Unit 09a : Advanced Hydrogeology

Chemical Reactions

Chemical Reactions

• A wide variety of chemical reactions can take place between gases, solutes and solids in groundwater systems:– Acid-base– Solution-precipitation– Complexation. – Redox– Hydrolysis– Isotopic processes

Hydrogen Ion Activity

• [H+] represents the activity of hydrogen ions in solution:

pH = - log[H+] = -log[H3O+]– since hydrogen ions exist in solution in the

hydrated form as H3O+ – this allows us to distinguish between hydrogen

ions and protons

• pH is a “master variable” controlling chemical systems.

• pH is controlled by acid-base reactions.

Acids and Bases

• An acid is a substance with a tendency to lose protons

• A base is a substance with a tendency to gain protons.

• Acids react with bases to transfer protons• In acid-base reactions, because no free

protons are produced, there must be two acid-base systems involved:

Acid1 + Base1 = Acid2 + Base2

Bicarbonate ReactionHCO3

- + H2O = H3O+ + CO32-

• A proton is transferred (donated) from the bicarbonate ion to the water molecule to create an hydrogen ion.

K = [H3O+ ] [CO32-] = [H+ ] [CO3

2-]

[HCO3- ] [H2O] [HCO3

- ] – assuming the activity of water is unity and abandoning our

“hydrogen ion” distinction.

• Remember the reaction is nevertheless an acid-base reaction with water as a base:

HCO3- = H+ + CO3

2-

Ammonia Reaction

H2O + NH3 = NH4+ + OH-

• Ammonia is a base that ionizes in water by accepting a proton.

• In this case, water is the proton donor and water acts as an acid.

• The concept of acid or base is simple: the proton donor is the acid the proton acceptor is a base.

• In the first reaction water accepted a proton from the bicarbonate ion. In this reaction water donates a proton.

Strong and Weak Acids

HA + H2O = H3O+ + A-

• The strength of an acid depends on its ability to drive the ionization reaction from left to right.

• Strong acids donate protons freely in spite of the fact that water is a weak base (proton acceptor).

Weak Acid-Base Reactions

• There are a few weak acid reactions that are important in groundwater systems:– dissociation of water– dissociation of carbonic acid (dissolution of

gaseous CO2)

– dissociation of silicic acid (dissolution of silicate minerals)

Dissociation of WaterH20 + H2O = H3O+ + OH-

• The equilibrium constant for this acid-base reaction, where water is both acid and base, is 10-14

Kw = [H+] [OH-]– again we assume [H2O] is unity and don’t bother to

include it explicitly in the equations

• Just as pH represents –log[H+], it is convenient to use pK to represent –log[K]

• For the dissociation of water pKw = 14– because [H+] = [OH-], the pH of pure water is 7.0

Carbonic Acid

CO2(g) + H20 = H2CO3* CO2(aq)

• The equilibrium constant pKCO2 for the solution of carbon dioxide in water to produce carbonic acid is 1.46.

H2CO3* = H+ + HCO3

-

• The first dissociation constant pK1 = 6.35

HCO3- = H+ + CO3

2-

• The second dissociation constant pK2 = 10.33

Carbonate Speciation Example p.1 • Calculate the distribution of mass between carbonate

species at pH 7 given [CO2]T = 10-3 M.• Step 1: Identify the species.

[CO2]T = [H2CO3*] + [HCO3

-] + [CO32-]

• Step 2: Calculate [H+] and [OH-]pH = 7 therefore [H+] = 10-7

[OH-] = Kw / [H+] = 10-14 / 10-7 = 10-7

• Step 3: Write [HCO3-] using K1

[HCO3-] = [H2CO3

*] (K1 / [H+])

= [H2CO3*] (10-6.35 / 10-7) = [H2CO3

*] (100.65)

• Step 4: Write [CO32-] using K2

[CO32-] = [HCO3

-] (K2 / [H+]) = [H2CO3*] (K1 / [H+]) (K2 / [H+])

= [H2CO3*] (100.65) (10-10.33 / 10-7) = [H2CO3

*] (10-2.68)

Carbonate Speciation Example p.2 • Step 5: Write [CO2]T in terms of [H2CO3

*] to find [H2CO3*]

[CO2]T = [H2CO3*] + [HCO3

-] + [CO32-]

[CO2]T = [H2CO3*] + [H2CO3

*] (100.65) + [H2CO3*] (10-2.68)

[CO2]T = [H2CO3*] ( 1 + 100.65 + 10-2.68 )

[CO2]T = [H2CO3*] (5.47)

[H2CO3*] = [CO2]T / (5.47)

[H2CO3*] = 10-3 / 5.47 = 10-3.74

• Step 6: Calculate [HCO3-]

[HCO3-] = [H2CO3

*] K1 / [H+] = 10-3.74(10-6.35 / 10-7) = 10-3.09

• Step 7: Calculate [CO32-]

[CO32-] = [HCO3

-] K2 / [H+] = 10-3.09(10-10.33 / 10-7) = 10-6.42

• The carbonate species [H2CO3*], [HCO3

-] and [CO32-] have

concentrations of 10-3.74, 10-3.09 and 10-6.42 M respectively at pH 7. Bicarbonate is the dominant ion.

Carbonate in Solution

-22

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

log

[C

]

H2CO3*

HCO3-

CO32-

OH-H+

H2CO3*

H+ OH-

HCO3-

CO32-

(CO2)T = 10-3 M

Carbonate in Solution

-22

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

log

[C

]

H2CO3*

HCO3-

CO32-

OH-

H+

H2CO3*

H+ OH-

HCO3-

CO32-

(CO2)T = 10-4 M

Carbonate System

• Plotted for (CO2)T = 10-3 M or about 100 mg/L

• Crossover points– pK1 = 6.35 where [HCO3

-] = [H2CO3*]

– pK2 = 10.33 where [HCO3-] = [CO3

2-]

• Carbonate speciesH2CO3

* is dominant for pH < 6.35

HCO3- is dominant for 6.35 > pH < 10.33

CO32- is dominant for pH > 10.33

Silicic Acid

• Carbon and silicon both form strong covalent bonds with oxygen.

• Silicic acid is also a weak acid like carbonic acid.

H2SiO3 = H+ + HSiO3-

• The first dissociation constant pK1 = 9.86

HSiO3- = H+ + SiO3

2-

• The second dissociation constant pK2 = 13.1

Silica in Solution

-22

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

log

[C

]

H2SiO3

HSiO3-

SiO32-

OH-H+

H2SiO3

H+ OH-

HSiO3-

SiO32-

(SiO2)T = 10-4 M

Silica System

• Plotted for (SiO2)T = 10-4 M or about 10 mg/L

• Crossover points– pK1 = 9.86 where [HSiO3

-] = [H2SiO3]

– pK2 = 13.1 where [HSiO3-] = [SiO3

2-]

• Silicate speciesH2SiO3 is dominant for pH < 9.86

HSiO3- is dominant for 9.86 > pH < 13.1

SiO32- is dominant for pH > 13.1

Alkalinity

• The Bjerrum plots (log[C] against pH) show the effect of pH alone on speciation but in “real world” systems there are additional factors to consider.

• The equilibria are influenced by strong bases added through the dissolution of carbonates and silicates.

• Alkalinity is the net concentration of strong bases in excess of strong acids with a pure water – CO2 system as a reference point (zero alkalinity).

Charge Balance

• When CO2 is dissolved in water (at a fixed

PCO2 ) the charge balance is:

[H+] = [OH-] + [HCO3-] + 2[CO3

2-]– the cations must balance the anions

• Adding a strong base (NaOH) and a strong acid (HCl) to the system will add Na+ and H+ cations and OH- and Cl- anions so the charge balance becomes:

[Na+] + [H+] = [OH-] + [HCO3-] + 2[CO3

2-] + [Cl-]

Alkalinity Defined

• The net excess contribution of ions from the strong base is the alkalinity (for Na+ > Cl-) given by:

[Na+] - [Cl-] = [OH-] + [HCO3-] + 2[CO3

2-] - [H+]• If we generalize to any base and any acid (rather

than NaOH and HCl) we can write:

Alkalinity = [i+]sb – [i-]sa

= [OH-] + [HCO3-] + 2[CO3

2-] - [H+]• For the pure water-CO2 reference system, the

alkalinity is zero.

Net Alkalinity and Net Acidity

• In most natural systems, the generation of net positive charges from dissolution of carbonates and silicates usually exceeds the contribution of negative charges from the ionization of strong acids. Most natural groundwaters are alkaline.

• When strong acids are present (for example from pyrite oxidation) groundwaters may display net acidity. Such cases are often associated with sulphide mineralization or contamination by acid rock drainage (ARD).

Mineral Dissolution

• Increasing alkalinity results from an increase in positive ions on the LHS of the alkalinity equation from mineral dissolution.

• An equal number of negative ions are added to the RHS to maintain neutrality and some of these ions come from ionization of H2CO3* to HCO3

- and H2SiO3 to HSiO3

-. • The equilibria for ionization of the weak acids removes

hydrogen ions as HCO3- and HSiO3

- are produced since the dissociation constant is invariant.

• In most natural groundwater systems, pH increases along the flow path as minerals are dissolved.

Solute Mass Loadings

• Water is an excellent solvent.• Mineral dissolution is primarily responsible for

the mass loadings in groundwater.• Other processes contribute to the reduction of

solute mass loadings. These include:– gas exsolution– volatilization– precipitation

Henry’s Law• Henry’s law does not strictly apply to gases (like CO2

and NH3) that react in solution.

• Very little [CO2]aq reacts and H2CO3 concentrations are very low such that [CO2]aq [H2CO3*]

• Henry’s law thus adequately approximates the distribution of CO2 between the aqueous and gaseous phases:

KH = PCO2

[CO2]aq

– KH is the Henry’s law constant with units of kPa.L.mol-1 or atm.L.mol-1.

Carbon Dioxide Solution

• Changes in PCO2 directly effect [CO2]aq and hence [H2CO3*]

• Changes in [H2CO3*] through the dissociation constants pK1 and pK2 influence [HCO3

-], [CO32-],

[OH-] and pH.

• Addition of CO2 to groundwater through the unsaturated zone increases [HCO3

-] and pH and enhances the ability of the solution to dissolve silicate minerals.

Volatilization

• Volatilization is the process of liquid or solid phase evaporation at a liquid-gas or solid-gas interface.

• The process of volatilization of solutes is controlled by Henry’s law. – NOTE: we are not discussing any non-aqueous

phase liquids.

• Volatilization can create problems in sampling. When samples have access to the atmosphere, loss of volatiles to the vapour phase can be significant.

Dissolution and Precipitation

• Dissolution and precipitation of solids are two of the most import processes controlling groundwater chemistry.

• Groundwater systems evolve towards the equilibrium state either from undersaturation (most natural systems) or oversaturation (some contaminated systems).

• In natural systems, dissolution proceeds and pH rises as waters evolve along the flow path.

• In contaminated systems, precipitation can remove metals as pH rises along the flow path.

Mineral Solubility

• Solubility reflects the extent to which the reactant (mineral) and products (ions and/or secondary minerals) are favoured in a dissolution-precipitation reaction.

• Because the activity of the reacting solid is taken to be unity, the magnitude equilibrium constant pK provides a relative measure of mineral solubility (in pure water).

• When other ions are present, absolute and relative solubilities can change.

Common Mineral SolubilitiesMineral IAP pK

Halite [Na+][Cl-] -1.54Sylvite [K+][Cl-] -0.98Quartz [H2SiO3] 4.00Gypsum [Ca2

+][SO42-] 4.62

Magnesite [Mg2+][CO32-] 7.62

Aragonite [Ca2+][CO32-] 8.22

Calcite [Ca2+][CO32-] 8.35

Na-Montmorillonite 9.10Kaolinite 9.40Siderite [Fe2+][CO3

2-] 10.70Brucite [Mg2+][OH-]2 11.10Ferrous Hydroxide [Fe2+][OH-]2 15.10Dolomite [Ca2+][Mg2+][CO3

2-]2 16.70Pyrrhotite [Fe2+][S2-] 18.10Spalerite [Zn2+][S2-] 23.90Galena [Pb2+][S2-] 27.50Gibbsite [Al3+][OH-]3 33.50

[Na+] [H4SiO4]4

[H4SiO4]2

Ionic Strength Effect

• Generally solubility increases with increasing ionic strength.

• The presence of other ions reduces the activity of ions involved in the reaction.

• This increases the number of ions needed in solution to achieve the equilibrium IAP.

Common Ion Effect

• When a solution contains the ion that is released when a solid dissolves, the presence of that ion means that less dissolution is required to reach the equilibrium IAP.

• This phenomenon decreases the solubility and is called the common ion effect.

Complexation

• A complex is an ion that forms by combining simpler anions, cations and molecules.

• The cation (or central atom) is typically a metal.

• The anion (or ligand) is almost any simple anion (halide, sulphate, carbonate, phosphate, etc)

• A simple complexation reaction involves a metal and a ligand:

Zn2+ + Cl- = ZnCl-

Complex Complexes

• Sometimes complexes combine with ligands and metals are distributed among a large number of cation complexes.

• For example, the hydrolysis of the trivalent chromium ion:

Cr3+ + OH- = Cr(OH)2+

Cr(OH)2+ + OH- = Cr(OH)2+

Cr(OH)2+ + OH- = Cr(OH)3

0

General Complexes

• Most reactions involving complexes are “fast” in a kinetic sense.

• A general complexation reaction involves a metal cation (M), b ligands (L) and c hydrogen ions (H):

aM + bL + cH = MaLbHc

• The stability constant for the complex KMLH is given by the association reaction:

KMLH = [MaLbHc] [M]a[L]b[H]c

or pKMLH = a.log[M] + b.log[L] + c.log[H] – log[MaLbHc]• The larger the value of KMLH, the more stable the complex

More about Complexes

• Most complexes involve a single cation:ZnCl+, Cr(OH)2+

• Polynuclear complexes are relatively unusual:

Cr3(OH)45+, Cu2(OH)2

2+

• Complexation facilitates the transport of potentially toxic metals such as Cd, Cu, Cr, Mo, Pb, and U. For example, U forms complexes with many ligands including F, CO3, SO4 and PO4.

Complex Speciation Example p.1

• A solution contains a trace of chromium (10-5 M) at a pH of 5. Determine the speciation among the Cr hydroxyl complexes given stability constants (where pK = -log K) pK2

= -10.0, pK1 = -18.3 and pK0 = -24.0• Step 1: Identify the species

[Cr]T = [Cr3+] + [Cr(OH)2+] + [Cr(OH)2+] + [Cr(OH)3

0]

• Step 2: Calculate [H+] and [OH-]

pH = 5 therefore [H+] = 10-5

[OH-] = Kw / [H+] = 10-14 / 10-5 = 10-9

• Step 3: Use the association reactions to find [C] for complexes

[Cr]T = [Cr3+] + K2[Cr3+][OH-] + K1[Cr3+][OH-]2 + K0[Cr3+][OH-]3

Complex Speciation Example p.2• Step 4: Solve for [Cr3+]

[Cr]T = [Cr3+] + K2[Cr3+][OH-] + Ki[Cr3+][OH-]2 + K0[Cr3+][OH-]3

[Cr]T = [Cr3+] (1 + K2[OH-] + Ki[OH-]2 + K0[OH-]3)

[Cr3+] = [Cr]T / (1 + K2[OH-] + Ki[OH-]2 + K0[OH-]3) [Cr3+] = 10-5 / (1 + 1010.10-9 + 1018.3.10-18 + 1024.10-27)

[Cr3+] = 10-5 / (1 +101 + 100.3 + 10-3) [Cr3+] = 10-5 / (1 + 10 + 1.995 + 0.001) = 10-5 / (12.996) = 10-6.12

• Step 5: Solve for [Cr(OH)2+]

[Cr(OH)2+] = K2[Cr3+][OH-] = 1010.10-6.12.10-9 = 10-5.12

• Step 6: Solve for [Cr(OH)2+]

[Cr(OH)2+] = Ki[Cr3+][OH-]2 = 1018.3.10-6.12.10-18 = 10-5.82

• Step 7: Solve for [Cr(OH)2+]

[Cr(OH)30] = K0[Cr3+][OH-]3 = 1024.10-6.12.10-27 = 10-9.12

• The chromium species [Cr3+], [Cr(OH)2+], [Cr(OH)2+], and

[Cr(OH)30] have molar concentrations of 10-6.12, 10-5.12, 10-5.82 and

10-9.12 M respectively at pH 5. Cr(OH)2+ is the dominant ion.

Major Ion Complexation

• When we calculated IAP/K ratios and saturation indices earlier, we used total concentrations.

• The true concentrations are significantly reduced by complexation.

• Even at relatively low ionic strength >0.02 M, the error in determining mineral saturations can be substantial if complexation is neglected.

• In seawater, for example, only 40% of the total SO4

exists as SO42-, 37% exists as NaSO4

+ and 19% as MgSO4

0

Metal Mobility

• Generally, in groundwater, metals are most mobile at low pH.

• Ignoring surface reactions, metal concentrations begin to decline when pH increases to the point where equilibrium is reached with a solid phase.

• The solid phases are usually metal-hydroxides, metal-sulphides or metal-carbonates.

Uranium Complexes

• Complexes can enhance the transport of metals at low concentrations.

• Uranium is a good example, forming many uranyl (UO2

2+) complexes with ligands including F-, CO3

2-, SO42- and PO4

3-.

• At pH 7, (UO2)(HPO4)22- is the dominant

uranyl species and increases the solubility of some uranium minerals by several orders of magnitude (Langmuir, 1978).

Surface Reactions

• When water containing a trace constituent is mixed with a disseminated solid and allowed to equilibrate, mass partitions between the solution and the solid surface.

S = (Co – C).V / Ms

where S is the mass sorbed on the surface (M/M); Co is the initial concentration in solution (ML-3) and C is the equilibrium concentration in solution (ML-3), V is the solution volume (L3) and Ms is the mass of solid (M).

Freundlich Isotherm• The function S(C)

describing the surface sorption for various equilibrium concentrations is called a sorption isotherm.

• Isotherms functions have no theoretical form and have be derived empirically.

S = K.Cn

is a form suggest by Freundlich where K and n are empirical constants.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Concentration

Mas

s S

orp

tio

n

n = 0.5

n = 0.2

n = 2

n = 1

n = 5

n = 0.1

Langmuir Isotherm

• The Langmuir isotherm has a more complex form:S = Q.K.C / (1 + K.C)

where Q is the maximum sorptive capacity of the surface and K is a partition coefficient.

• The value of K controls the extent of sorption.

• The Langmuir isotherm limits the maximum mass sorption through the parameter Q. 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Concentration

Mas

s S

orp

tio

nK = 0.5

K = 20

Linear Isotherm• The Freundlich isotherm

for n =1 is called a linear isotherm

S = Kd.C

where Kd is the distribution coefficient.

• This special case has been widely used to represent sorption of metals.

• Finding a single value Kd to characterize the sorption process has proved difficult and more complex models are demanded by experience.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Concentration

Mas

s S

orp

tio

n

Surface Reactions

• For surface reactions, it is possible to account for the properties of the solution and the solid surfaces.

• Cation exchange is the best-known surface reaction.• The process is driven by electrostatic attraction

between charged cations and the surface charge on clay mineral and oxide/hydroxide surfaces.

• Clay mineral surfaces have significant negative fixed charges due lattice substitutions and broken bonds at the edges of the minerals.

• Cations bind to the surfaces to balance the charge.

Cation Exchange Capacity

• Cation exchange capacity (CEC) describes the quantity of exchangeable cations sorbed onto a surface.

• CEC has units of meq per 100 g of sample.

• CEC varies from one mineral to another but is strongly related to surface area.

Clay Mineral CEC

Mineral CEC(meq/100g)

Surface Area(m2/g)

Kaolinites* 5-15 15

Illites 25 80

Chlorites 10-40 80

Vermiculites 100-150 100

Montmorillonites 80-100 800

Cation Affinity• Clay minerals exhibit a preference for specific ions

occupying exchange sites:

Li+ < Na+ < H+ < K+ < NH4+ < Mg2+ < Ca2+ < Al3+

• In general, cation affinity for exchange sites increases with ionic charge.

• At high concentrations, ion hydration and complexation can influence cation affinities.

• In general, monovalent ions have hydration energies of around 100 kcal/mol compared with 400-500 kcal/mol for divalent ions and >1000 kcal/mol for Al3+ and Fe3+

Exchange Reactions• The general form of a cation exchange reaction is:

nMX + mNn+ = nMm+ + mNX where M and N are metal cations with charges m+ and n+

respectively and MX and NX are the corresponding metals sorbed on the solid phase.

• For example:

Na-clay + K+ = Na+ + K-clayn1Ca2+ + n2Mg2+ + n3Fe2+ + 2(n1+n2+n3)Na-clay =

2(n1+n2+n3)Na+ + Ca-Mg-Fe-clay

Ca2+ + 2Mg2+ + Fe2+ + 8Na-clay = 8Na+ + Ca-Mg-Fe-clay

Other Sorption Reactions

• A second group of sorption reactions involve solids whose surface charge depends on groundwater composition.

• Hydrated metal oxides and hydroxides (Si, Al, Fe) and kaolinites are the most important solids in this group.

• Surfaces typically carry a positive charge at low pH but become negatively charged cation exchangers at higher pH.

Oxides and Hydroxides

XOH = H+ + XO-

XOH + H+ = XOH2+

• At low pH, XOH2+ is the dominant

surface species whereas at higher pH, XO- is dominant.

• The neutral point for a surface in terms of pH is called the isoelectric point

Isoelectric Points

Mineral

Quartz

Kaolinite

Hematite

Magnetite

Goethite

Corundum

Gibbsite

pH

2.0 - 3.5

1.8 - 4.6

5.0 - 9.0

6.5

6.0-7.0

9.1

~9

Redox

• The hydrogen ion activity (H+) and the availability of electrons (e-) are the master variables of groundwater reactions.

• Unlike H+ ions, electrons are not 'free forming'; they are contained within atoms or molecules. Electrons are only transferred between species.

• Redox, short for reduction-oxidation, is the termed used to denote reactions involving the transfer of electrons.

Microorganisms

• Oxidation-reduction reactions differ from many other reactions because they are frequently mediated by microorganisms.

• The role of the microorganisms is usual to act as a catalyst and increase the rate of reaction.

• Microbial films on grains and fracture surfaces use redox reactions as a source of energy.

Redox Reactions

• Oxidation-reduction reactions involve electron (e-) exchange

• An element changes oxidation state• There are no free electrons an electron

transfer occurs• The process is described by paired (or

coupled) half-reactions involving oxidation and reduction together.

Oxidation Number

• The oxidation number of an element indicates the number of electrons lost, gained, or shared as a result of chemical bonding.

• The change in the oxidation state of a species lets you know if it has undergone oxidation or reduction.

• Oxidation can be defined as "an increase in oxidation number".

• Reduction can be defined as "a decrease in oxidation number".

Simple Reaction

2Na + Cl2 > 2NaCl

• The sodium starts out with an oxidation number of zero (0) and ends up having an oxidation number of +I. It has been oxidized from a sodium atom to a positive sodium ion.

• The chlorine also starts out with an oxidation number of zero (0), but it ends up with an oxidation number of -I. It, therefore, has been reduced from chlorine atoms to negative chloride ions.

Oxidizing and Reducing Agents• The substance bringing about the oxidation of

the sodium atoms is the chlorine, thus the chlorine is called an oxidizing agent.

• The substance bringing about the reduction of the chlorine is the sodium, thus the sodium is called a reducing agent.

• Oxidation is ALWAYS accompanied by reduction.

• Reactions in which oxidation and reduction are occurring are called Redox reactions.

Oxidation Numbers (I)• Oxidation numbers are often written as roman numerals.

Example: Cr(VI), Mn(IV), Fe(III)

• The oxidation number of an atom in the elemental state is zero. Example: Cl2, Al and N2 are all 0

• The oxidation number of a monatomic ion is equal to its charge. Example: In the compound NaCl, the Na+ has an oxidation number of +I and the Cl- has an oxidation number of -I.

• The algebraic sum of the oxidation numbers in the formula of a compound is zero. Example: the oxidation numbers in the NaCl above add up to 0

Oxidation Numbers (II)• The oxidation number of hydrogen in a compound is +I, except

when hydrogen forms compounds called hydrides with active metals, and then it is -I. Example: H is +I in H2O, but -I in NaH (sodium hydride).

• The oxidation number of oxygen in a compound is -II, except in peroxides when it is -I, and when combined with fluorine. Then it is +II. Example: In H2O the oxygen is -II, in H2O2 it is -I.

• The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is the charge on that ion. Example: in the sulphate ion, SO4

2-, the oxidation numbers of the sulphur and the oxygens add up to -II. The oxygens are -II each, and the sulphur is +VI.

Oxidation States

+VI

+V

+IV

+III

+II

+I

0

-I

-II

-III

-IV

Sulphur

SO42-

SO32-

S

FeS2

H2S,FeS

Nitrogen

NO3-

NO2-

N2

NH4+

Iron

Fe(OH)3, FeO(OH)

Fe(OH)2

Fe

Carbon

CO2, HCO3-

C

CH2O,CH3OH

CH4

Oxidation and Reduction• When an oxidation or reduction reaction is written

independently; for example, the reduction of CO2

CO2 + 4e- + 4H+ = CH2O + H2O

or for the oxidation of H2O

2H2O = O2 + 4e- + 4H+

a ‘free’ electron (e-) is written in the equation. • An overall redox reaction will never have an (e-)

shown:

CO2 + H2O = CH2O + O2

Half Reactions

• Reduction of elemental oxygen:

½O2 + 2H+ + 2e- = H2O

• Oxidation of ferrous iron:

Fe2+ = Fe3+ + e-

• Combining the two half-reactions gives a balanced redox reaction:

2Fe2+ + ½O2 + 2H+ = 2Fe3+ + H2O

Electron Donors• By far the most prevalent electron donor in

the shallow subsurface is organic carbon.

CH2O + H2O = CO2 + 4e- + 4H+

• This half reaction is what supplies energy to microorganisms within soils.

• The electron donor (organic carbon) is the reducing agent and is oxidized to CO2.

Inorganic Electron Donors

• Common electron donors that participate in chemical redox couples in groundwaters include:

• Mn(II) = Mn(IV) + 2e-

• Fe(II) = Fe(III) + e-

• S2- = SO42- + 8e-

• As(III) = As(V) + 2e-

Redox Chain

• Groundwaters follow a chain of redox reactions during infiltration that involve consuming organic matter

• Each redox step is generally controlled by the availability of an oxidant (bacterial catalysts are ubiquitous)

• The redox state of the groundwater is usually controlled by the dominant redox pair

• Redox state affects the mobility (solubility) of redox-sensitive metals

Common Redox Pairs

Reduced Form

H2S hydrogen sulphide

NH4+ ammonium

CH4 methane

Fe2+ ferrous iron

Mn2+

As3+

Oxidized Form

SO42- sulphate

NO3- nitrate

CO2 carbon dioxide

Fe3+ ferric iron

Mn4+

As5+

Electron Activity

• For equilibrium reactions:

Ox + n e- = Red

K = [Red] / [Ox][e-]n

• For the electron transfer reaction:

pE = (1/n) ( log K - log [Red] / [Ox] )

pE is analogous to pH such that pE = -log[e-]• If reactions are always written such that n=1:

pE = log K - log [Red] / [Ox]

pE = pEo - log [Red] / [Ox]

Expressions for pE• For example: ½O2 + 2H+ + 2e- = H2O

pE = ½ ( log K - log 1 / PO21/2[H+]2 )

pE = ½ ( log K + log PO21/2[H+]2 )

pE = ½ ( log K + log PO21/2 + 2 log [H+] )

pE = ½ ( log 1041.55 + ½ log PO2 - 2 pH )

pE = 20.78 + 1/4 log PO2 - pH

• For example: Fe2+ = Fe3+ + e-

pE = log K + log [Fe3+] / [Fe2+]

pE = log 1012.53 + log [Fe3+] / [Fe2+]

pE = 12.53 + log [Fe3+] / [Fe2+]

Electrode Potential

• While pE is a convenient way to 'view' redox reactions, it is not real.

• Since electrons do not exist in solution we need a separate way of measuring their reactivity in a system.

• This is done with electrode potential: the potential for an electron to participate in a reaction measured as a voltage relative to a reference (hydrogen) electrode.

pE and Eh• pE, represents the negative logarithm of the electron

activity• Eh is the thermodynamic redox potential expressed in

terms of millivolts.

• pEo = 59.2 Eho (at 25oC)

• For measurement purposes, the Eh of a system may be defined as the potential developed at an inert metallic electrode expressed relative to the standard hydrogen electrode in a reversible redox system.

• Eh is dependent on pH and so the pH at which any measurement of Eh is made must be stated.

Redox State of Soils• Soils have been broadly classified into three redox

states based on their pE (or Eh) values:• Oxic: pE > 7 Eh > 400 mV• Suboxic pE 2-7 Eh < 100-400 mV• Anoxic pE < 2 Eh < 100 mV• This classification ignores pH.• A better classification associates suboxic zones with

the reduction of Fe and Mn oxides, but not the reduction of sulphate and anoxic zones with sulphate reduction.

pE and Eh in Natural Systems

• Interpreting the Eh or pE of a system, especially a natural system, is extremely difficult.

• Measurements rely on the assumptions that the redox system is reversible, i.e. the kinetics of the relevant reaction are fast, and that the system is in equilibrium.

• Neither of these assumptions are likely to be true in most cases.

• Additionally, any number of reactions could contribute to the measured value, further complicating interpretation.

Free Oxygen• The presence of O2 increases the redox potential of a

system and makes it more oxidized.• Coupling oxygen’s reduction half-reaction (Eho =

1229 mV) with any redox couple having a lower Eh value will result in an energetically favourable redox reaction.

• In the shallow subsurface, micro-organisms control the redox potential along with the redox couple of the electron acceptor.

• Oxygen is the preferred acceptor because it is most easily reduced to water of the available acceptors.

Electron Acceptors

• Alternate electron acceptors, in order of preference (based on Eho values) are:

• Mn(III) > Mn(II)• NO3- > N2 (or other reduced N forms)• Mn(IV) > Mn(II)• Fe(III) > Fe(II)• SO4

2- > S2- (or H2S)• The order of preference is due to the redox

potentials for the half-reactions.

Redox PotentialsHalf Reaction Eho (mv) pEo

Mn3+ + e- = Mn2+ 1510 -25.51

MnO(OH)(s) + 3 H+ + e- = Mn2+ + 2 H2O 1450 -24.49

0.2 NO3- + 1.2 H+ + e- = 0.1 N2(g) + 0.6 H2O 1245 -21.03

0.5 MnO2 + 2 H+ + e- = 0.5 Mn2+ + H2O 1230 -20.78

0.25 O2(g) + H+ + e- = 0.5 H2O 1229 -20.76

Fe(OH)3(s) + 3 H+ + e- = Fe2+ + 3 H2O 1057 -17.85

0.5 NO3- + H+ + e- = 0.5 NO2

- + 0.5 H2O 834 -14.09

Fe3+ + e- = Fe2+ 742 -12.53

0.5 O2(g) + H+ + e- = 0.5 H2O2 682 -11.52

0.125 SO42- + 1.25 H+ + e- = 0.125 H2S + 0.5 H2O 303 -5.12

0.167 N2(g) + 1.333 H+ + e- = 0.333 NH4+ 279 -4.71

0.125 CO2(g) + H+ + e- = 0.125 CH4 + 0.25 H2O 169 -2.85

H+ + e- = 0.5 H2(g) 0 0.00

pH and Eh Stability Fields

Eh

pH0 2 4 6 8 10 12 14

+1.0

+0.8

+0.6

+0.4

+0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0hydrogen

oxygen

water

pH and Eh• As groundwaters become reduced, their pH tends to move toward

neutrality. • When the pH is initially low, H+ consumption in the reduction

reactions increases the pH.• For example:

MnO2(s) + 4H+ + 2e- = Mn2+ + 2H2O

• If the pH is initially basic, then precipitation of metal ions such as Fe2+ and Mn2+ as hydroxides, carbonates, or sulphides tends to lower the pH.

• For example:

Fe2+ + 2H2O = Fe(OH)2(s) + 2H+

Fe2+ + HCO3 = FeCO3(s) + H+

Redox Kinetics

• Redox reactions tend to be “slow” because:

– numbers of microorganisms are small

– reactants are not easily metabolized

• Large pEo values make reactions essentially irreversible (hard to get [Red]/[Ox] to exceed pEo)

– dissolved oxygen will continue to oxidize organic carbon until all organic matter is destroyed

• pE can be calculated from many couples and is rarely the same because of disequilibrium (ie Fe(II)/Fe(III) may give a different pE to S2-/SO4

2-)

Dominant Couples

• Constant Eh (or pE) exists when the concentration of one of the couples is much greater than the other:

CH2O + O2 = CO2 +H2O

• The O2/H2O couple dominates CH2O/CO2 in natural systems such that the dissolved oxygen concentration only changes marginally to oxidize all the organic matter.

Sulphate/Sulphide System

• The sulphate/sulphide couple controls the mobility of many metals.

• (ST) exists as SO42-, H2S, HS-, S2- with concentration

controlled by equilibrium relationships (like carbonate and silicate systems).

• In sulphate the oxidation state of S is S(+VI) and in the other species S(-II).

• In oxic environments (Eh > 400 mV) SO42- dominates

• Under anoxic conditions (Eh < 100 mV) the S(-II) species prevail.

Redox and Metals Mobility

• When SO42- dominates metal concentrations are high

because there are no significant solubility constraints for Fe, Ni, Co, Cu, Zn etc

• Under anoxic conditions where S2- is present almost all the sulphur and metals are precipitated because of the very low solubilities of metal sulphides.

• pH changes the system significantly and Eh-pH diagrams are useful to examine metal mobilities in aqueous solutions.

• For example, Fe3+ is stable in solution only at high Eh >790 mV and low pH < 3.

Iron Stability Fields

Eh

pH0 2 4 6 8 10 12 14

+1.0

+0.8

+0.6

+0.4

+0.2

0.0

-0.2

-0.4

-0.6

-0.8

-1.0

Iron Oxides

Iron Sulphides

Fe3+

Fe2+