Lecture No. 2. Chemical Fundamentals

Recommended Reading: Berner and Berner, pages 141-171.

The major process that releases contaminants from natural and anthropogenic materials is weathering.One of the most important reactions is the weathering of sulfide minerals, because sulfides commonly contain metals and metalloids and their oxidation releases metals and can form acid.Let's look at some examples of these reactions.

Formation of Sulfate Minerals and Acid

When sulfide minerals come into contact with air and water, they oxidize. The sulfide normally is oxidized to sulfate. Sulfides themselves do not produce acids, but they do liberate soluble metals and sulfate. Examples:

Chalcocite: Cu₂S + 2 1/2 O₂ + 2 H⁺ --> 2 Cu²⁺ + SO₄^2- + H₂O

Covellite: CuS + 2 O₂ --> Cu²⁺ + SO₄^2-

Arsenopyrite: FeAsS + 5 1/2 O₂ + 3 H₂O --> 2 Fe²⁺ + 2 H₃AsO₃ + 2 SO₄^2-

In these examples, chalcocite consumes acid, because copper is oxidized from Cu (I) to Cu (II); covellite oxidation is a completely neutral reaction, neither producing nor consuming acid; and arsenopyrite oxidation produces the very weak arsenous acid.

Minerals containing the disulfide group, pyrite and marcasite, do produce acid.

FeS₂ + 3 1/2 O₂ + 3 H₂O --> Fe²⁺ + 2 SO₄^2- + 2 H⁺

Furthermore, oxidation of iron from Fe (II) to Fe(III) produces more acid at common pH values:

2 Fe²⁺ + 1/2 O₂ + 5 H₂O --> 2 Fe(OH)₃ + 4 H⁺

since Fe³⁺ removes hydroxide ions from water. Thus oxidizing one mol of pyrite (or marcasite) produces 4 mols of acid:

2 FeS₂ + 7 1/2 O₂ + 7 H₂O --> 2 Fe(OH)₃ + 4 SO₄^2- + 8 H⁺

Other sources of acid include oxidation of iron liberated from other minerals, such as arsenopyrite and pentlandite ((Fe, Ni)₉S₈), and oxidation of arsenic (III) to arsenic (V), converting the weak arsenous acid to the somewhat stronger arsenic acid:

H₃AsO₃ + 1/2 O₂ --> H₂AsO₄^- + H⁺

Because pyrite is commonly present in sulfidic rocks at a fairly high concentration, and because it is the principal source of reduced iron in most rocks, pyrite is the source of most acid mine drainage.

Many of the reactions above are a combination of acid-base and oxidation-reduction reactions.Let's look at some rules for balancing these reactions.

Balancing Oxidation-Reduction Equations in Aqueous Systems

This can be done in 5 steps:

1. Establish what the reactants and products are.
2. Find the oxidation states of reactant and product elements.
3. Balance electrons.
4. Balance elements not subject to redox reactions. Use H₂O, H⁺, or OH^– as appropriate to balance hydrogen and oxygen.

• Under acidic conditions, H is supplied by H⁺ and O is supplied by H₂O.
• Under basic conditions, H is supplied by H₂O, and O is supplied by OH^–.
• Under neutral conditions, both H and O are supplied by H₂O.
• Make sure that both H⁺ and OH^– are not present together on the same side of an equation! If both are present, combine them to make H₂O until one or the other is used up.
• If there are H⁺, OH^–, or H₂O on both sides of the equation, eliminate the excess until they are only on one side. (We only put as much water into the equation as we have to.)
• Hint: if oxygen is reduced to the –2 state and is not making CO₂, FeOOH, or some other oxide, make water with it. This may require H⁺ to balance the equation.
• Make sure your equation is consistent with the original conditions. If a reaction takes place in acidic solution and appears to produce hydroxide, add a hydrogen ion on the left and convert the hydroxide to a water on the right. Similarly, if a reaction takes place in basic solution and appears to produce a hydrogen ion, add a hydroxide on the left and change the hydrogen ion on the right to a water.

5. Check the charge balance and element balance.

Example 1: The air oxidation of pyrite under near-neutral conditions.

1. Establish reactants and products. Pyrite and oxygen are reacting to produce soluble iron (II) ( Fe²⁺ ) and sulfate. Since the solution pH is near neutral, sulfate is completely ionized. (Below pH 2, HSO₄^– would be present; more on this later.)

FeS₂ + O₂ --> Fe²⁺ + SO₄^2–

2. Oxidation states of reactants and products. Iron is in the +2 oxidation state before and after the reaction. Sulfur is oxidized from –1 to +6, while oxygen is reduced from 0 to –2. Each sulfur atom thus gives up 7 electrons, while each oxygen atom gains two electrons.

3. Balance electrons. Each disulfide ion gives up 14 electrons; this requires 7 oxygen atoms. But oxygen is present as a diatomic molecule, so this amounts to 3 1/2 oxygen molecules.

FeS₂ + 3 1/2 O₂ --> Fe²⁺ + 2 SO₄^2–

4. Balance elements not subject to redox reactions. Since there are 7 oxygen atoms on the left and 8 on the right, we need an additional source of already-reduced oxygen to make up the difference. Since we have stipulated near-neutral conditions, H₂O will be our source of O^2–.

FeS₂ + 3 1/2 O₂ + H₂O --> Fe²⁺ + 2 SO₄^2- + 2 H⁺

Note that breaking up water to supply the O^2– for sulfate produces 2 H⁺ ions.

5. Check the charge balance and element balance. The net charge on both sides is zero. On each side there are one Fe, 2 S, 2 H, and 8 O atoms.

Note that, once the equation is balanced this way, we can predict that the reaction will produce acid.

Example 2: Air oxidation of dissolved sulfide to sulfur under mildly basic conditions.

1. Establish reactants and products. Under mildly basic conditions, sulfide is present as HS^–. Reactants are HS^– and O₂; assume for the moment that the products are S and H₂O.

HS^– + O₂ --> S + H₂O

2. Oxidation states of reactants and products. Sulfur is oxidized from the –2 state to the 0 state; oxygen is reduced from the 0 state to the –2 state.

3. Balance electrons. Each sulfur gives up 2 electrons and each oxygen accepts 2; thus we have the same number of S atoms as O atoms. The 1 HS^– ion requires 1/2 O₂ molecule.

HS^– + 1/2 O₂ --> S + H₂O

4. Balance elements not subject to redox reactions. There is only one H on the left, and there are 2 on the right. But we cannot add H⁺ on the left, because this is a basic solution. Adding a water molecule on the left would not help:

HS^– + 1/2 O₂ + H₂O --> S + OH^– +H₂O

Instead, we let the product containing oxygen be hydroxide ion (which is consistent with the given basic conditions):

HS^– + 1/2 O₂ --> S + OH^–

5. Check charge balance and element balance. The net charge is –1 on each side. On each side there are 1 S, 1 H, and 1 O atom.

Here's a more complex example from Jim Castro who did his Ph.D. on arsenic geochemistry here at U of M a couple of years ago:

Air oxidation of arsenopyrite under near-neutral conditions.

Arsenopyrite is FeAsS. We know that iron is in the +2 oxidation state, since Fe (III) is unstable in the presence of reduced sulfur. But what are the oxidation states of the sulfur and arsenic?

Elemental S has 6 valence electrons, and elemental As has 5. The combination or S and As has a net –2 charge (because the Fe has a net +2 charge), so there are 6+5+2=13 valence electrons in all for AsS^2–. The only arrangement of covalent bonds between As and S that will not violate the octet rule is one with a single bond between the atoms. Since sulfur is more electronegative than arsenic, we give a full octet of electrons to sulfur. This leaves arsenic with 7 valence electrons.

Now we count the electrons. Sulfur has 6 electrons in nonbonding orbitals, plus 2 that it shares with arsenic; 6 + ¸ (2) = 7. Thus sulfur has 7 valence electrons, or one more than a neutral sulfur atom, so we assign an oxidation state of –1 to sulfur. Arsenic has 5 nonbonding electrons plus a shared pair; 5 + ¸ (2) = 6. Thus arsenic has 6 valence electrons. Elemental arsenic has 5, so we assign a –1 oxidation state to As too.

We are going to assume that the oxidation proceeds only to the extent of raising S to the +6 state, raising As to its lowest oxidation state that is stable in the presence of oxygen (+3), and leaving Fe in the +2 state. The products thus will be Fe²⁺, SO₄^2–, and H₃AsO₃. (Arsenous acid is weak enough so that it remains substantially unionized under neutral or acidic conditions.) We have now accomplished step 1.

1. Establish reactants and products. Arsenopyrite plus oxygen produces Fe²⁺, SO₄^2–, and H₃AsO₃.

FeAsS + O₂ --> Fe²⁺ + SO₄^2– + H₃AsO₃

2. Oxidation states of reactants and products. On the left side, As and S are each in the –1 state, while oxygen is in the 0 state. On the right, As is in the +3 state, S is in the +6 state, and O is in the –2 state.

3. Balance electrons. Each AsS^2- group gives up 11 electrons (7 from S, 4 from As), so this requires 5¸ oxygen atoms. We will not split atoms (requires a DOE license), so we will say 2 AsS^2- groups require 11 oxygen atoms, or 5¸ O₂ molecules.

2 FeAsS + 5¸ O₂ --> 2 Fe²⁺ + 2 SO₄^2– + 2 H₃AsO₃

4. Balance elements not subject to redox reactions. There are 11 oxygen atoms on the left and 14 on the right; this requires 3 O^2– to be supplied by water. There are no hydrogen atoms on the left and 6 on the right; this will require 6 H⁺ to be supplied by water. As it happens, 3 water molecules supply just the right amount of H⁺ and O^2–.

2 FeAsS + 5¸ O₂ + 3 H₂O --> 2 Fe²⁺ + 2 SO₄^2- + 2 H₃AsO₃

5. Check the charge balance and element balance. The net charge is zero on both sides. On each side there are 2 Fe, 2 S, 2 As, 14 O, and 6 H atoms.

Measurement Fundamentals

A. pH is a measure of H⁺ activity.

The hydrated hydrogen ion (or hydrated proton) is sometimes written as H₃O⁺. However, there is no reason to believe that the proton is associated with just one water molecule, except in the case of solid oxonium salts. In fact, the proton in solution is associated with a fairly large number of water molecules. We will just write it as H⁺ and understand that it is hydrated, just as all other dissolved ions are hydrated.

pH ¼ – log₁₀ {H⁺}

That is, the pH is defined as the negative base ten logarithm of the hydrogen ion activity expressed as moles per liter.

[Strictly speaking, ion activity is normally expressed as molality, or mols per kilogram of water. We will be working with dilute solution, in which 1.00 kg of water makes 1.00 L of solution. Therefore we can use molarity without introducing significant error.]

The pH is extremely important to the mobility of contaminants, especially metals and metalloids. Does anybody have an idea why pH is important?

It is not a simple solubility effect in most cases. Adsorption-desorption equilibria are also important. We will get into the reasons for this later. In any case, low pH correlates closely with high dissolved metals and low metal content in the sediment. High pH correlates with low dissolved metals and high sediment metals. This is true for Fe, Cu, Cd, Mn, Zn, Pb, Hg, and many other metals.

B. Oxidation-Reduction (Redox) Chemistry

E_h (or E_H) ¼ redox potential

pe ¼ –log₁₀ {e^–}

The pe value is used by analogy with pH. The quantity {e^–} is considered the electron activity in units of molarity. A one-molar electron activity would have a pe of zero.

What does all this mean?

The E_h is a measure of the free energy of a chemical reaction. Free energy (in kJ/mol) can be converted to an electrical potential, E⁰, by the equation

D G⁰ = – n F E⁰

where D G⁰ is the molar free energy of the reaction under standard conditions (which generally means 1 M concentrations of all reactants at one atmosphere pressure and at 25ÐC). E⁰ is referred to as the standard electromotive force or standard potential of the reaction and is measured in volts. If a reaction proceeds forward spontaneously, D G is negative (the system gives up energy or does work), and E is positive. F is the Faraday constant, 96487 coulombs/equivalent, and n is the number of equivalents (mols of electrons) involved in a reaction. For example, in the oxidation of sulfur from sulfide to sulfate, one mol is 8 equivalents of sulfur.

Under other than standard conditions, D G is the free energy and E (or E_h) the electromotive force of the reaction, and

D G = –nFE

For a chemical reaction that can be written as

A + B --> C + D

The equilibrium constant may be written as

where {C} is the activity of a chemical species C. Under non-equilibrium conditions, we will define the equilibrium constant expression as Q:

The value of DG under non-standard conditions (different concentrations, different temperatures) can be calculated by the expression

where R is the gas constant (8.3143 J K^–1 mol^–1) and T is the absolute temperature (K). At equilibrium, D G = 0 and Q = K, so the expression reduces to

This means the equilibrium constant can be calculated from the free energy of the reaction, or vice versa.

Since the redox potential of the reaction can be calculated from the free energy, this means we can also calculate E_h if we know E⁰:

Many redox potentials are calculated using half reactions, e.g.,

Fe^{2+ --}> Fe³⁺ + e^–

E⁰ = –0.771 V

and

2 H₂O (liq) --> O₂ + 4 H⁺ + 4 e^–

E⁰ = – 1.229

The rules for manipulating these equations are fairly simple:

1. If an equation is written backwards, the sign of the potential changes.
2. If the entire equation is multiplied by a constant (e.g., everything is doubled), the potential is unchanged. (This is because the units are volts — joules per coulomb — so doubling the coulombs does not change the volts.)

We can combine the half-reactions above to get the reaction potential for the oxidation of Fe²⁺ to Fe³⁺ by oxygen in acid solution.

First we double the equation for the iron half-reaction:

2 Fe²⁺ --> 2 Fe³⁺ + 2 e^–

E⁰ = -0.771 V

Then we reverse and halve the oxygen half-reaction equation:

1/2 O₂ + 2 H⁺ + 2 e^– --> H₂O

E⁰ = + 1.229 V

Combining the two, we get

2 Fe²⁺ + 1/2 O₂ + 2 H⁺ --> 2 Fe³⁺ + H₂O
E⁰ = + 0.458 V

Note that this relationship is valid at 1 atmosphere oxygen pressure and at pH 0 (1 M acid).

Half-reaction redox potentials are listed in tables in various handbooks.

There is another application for the half-reaction. If we consider a half-reaction as a complete reaction, with the electron as a solute, we can work out what (fictitious) electron activity corresponds to a given E_h. Consider the iron half-reaction again, and write it as an oxidation:

Fe²⁺ --> Fe³⁺+ e^–

E⁰ = –0.771 V

Assume that Fe²⁺ and Fe³⁺ are at unit concentrations and that the electron activity is sufficient to keep the system at equilibrium (so E⁰ = 0 when pe = pe⁰). Then

Converting to base 10 logarithms,

If we let T = 10 C (= 283 K), and note that the Fe²⁺ and Fe³⁺ activities cancel out, then

E_h = –0.05615 log₁₀ {e^–} = 0.05615 pe

if E_h is expressed in volts, or

pe = 17.81 E_h

at 10 C. More generally,

pe = (5040.1/T) E_h

Note that pe is a fiction; reduction is not accomplished by solvated electrons, but rather by molecules or ions that readily give up electrons. However, it is a handy fiction. In practical terms, pe and E_h are easily convertible parameters that describe the redox state of a reaction. Many reaction equilibria can be described in terms of the sum of pH and pe. (More about that later.)

Note that E_h is tied to a specific reaction. When people speak of "system E_h" they are assuming that all potential reactions are at equilibrium. This is in general not true, although in a few rare instances (e.g., simple systems at low pH) it may be true. But "system E_h" is a dangerous term to throw around.

E_h is customarily measured using a platinum electrode. It only measures voltages that are generated by chemical species that react readily on the platinum surface. This leaves out many reactions, such as air oxidation of solid minerals.

Additional notes on redox chemistry

The formula D G = nFE is just a way to convert units from D G (J/mol) to E (volts, or J/coulomb).

If you are not sure of the oxidation state of an element in a compound, you can usually work it out from the formula and from the following principles. In substances important in environmental geochemistry,

• O in compounds is always in the –2 state. (Peroxides are not important.)
• H in compounds is always in the +1 state. (Hydrides are not important.)
• C in inorganic compounds (CO₂, carbonates) is in the +4 state. (Carbon monoxide is not important.)

To previous lecture: Lecture No. 1. Introduction to Environmental Geochemistry

To next lecture: Lecture No. 3. Chemical Fundamentals II.

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