remember (commonly
encountered redox-active species in natural systems)
• S: +6 (SO4-2), +4 (SO2), 0 (S°), -1 (FeS2), -2 (H2S)
• Fe: +3 (ferric, as in Fe2O3), +2 (ferrous, as in FeCO3)
• Mn: +7 (MnO4-), +4 (MnO2), +3 (Mn3O4), +2 (MnCO3)
• N: +5 (NO3-), +2 (NO), +1 (N2O), 0 (N2), -3 (NH3)
• Don’t forget: valence numbers for atoms in a molecule
MUST sum to the overall charge of the molecule.
Balancing Redox Reactions
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First, determine what is being oxidized and what is being reduced.
Separate the species that is being oxidized into its own half-reaction
• EX: iron metal oxidation: Fe0 (reduced), Fe2O3 (oxidized)
• Balance H using H+, O using H2O, then add as many electrons as are
needed to establish charge balance
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Make a reaction for the species being reduced
• Initial (oxidized) and final (reduced) species
• EX: oxygen is reduced to water: O2 (oxidized), H2O (reduced)
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Add the two balanced half-reactions together
Electrons should cancel out
End result for overall reaction couple should be mass balanced,
charge balanced
Directionality of Redox Reactions
• Redox reactions can proceed to the left or right, just as
any other chemical reaction
• Directionality depends on thermodynamic driving force
• ∂Gr must be negative for overall reaction
• DON’T CONFUSE ∂Gºr with ∂Gr
• Directionality depends on ∂Gr under the particular conditions of
interest (i.e. activities of all reactants are not usually at 1 mol/L)
Directionality of Redox Reactions
• EX: Zn0 + Cu2+ = Zn2+ + Cu0
∂Gºr < 0
• Thus, under reference conditions (all dissolved species at
1 mol/L) the reaction is energetically favored to the right
• Zn is the stronger electron donor
• Cu2+ is the stronger electron acceptor
• With enough data, one could develop a ranking system of
electron donor / acceptor “strength”, or potential
Comparing Redox “Potentials”
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Redox half-reactions have a particular “strength”, or potential… the
energetic driving force under standard conditions
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Can be measured experimentally
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Isolate two half-reactions in separate vessels…
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Allow communication between the two vessels only in terms of
electron flow (through a conductor)
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Measure electrical potential (Volts), or electromotive force, separating
the two systems when they are far from equilibrium
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An electrochemical cell…
Electrochemical Cells
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Setup
• Two reaction vessels containing water, reactants of each half-reaction
• Link vessels using a platinum wire connected to a voltmeter
• Use a separate salt bridge or platinum conductor to form a circuit
voltmeter
Vessel 1:
Cu2+ and
solid Cu
in water
salt bridge
Vessel 2:
Zn2+ and
solid Zn
in water
• EMF diminishes as system approaches equilibrium
• works like a battery (actually, this IS a battery…)
Electrochemical Cells
• Vessel 1:
• Vessel 2:
• Overall:
Cu2+ + 2e- = Cu
Zn = Zn2+ + 2eZn + Cu2+ = Zn2+ + Cu
∂Gºr = -212.8 kJ
voltmeter
Vessel 1:
Cu2+ and
solid Cu
in water
salt bridge
Vessel 2:
Zn2+ and
solid Zn
in water
• At a starting (setup) condition where [Zn2+] = [Cu2+] = 1 mol/L…
• The measured EMF at that moment is proportional to the ∂Gºr
Nernst Equation
• Relationship between EMF and ∂G is always the same…
∂Gr = nFE
∂Gºr = nFEº
• Where
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∂Gr is the Gibbs free energy of reaction
E is the EMF in Volts (Eº = E under reference conditions)
F is Faraday’s constant = 96,489 Coulombs/mole en is the number of electrons transferred in the coupled redox
reaction
• For the Zn-Cu example, Eº = 1.10 V
• This is for a whole, coupled reaction…
…what about for each half-reaction?
Standard Hydrogen Electrode
• There is no way to get each half-reaction Eº value,
unless…
• Every measured EMF is standardized to a single, agreedupon “standard” half-reaction, which serves as a frame of
reference
• Standard Hydrogen Electode (SHE)
• H2(g) = 2H+ + 2e• A Pt electrode in contact with a solution of 1 mol/L H+ (pH = 0)
through which is bubbled H2(g)
• SHE is assigned an arbitrary value of zero (EºSHE = 0 V)
• Everything is measured against SHE
• EX: for Zn oxidation…
Zn + 2H+ = Zn2+ + H2
Eº = -0.763
Calculating Eº
• Example: Fe3+ + e- = Fe2+
• First, calculate ∂Gºr :
What is Eº?
∂Gºr = ∂Gºf(Fe2+) - ∂Gºf(Fe3+)
∂Gºr = (-82.88) - (-8.56)
∂Gºr = -74.32 kJ
• ∂Gºr for the electron (e-) = 0 by convention
• Eº = ∂Gºr / nF
= (-74,320 J) / (1 e-)(96,489 C/mol e-)
= -0.778 V
Non-Standard State Electrode Potentials
• Nernst Equation:
• Also…
∂Gºr = nFEº
• So…
nFE = nFEº + 2.303(RT) log Q
E = Eº + 2.303 (RT/nF) log Q
• At equilibrium
Eº = 2.303 (RT/nF) log K
Eº = (0.0592 / n) • log K
• Or for any system:
Eh = Eº + (0.0592 / n) log Q
∂Gr = ∂Gºr + 2.303RT log Q
(at 25º C, 1 bar)
Eh is used to refer to the system being in reference to the SHE
Eh in the Environment
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Eh is often used as a measure of overall oxidation state in environmental
systems
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“Eh” probes can be purchased, are often employed in environmental
monitoring work
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Eh probes measure redox potential with Pt electrode, referenced to an
internal electrode using calomel (HgCl2) [Hg2+ + 2e- = Hg]
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Eh probes recalculate the measured redox potential to a value referenced to
SHE, and read out a value.
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Qualitative Only!
Calomel Eh electrodes cannot measure “system Eh”, because many redox
couples are far from equilibrium in natural systems… readings should only be
used as a qualitative indicator of oxidizing versus reducing conditions.
Electron Activity
• Can also think of the redox state of a system in terms of
the activity of the electron
• pE
pE = -log(ae-)
• Similar to pH…
• For the example of Fe3+ + e- = Fe2+…
K = aFe2+/(aFe3+)(ae-)
log K = log aFe2+ - log aFe3+ - log aelog K = log aFe2+ - log aFe3+ + pE
pE = log K + log (aFe3+ / aFe2+)
Remember
At 25º C and 1 bar:
Eh = 0.059 pE
pE - pH Diagrams
• One of the most useful ways of predicting the stability of
species as a function of pH and redox potential…
• Plot the stability fields of dominant species as pE v. pH
• Only half-reactions are plotted
• Limits to pE in natural systems: stability of WATER
• H2O = 0.5O2(g) + 2H+ + 2e• Maximum fugacity of O2 is 1 atmosphere at Earth’s surface, in the
presence of liquid water
• 2H+ + 2e- = H2(g)
• Maximum fugacity of H2 is 1 atmosphere at Earth’s surface, in the
presence of liquid water
Limits of pE-pH
• H2O = 0.5O2(g) + 2H+ + 2e∂Gºr = 237.14 kJ
Eº = 1.23 V
log K = -41.5 = 0.5 log aO2 - 2 pH - 2 pE
pE = 20.775 – pH
(at fO2 = 1 atm… upper limit at Earth surface)
• 2H+ + 2e- = H2(g)
∂Gºr = 0 kJ
Eº = 0 V
log K = 2 pE + 2 pH
pE = - pH
(at fH2 = 1 atm… lower limit at Earth surface)
pE-pH Diagrams
20
O2
15
FORBIDDEN ZONE
H 2O
10
pE
ALLOWED CONDITIONS
IN THE PRESENCE OF LIQUID WATER
AT 1 ATM TOTAL PRESSURE
5
0
-5
H 2O
FORBIDDEN ZONE
-10
2
3
4
H2
5
6
pH
7
8
9
10
Building pE-pH Diagrams
• Example:
2Fe2+ + 3H2O = Fe2O3 + 6H+ + 2e-
• ∂Gºr = (-742.8) - (3*[-237.14]) - (2*[-82.88])
= 134.38 kJ
• log K = -(134.38)/5.708 = -23.54
• -23.54 = - 6 pH - 2 pE - 2 log aFe2+
• 2 pE = - 2 log aFe2+ + 23.54 - 6 pH
pE = - log aFe2+ + 11.77 - 3 pH
• At a given total activity of Fe2+, this is a straight line in pEpH space
Building pE-pH Diagrams
• Example:
2Fe2+ + 3H2O = Fe2O3 + 6H+ + 2e-
• pE = - log aFe2+ + 11.77 - 3 pH
• Assume that the activity of ferrous iron is 1 mM…
• At pH = 2,
pE = - log(0.001) + 11.77 - 3(2)
= 3 + 11.77 - 6 = 8.77
• At pH = 10,
pE = - log(0.001) + 11.77 - 3(10)
= 3 + 11.77 - 30 = -15.23
pE-pH Diagrams
20
FORBIDDEN ZONE
15
10
pE
Hematite
5
Fe2+
0
-5
FORBIDDEN ZONE
-10
2
3
4
5
6
pH
7
8
9
10
Building pE-pH Diagrams
• Example:
3Fe2+ + 4H2O = Fe3O4 + 8H+ + 2e-
• ∂Gºr = (-1012.9) - (4*[-237.14]) - (3*[-82.88])
= 184.3 kJ
• log K = -(184.3)/5.708 = -32.29
• -32.29 = - 8 pH - 2 pE - 3 log aFe2+
• 2 pE = - 3 log aFe2+ + 32.29 - 8 pH
pE = - 1.5log aFe2+ + 16.15 - 4 pH
• Again, at a given total activity of Fe2+, this is a straight line
in pE-pH space
Building pE-pH Diagrams
• Example:
3Fe2+ + 4H2O = Fe3O4 + 8H+ + 2e-
• pE = - 1.5log aFe2+ + 16.15 - 4 pH
• Assume that the activity of ferrous iron is 1 mM…
• At pH = 2,
pE = 12.65
• At pH = 10,
pE = 0.65
pE-pH Diagrams
20
FORBIDDEN ZONE
15
Magnetite
10
Fe2+
pE
Hematite
5
Fe2+
0
-5
FORBIDDEN ZONE
-10
2
3
4
5
6
pH
7
8
9
10
pE-pH Diagrams
20
FORBIDDEN ZONE
15
Magnetite
Hematite
10
Fe2+
pE
Hematite
5
?
Magnetite
6
pH
7
Fe2+
0
Fe2+
-5
FORBIDDEN ZONE
-10
2
3
4
5
8
9
10
pE-pH Diagram: Fe(1 mM),H2O
(Fe2O3)
(Fe3O4)
pE-pH Diagram: S(1 mM),H2O
pE-pH Diagram: As(1 mM),H2O
Solubility of salts
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Dissolution of mineral solids into aqueous cations and anions
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Example:
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SrSO4 == Sr2+ + SO42-
SrSO4 (celestite) dissolution
log K = -6.5
Solubility of salts
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Dissolution of mineral solids into aqueous cations and anions
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Example:
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SrSO4 == Sr2+ + SO42-
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log K = log[Sr2+] + log[SO42-] – log[SrSO4] = -6.5
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In an aqueous solution how much celestite would dissolve at
equilibrium?
SrSO4 (celestite) dissolution
log K = -6.5
Solubility of salts
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Dissolution of mineral solids into aqueous cations and anions
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Example:
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SrSO4 == Sr2+ + SO42-
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log K = log[Sr2+] + log[SO42-] – log[SrSO4] = -6.5
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In an aqueous solution how much celestite would dissolve at
equilibrium?
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k = [Sr2+][SOr2-] / [SrSO4] = [Sr2+][SOr2-] / 1 = [x][x] = x2
SrSO4 (celestite) dissolution
log K = -6.5
Solubility of salts
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Dissolution of mineral solids into aqueous cations and anions
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Example:
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SrSO4 == Sr2+ + SO42-
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log K = log[Sr2+] + log[SO42-] – log[SrSO4] = -6.5
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In an aqueous solution how much celestite would dissolve at
equilibrium?
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k = [Sr2+][SOr2-] / [SrSO4] = [Sr2+][SOr2-] / 1 = [x][x] = x2
x = (k)1/2 = [10-6.5]1/2 = 0.0005623 = 5.62 x 10-4 mol/L
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[Sr2+] = [SOr2-] = 5.62 x 10-4 mol/L
SrSO4 (celestite) dissolution
log K = -6.5
Saturation and supersaturation
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SrSO4 == Sr2+ + SO42-
log K = -6.5
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[Sr2+] = [SOr2-] = 5.62 x 10-4 mol/L at equilibrium, at which condition
an aqueous solution is saturated with respect to celestite
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Any more [Sr2+] or [SOr2-] added to solution would create a state of
supersaturation
Saturation and supersaturation
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SrSO4 == Sr2+ + SO42-
log K = -6.5
•
[Sr2+] = [SOr2-] = 5.62 x 10-4 mol/L at equilibrium, at which condition
an aqueous solution is saturated with respect to celestite
•
Any more [Sr2+] or [SOr2-] added to solution would create a state of
supersaturation
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If [Sr2+] = 0.03 mol/L and [SOr2-] = 0.03 mol/L...
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Ion Activity Product (IAP)
= [Sr2+][SO42-]
= [0.03][0.03] = 9.00 x 10-4
Saturation and supersaturation
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SrSO4 == Sr2+ + SO42-
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[Sr2+] = [SOr2-] = 5.62 x 10-4 mol/L at equilibrium, at which condition
an aqueous solution is saturated with respect to celestite
•
Any more [Sr2+] or [SOr2-] added to solution would create a state of
supersaturation
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If [Sr2+] = 0.03 mol/L and [SOr2-] = 0.03 mol/L...
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Ion Activity Product (IAP)
ksp = 10-6.5 = 3.16 x 10-7
log K = -6.5
= [Sr2+][SO42-]
= [0.03][0.03] = 9.00 x 10-4
Saturation and supersaturation
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SrSO4 == Sr2+ + SO42-
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[Sr2+] = [SOr2-] = 5.62 x 10-4 mol/L at equilibrium, at which condition an
aqueous solution is saturated with respect to celestite
•
Any more [Sr2+] or [SOr2-] added to solution would create a state of
supersaturation
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If [Sr2+] = 0.03 mol/L and [SOr2-] = 0.03 mol/L...
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Ion Activity Product (IAP)
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[9.00 x 10-4] / ksp = [9.00 x 10-4] / [3.16 x 10-7] = 2,846
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Saturation Index = ratio of IAP to ksp
log K = -6.5
= [Sr2+][SO42-]
= [0.03][0.03] = 9.00 x 10-4
ksp = 10-6.5 = 3.16 x 10-7
Saturation Index
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Saturation Index = ratio of IAP to ksp
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If the SI > 1, a solution is supersaturated with respect to that salt
If the SI < 1, a solution is undersaturated with that salt
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For the example of a solution of [Sr2+] = 0.03 mol/L and [SOr2-] = 0.03
mol/L, SI = 2,846... A highly supersaturated solution with respect to
celestite
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Effect of activity on saturation
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Saturation Index = ratio of IAP to ksp
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If the SI > 1, a solution is supersaturated with respect to that salt
If the SI < 1, a solution is undersaturated with that salt
•
For the example of a solution of [Sr2+] = 0.03 mol/L and [SOr2-] = 0.03
mol/L, SI = 2,846... A highly supersaturated solution with respect to
celestite
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Taking account of activity: First must calculate the Ionic strength
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I = 1/2 Σmizi2
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I = 1/2 * [(0.03*(22) + (0.03*(22)] = 0.12 mol/L
Calculating activity coefficients
•
I = 1/2 * [(0.03*(22) + (0.03*(22)] = 0.12 mol/L
Az 2 I 1/ 2 )
(
− log γ =
(1+ aBI1/ 2 )
Calculating activity coefficients
•
I = 1/2 * [(0.03*(22) + (0.03*(22)] = 0.12 mol/L
Az 2 I 1/ 2 )
(
− log γ =
(1+ aBI1/ 2 )
•
•
Values of A and B are temperature dependent (table 10.3 in Faure)
For 25º C, A = 0.5085 and B = 0.3281
Calculating activity coefficients
•
I = 1/2 * [(0.03*(22) + (0.03*(22)] = 0.12 mol/L
Az 2 I 1/ 2 )
(
− log γ =
(1+ aBI1/ 2 )
•
•
Values of A and B are temperature dependent (table 10.3 in Faure)
For 25º C, A = 0.5085 and B = 0.3281
•
•
Values of a depend on the ion
For [Sr2+], a = 5.0 and for [SOr2-], a = 4.5
Calculating activity coefficients
•
I = 1/2 * [(0.03*(22) + (0.03*(22)] = 0.12 mol/L
Az 2 I 1/ 2 )
(
− log γ =
(1+ aBI1/ 2 )
•
•
Values of A and B are temperature dependent (table 10.3 in Faure)
For 25º C, A = 0.5085 and B = 0.3281
•
•
Values of a depend on the ion
For [Sr2+], a = 5.0 and for [SOr2-], a = 4.5
•
-log γSr++ = (0.5085*(22)*(0.122)/(1 + 5.0*0.3281*(0.122) = 0.0287
•
-log γSO4-- = (0.5085*(22)*(0.122)/(1 + 4.5*0.3281*(0.122) = 0.0287
Calculating activity coefficients
•
-log γSr++ = (0.5085*(22)*(0.122)/(1 + 5.0*0.3281*(0.122) = 0.0287
•
-log γSO4-- = (0.5085*(22)*(0.122)/(1 + 4.5*0.3281*(0.122) = 0.0287
•
log γSr++ = log γSO4-- = -0.0287
•
γSr++ = γSO4-- = 0.936
Calculating activity coefficients
•
-log γSr++ = (0.5085*(22)*(0.122)/(1 + 5.0*0.3281*(0.122) = 0.0287
•
-log γSO4-- = (0.5085*(22)*(0.122)/(1 + 4.5*0.3281*(0.122) = 0.0287
•
log γSr++ = log γSO4-- = -0.0287
•
γSr++ = γSO4-- = 0.936
•
aSr++ = (γSr++)*[Sr2+] = (0.936)*(0.03) = 0.0281 mol/L
•
IAPcelestite = (0.0281)*(0.0281) = 7.90 x 10-4
•
SIcelestite = 7.90 x 10-4 / 3.16 x 10-7 = 2,498
Aqueous Complexation
•
Metals in natural waters tend to form coordinative associations with
available anions and ligands
•
Chemical Speciation
•
•
the chemical or physical form in which an element is actually present
a particular form for an element is called its chemical species
•
Due to complexation, the bulk concentration of a dissolved metal can
differ dramatically from its “true” activity… and not simply because of
activity coefficients.
•
The chemical behavior of an element in solution is critically
dependent on its speciation
Metal Complexation
•
Even in pure water, dissolved metal ions exist in a complexed state…
•
Hydration sphere - any dissolved ion will tend to be surrounded by a
group of electrostatically-coordinated water molecules
• The number of waters surrounding a metal ion is dependent on the metal’s
coordination number (CN)
• CN describes the number of coordination points at which complexing
anions/ligands can “dock” with the cation
• CN values are usually 1, 2, 3, 4, 6, or 8, and are specific to a particular
metal ion
• CN is dependent on ionic radius and ion charge
• EX: Zn++ ions normally have a CN of 6
• EX: UO2++ ions normally have a CN of 8
• EX: H+ ions have a CN of 1
Coordination Geometry
• CN = 2, 3 Polar, Triangle
• CN = 4
2
Tetrahedron, or square planar
4
• CN = 6
4
Octahedral
6
• CN = 8
Cubic
8
3
Ion Pairs and Complexes
• Metal-ligand complexes form when a water in the
hydration sphere is replaced by an electron-donor ligand
• Two end-member types of arrangements can form…
• Ion pair
• ions of opposite charge, attracted due to electrostatic interactions…
• metal ion or ligand or both retain a complete hydration sphere…
• metal ion and ligand are separated by one or more water molecules
• sometimes called “outer-sphere” complexes
• Complexes
• form largely as a result of covalent forces
• fairly stable coordinative structure, relative to an ion pair
• metal and interacting ligand are immediately adjacent to each other
• “inner sphere” complexes
Ligands
• A wide variety of ligands occur in natural systems
• Hydroxide (OH-): in all natural waters at higher pH
• Complexation with hydroxide is called “hydrolysis”
• In most natural waters (pH < 8) hydrolysis involves reaction of a cation with
H2O to produce a metal-hydroxide complex and H+
• Halogens (F-, Cl-, Br-, I-): most common is chloride
• Form strongly ionic bonds with alkalis, alkaline earth metals
• Slightly more covalent bonds with transition metals
• Inorganic oxyanions (CO32-, SO42-, PO43-, AsO43-, etc..)
• Ubiquitous in natural waters, most common are carbonate, sulfate…
• Organic ligands
• Low molecular-weight organic acids (acetate, oxalate, citrate, etc…) produced
by plants, bacteria, fungi
• Humic, fulvic acids… high molecular weight multiprotic ligands
Complexation Reactions
•
Represented like any normal chemical reaction…
•
EX:
Pb2+ + Cl- = PbCl+
K = [PbCl+]/[Pb2+][Cl-]
log K1 = log[PbCl+] - log[Pb2+] - log[Cl-] = 1.44
•
EX:
PbCl+ + Cl- = PbCl20
K = [PbCl20]/[PbCl+][Cl-]
log K2 = log[PbCl20] - log[PbCl+] - log[Cl-] = 2.00
•
Beta (β) notation
• The reaction forming PbCl20 in one step, from Pb2+ and 2Cl-, is…
log K1 + log K2 = log β2
• Where
log K1 = log β1
β2 = K1K2
βi = K1K2K3….Ki
Metal Hydrolysis
•
Most metals will hydrolyze at circumneutral to basic pH values
•
M2+ + OH- = MOH+
or
M2+ + H2O = MOH+ + H+
•
M2+ + 2(OH-) = M(OH)20
or
M2+ + 2H2O = M(OH)20 + 2H+
•
M2+ + 3(OH-) = M(OH)3-
or
M2+ + 3H2O = M(OH)3- + 3H+
•
•
M2+ + 4(OH-) = M(OH)42-
or
M2+ + 4H2O = M(OH)42- + 4H+
•
•
Tendency is toward greater degrees of hydrolysis with increasing pH
Higher-valence metals will tend to hydrolyze at lower pH than lowervalence metals
Trend is NOT toward making a neutral complex… but toward
hydrolyzing all available coordination positions… this is how you can
get negatively-charged metal hydrolysis products
•
etc…
Transition Metals: Hydrolysis of Fe(III)
Transition Metals: Hydrolysis of Cu(II)
Heavy Metals: Hydrolysis of Pb(II)
Lanthanides: Hydrolysis of La(III)
Actinides: Hydrolysis of U(VI)
Alkaline Earths: Hydrolysis of Ca(II)
Classification of Metal Cations
• Metal ions can be classified broadly into categories of
complexation behavior, based on electron configuration
• A-type metal cations
• Electron configuration of a noble gas (filled d-orbital)
• Low polarizability (electron distribution is homogeneously
spherical)… “hard spheres”
• EX: H+, Li+, Na+, K+, Be2+, Mg2+, Ca2+, Sr2+, Al3+, Sc3+, La3+ (REE),
Si4+, Ti4+, Zr4+, Th4+
• Prefer to form bonds with F- and O-donor ligands (carboxyl, etc..)
• Weak to little bond formation with Cl-, I-, S-donor ligands (sulfide)
• Form sparingly soluble precipitates with OH-, CO32-, PO43-
Classification of Metal Cations
• B-type metal cations
• Electron number corresponds to Ni0, Pd0 and Pt0 (10 or 12 outer
shell electrons)
• Low electronegativity, high polarizability, “soft spheres”
• EX: Cu+, Ag+, Au+, Tl+, Ga2+, Zn2+, Cd2+, Hg2+, Pb2+, Sn2+, Tl3+,
Au3+, In3+, Bi3+
• Coordinate with bases containing I, S, or N donor atoms
• More stable I- and Cl- complexes than F• Form insoluble complexes with sulfide
Classification of Metal Cations
• Transition metal cations
• One to nine outer shell electrons
• Not spherically symmetric
• EX: V2+, Cr2+, Mn2+, Fe2+, Co2+, Ni2+, Cu2+, Ti3+, V3+, Cr3+, Mn3+,
Fe3+, Co3+
• Intermediate in properties between A and B types
• Behavior is highly individualistic, depending on electron config.
• Complex stability tends to increase in the rough order:
Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+
Question 1:
1. Construct an Eh versus pH diagram of Eu at 25º C and 1 bar, taking account of the
chemical species Eu +3, Eu+2 and Eu(OH)3,c only. Use a pH range of 2 to 10, an Eh
range of -1.0 to +1.0 Volts, and total Eu concentration of 1 micromole/L. Use the
following values to construct your diagram:
Eu(OH)3(s)
dGºf = -1194.5 kJ/mol
Eu3+
dGºf = -564.0 kJ/mol
Eu2+
dGºf = -540.2 kJ/mol
Assume activities equal concentrations. Your resulting diagram should be accurate,
clean, clearly labeled and visually non-ambiguous.
Include on your figure titles for the x and y axes, units along both axes, reaction lines
for the upper and lower limits to the stability of water, and the appropriate labeled
stability fields of all pertinent europium chemical species. (1 pt)
Diagram includes all appropriate reaction lines, accurately, with clean, clear stability
fields for all species. (3 pt)
Question 2:
Write a balanced, coupled reaction describing a reaction of silicon metal with magnesite to
produce enstatite (orthorhombic) and methane gas. (1 pt)
Determine the log K for this reaction at 298.15 K and 1 bar pressure. (1 pt)
Use thermodynamic values solely from Appendix 5 in the Misra text. If multiple values are
available for a species, use the first listed value.
HINT: In setting up this problem, first define the oxidation half-reaction and the reduction
half-reaction separately, then combine them to produce the overall reaction.
FURTHER HINT: It's okay to use Mg++ in the half-reactions.
NOTE: There should be no aqueous Si-containing chemical species in your balanced
equation. None at all.
Question 3:
Calculate the value of the activity coefficient (γ, gamma) applicable to Fe 2+ in aqueous
solution with an ionic strength ranging from 0.0 to 1.0, by setting up a spreadsheet or
other calculation where values at 1% increments (0.00, 0.01, 0.02,....) are used to
calculate gamma values through the entire range of ionic strength. Perform this
calculation using the Debye-Huckel limiting law equation, the extended D-H equation,
the Truesdell-Jones equation, and the Davies equation. Plot these curves (WITHOUT
DATA POINTS - you don't have data, you have calculations) on a diagram with the xaxis being ionic strength ranging from 0.00 to 1.00. Axis scale markers do not need to
be at 1% increments; that would look terrible and would be unreadable. Parameters for
calculations are found in tables 7.1, 7.2 and 7.3 of the textbooks.
The grading rubric is as follows:
Graph includes complete, accurate curves for ferrous iron ion activity coefficients as a
function of ionic strength from 0.00 to 1.00 for the four mentioned methods of
calculating activity coefficients. (4 pts)
Graph is readable, professional-looking and includes axis labels, appropriate scale
ranges, scale labels, and curves designated by labels, color, or line types. (1 pt)
Personally, I like to draw and paint by hand with physical media, but I don't usually
make diagrams that way. If you want to draw yours, be careful to make it look like a
professional graphic, not a scrawled pencil sketch on lined paper photographed in poor
lighting.
Graph includes individual data points along the calculated curves. (-5 pts) Never do
that. Data points are measured, a calculated curve itself has no discrete, specific points
along the curve that are individually any more or less real than any other.
Question 4:
Using thermodynamic values from Misra appendix 5, NOT R&H95, write reactions that
solubilize hematite (Fe2O3) to Fe3+, FeOH2+, Fe (OH)2+ and Fe(OH)3(aq), then construct a
solubility diagram depicting hematite solubility relations as a function of (x-axis) pH from 0
to 10, and (y-axis) activity of aqueous Fe(III) species in the system.
Diagram includes all reaction lines for all species. (1 pt)
Diagram includes correctly calculated values for reaction lines. (1 pt)
Diagram includes readable labels for hematite and all aqueous Fe(III) species, both axes, and
axis numerical values. (1 pt)
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