Chemistry of the Environment
Third Edition
Thomas G. Spiro
Kathleen L. Purvis-Roberts
William M. Stigliani
University Science Books
www.uscibooks.com
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CHAPTER 15
Oxygen and Life
Preview
We now turn to water as the medium that supports life. All organisms require
water, and a large fraction of them make their home in rivers, lakes, and the
oceans. Life started in the ocean and occupied dry land only later. Moreover,
biological processes have a profound influence on the chemistry of natural
waters, and indeed of the entire globe. Were it not for the evolution of photosynthetic organisms, first in the ocean, and then on land, the atmosphere would
be devoid of oxygen. The profound influence of oxygen on the chemistry of
the atmosphere was considered at length in Part II. Oxygen is also the dominant actor in the chemistry and biochemistry of the hydrosphere. The limited
availability of O2 in H2O sets the boundary between aerobic and anaerobic
life, with crucial consequences for water quality and the health of ecosystems.
Topics include the following:
• Energy from redox reactions.
• Biological oxygen demand.
• Oxidation–reduction reactions in water systems.
• Relationship between pE and pH.
• Ecological aspects of water, including nutrients, eutrophication, and
anoxia.
15.1 Redox Reactions and Energy
Life is powered by redox reactions, chemical processes in which electrons
are transferred from one molecule to another, with the release of energy.
­Organisms have evolved machinery, made up of proteins and membranes,
which channels this energy into the biochemical pathways that support vital
functions.
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Chapter 15. Oxygen and Life
In an aerobic environment, the most important biological redox process is
respiraton,
(CH2O) + O2 → CO2 + H2O
(15.1)
which we encountered previously as part of the global carbon cycle (Section
6.7). In this case, carbohydrate molecules provide electrons for the reduction
of oxygen. All higher life forms obtain their energy via respiration. However,
many other redox processes are utilized by bacteria. Indeed, bacteria have
evolved to exploit just about any redox process that is available in nature.
Anyplace where a supply of oxidizable molecules coexist with molecules
­capable of oxidizing them, it is a good bet that bacteria are present that can
utilize the potential redox reaction. The oxidation of iron sulfide (FeS2) by
thiobacillus ferrooxidans in the previous discussion of acid mine drainage
is a good example (Section 14.5e).
15.2 Biological Oxygen Demand
Wherever oxygen is present, respiration provides life-supporting redox energy, but in liquid water oxygen can easily become depleted. The solubility of
O2 in H2O is only 9 mg L−1 (~0.3 mm) at 20°C, and less at higher temperatures. (Higher temperature increases the tendency of molecules to escape into
the gas phase, and therefore diminishes solubility for all gases.) The oxygen
supply can be replenished by contact with the air, as in rapidly flowing streams.
But in standing water or in waterlogged soils, the diffusion of oxygen from the
atmosphere is slow relative to the speed of microbial metabolism, and the
oxygen is used up.
Given the centrality of oxygen to metabolism, a parameter called biological oxygen demand (BOD) has been defined to measure the reducing power of
water containing organic carbon. Biological oxygen demand is the number of
milligrams of O2 required to carry out the oxidation of organic carbon in 1 L
of water. Values for various industrial wastes and municipal sewage are given
in Table 15.1.
Table 15.1 Typical BODs for Various Processes
Type of Discharge
BOD (mg O2 L−1 Wastewater)
Domestic sewage
All manufacturing
Chemicals and allied products
Paper
Food
Metals
165
200
314
372
747
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15.3 Oxidation Levels and Water
401
Worked Problem 15.1 Biological Oxygen Demand
Q. What is the BOD of H2O in which 10 mg of sugar (empirical for-
mula CH2O) is dissolved in 1 L? How does this compare with the O2
solubility at 20°C?
A. Since each mole of CH2O requires 1 mol of O2 [Eq. (15.1)], we
divide 10 mg by the molecular weight of CH2O (30 g), to obtain the
required number of moles and then multiply by the molecular weight
of O2 (32 g mol−1), to obtain the number of milligrams:
BOD = 10 mg × 32 g/30 g = 10.7 mg L−1
This value exceeds the O2 solubility (9 mg L−1) by ~20%.
15.3 Oxidation Levels and Water
Many elements can exist in multiple oxidation states, depending on the number of electrons added to or removed from the valence shell of the atoms. In
an aqueous world, the stability of these different oxidation states depends on
the properties of water. Thus we are familiar with Na+, and Mg2+ ions, because
sodium and magnesium have one and two electrons, respectively, in their valence shells, which are easily removed when water molecules are available to
stabilize the resulting ions (Fig. 13.5). All metals form positive ions in water,
and in the case of transition metals multiple oxidation states are available; for
example, iron can exist in water as Fe3+ or Fe2+.
Nonmetals, being electronegative elements, readily attain negative oxidation levels, depending on the number of electrons that their valence shells can
accommodate. Thus the lowest oxidation levels attainable by F, O, N, and C
are −I, −II, −III, and −IV, respectively. We use Roman numerals to denote the
oxidation number, in order to distinguish them from the actual charge. Although Cl− ions exist as such in water, O2− ions do not. The O2− proton affinity
is high enough that it is completely converted to OH− (or to H2O, depending
on the pH). The lowest oxidation levels for N(−III) and C(−IV) are represented by NH3 (or NH4+) and CH4.
Positive oxidation levels are also accessible to the non-metals because of
the stabilization available through bonding to oxide ions. Thus C, N, S, and Cl
are in their maximum oxidation states, +IV, +V, +VI, and +VII, respectively,
when surrounded by oxide: CO2 (or CO32−), NO32−, SO42−, and ClO4−. The actual charges on the central atoms in these molecules are f+4, +5, +6, or +7,
since electrons are shared in the polar but covalent bonds with the O atoms.
Nevertheless, the oxidation state is crucial in determining the possibilities for
redox chemistry. Table 15.2 provides a review of assigning oxidation numbers
to atoms in a molecule. For example, eight electrons must be removed from N
in order to convert NH3 to NO32−.
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Chapter 15. Oxygen and Life
Table 15.2 Assigning Oxidation Numbers
1.The oxidation number of an element that is not combined with another element
is 0.
2.The overall charge of a molecule is equal to the sum of the oxidation numbers
of the atoms in the species.
3.The oxidation number of hydrogen is +1 when combined with nonmetals and
−1 when combined with metals.
4.The oxidation numbers of alkali metals and alkaline earth metals is equal to
their group number (1 and 2, respectively).
5.The oxidation number for halogens is −1, unless the halogen atom is in
combination with oxygen or another halogen higher in the group. The oxidation
number of fluorine is always −1.
6.The oxidation number of oxygen is usually −2. Exceptions include when it
forms compounds with fluorine (see item 5 above), or is found as peroxides
(O22−), superoxides (O2−), and ozonides (O3−).
Table 15.3 Balancing Redox Equations
1.Identify the species being oxidized and reduced from the change in their
oxidation numbers.
2.Separate out the (unbalanced) oxidation and reduction half-reactions.
3.Balance all of the elements in each half-reaction, except O and H.
4.If the reaction occurs in an acidic solution, balance O by using H2O and then
balance H by using H+. If in basic solution, balance O with H2O, then balance H
by adding H2O to the side of each half-reaction that needs an H, then add HO−
to the other side to balance the equation.
5.Balance electrical charges by adding electrons (to the left side for reduction
reactions and to the right side for oxidation reactions) until the charges on the
two sides of the arrow are equal.
6.Multiply all species in either one or both half-reactions by factors that provide
equal numbers of electrons in both half-reactions.
7.Add the two half-reactions together, and make sure that the number of charges
and atoms balance each other out.
In the case of the respiration reaction (15.1), carbon in (CH2O) is in the
oxidation state 0 (the rules are that oxygen counts for −2, and H counts for +1
in determining the oxidation state, of the remaining atoms); four electrons are
transferred to O2 in converting (CH2O) to CO2. Table 15.3 provides a review
of balancing redox reactions in acidic and basic media.
Worked Problem 15.2 Calculating the Oxidation State and
Balancing Redox Equations
Q. What is the oxidation state of N in the nitrite ion (NO2−)? See
Table 15.2 for a review of assigning oxidation numbers.
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15.4 Reduction Potentials
403
A. Since O counts as −2 (−4 total for two O atoms), and there is an
overall −1 charge, N must have an effective charge of +3. The oxidation level is III.
Q. Write a balanced chemical equation for the reduction of NO2− to
NH3 by H2. See Table 15.3 for a review of balancing redox reactions.
A. First balance the number of electrons transferred from oxidant to
reductant. Since the N oxidation number changes from III to −III, six
electrons are transferred. The H changes from oxidation number 0 to
I, so six H atoms, or three H2 molecules, are required to receive the
electrons.
NO2− + 3 H2 → NH3
Since the reaction is in H2O, it is permissible to add H2O, H+, or HO−
to either side of the reaction, as needed. Seeing that nitrite had two O
atoms, we balance these by adding two H2O molecules to the righthand side.
NO2− + 3 H2 → NH3 + 2 H2O
The total H count on the right-hand side is now seven, which we balance by adding one H+ to the left-hand side. This addition also balances the charge.
NO2− + 3 H2 + H+ → NH3 + 2 H2O
15.4 Reduction Potentials
All redox reactions can be divided, at least conceptually, into two reduction
half-reactions, one proceeding forward and the other in reverse. For example,
the oxidation of hydrogen by oxygen,
2 H2 + O2 → 2 H2O
(15.2)
O2 + 4 e− + 4 H+ → 2 H2O
(15.3)
4 H+ + 4 e− → 2 H2
(15.4)
can be divided into
and
Subtracting half-reaction (15.4) from (15.3) gives the overall reaction (15.2).
These half-reactions can actually be carried out at the electrodes of a
­hydrogen–oxygen fuel cell, as discussed in Section 11.2. A potential ­difference
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Chapter 15. Oxygen and Life
is developed between the oxygen electrode and the hydrogen electrode, allowing a current to flow through the external circuit. For the hydrogen–oxygen­
fuel cell, this potential difference approaches 1.24 V, at the standard temperature of 25°C, when the gases are at 1-atm pressure, and the electrodes behave
reversibly, that is, when the reactants and products are at equilibrium with the
electrodes (implying rapid electron-transfer rates).
The potential difference (ΔE) is the energy of the electrochemical cell per
unit of charge delivered. [Specifically 1 V = 1 J C−1, where V = volt, J = joule,
and C (coulomb) is the unit of charge]. Here ΔE is related to the free energy
of the cell reaction by the relation
ΔG = −nFΔE
(15.5)
where F (the Faraday) is the amount of charge in 1 mol of electrons (96,500
C), and n is the number of electrons transferred in the reaction.
Note that Eq. (15.5) sets the convention that ΔE is positive when ΔG is
negative.
Thus in reaction (15.2), 4 electrons are transferred from 2 H2 to O2, and
ΔG = −4 mol e− × 96,500 C mol−1 e− × 1.24 J C−1 = −479,000 J, or −479 kJ.
(Recall that this value, in combination with the entropy of the reaction gives a
theoretical energy conversion efficiency of 80% for the H2/O2 fuel cell, Section
11.2)
Numerous electrode combinations are possible in electrochemical cells,
and it is convenient to specify a standard potential (E°) for each electrode by
referencing it to the hydrogen electrode, whose standard potential is defined
as zero. Thus E° = 1.24 V for the oxygen electrode, represented by halfreaction­(15.3). The standard conditions for E° are unit activities (partial pressure or molar concentration) of the reactants and products, at 25°C.
There are many half-reactions whose electrode potential cannot actually
be measured, because the electron-transfer reaction at an electrode is too slow.
These potentials can nevertheless be calculated from the free energy of appropriate redox reactions. For example, the formation of NO from N2 and O2,
whose thermodynamics was considered in Section 4.2a, is a redox reaction:
O2 + N2 → 2 NO
(15.6)
which can be divided into the half-reactions
O2 + 4 e− + 4 H+ → 2 H2O
(15.7)
2 NO + 4 e− + 4 H+ → N2 + 2 H2O
(15.8)
and
From the free energy of the overall reaction (Section 4.2), 173.4 kJ, we
obtain a cell potential of −0.45 V [using Eq. (15.5)]. Then, knowing that the
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15.5 Natural Sequence of Biological Reductions
405
standard potential of the oxygen electrode is 1.24 V, we can readily calculate that the standard potential for half-reaction (15.8) is 1.69 V [1.24 V −
(−0.45 V)], even though it is impossible to measure this potential directly
because the electron transfer between the electrode and the NO and N2 molecules is too slow to establish a reversible potential.
15.5 Natural Sequence of Biological Reductions
When water is depleted of oxygen, organisms that depend on aerobic respiration cannot survive, and anaerobic bacteria take over. These bacteria utilize
oxidants other than O2. These alternative oxidants are less powerful than O2,
and cannot produce as much energy. Nevertheless, bacteria are quite capable
of surviving on lower energy processes; in doing so, they can fill ecological
niches that are not available to aerobic organisms. The oxidizing power of
anaerobic environments in the biosphere is mainly controlled by five molecules. In decreasing order of energy produced, they are nitrate (NO3−), manganese dioxide (MnO2), ferric hydroxide (Fe(OH)3), sulfate (SO42−) and, in the
absence of other oxidants, carbohydrate can oxidize itself (disproportionation), producing CO2 and methane itself.
The oxidizing power of a molecule depends on the specific reaction being
carried out, and is measured as the reduction potential associated with the
reduction of the oxidant. Important reactions are listed in Table 15.4 for the
Table 15.4 Thermodynamic Sequence for Reduction of Important
Environmental Oxidants at pH 7.0 and 25°C
Reaction
Eh(V)a
Disappearance of O2
O2 + 4 H+ + 4 e− ↔ 2 H2O
0.812
Reduction of NO3− to N2
NO3− + 6 H+ + 5 e− ↔ 1/ 2 N2 + 3 H2O
0.747
Reduction of MnO2 to Mn2+
MnO2 + 4 H+ + 2 e− ↔ Mn2+ + 2 H2O
0.526
Reduction of Fe3+ to Fe2+
Fe(OH)3 + 3 H+ + e− ↔ Fe2+ + 3 H2O
−0.047
Formation of H2S
SO42− + 10 H+ + 8 e− ↔ H2S + 4 H2O
−0.221
Formation of CH4
CO2 + 8 H+ + 8 e− ↔ CH4 + 2 H2O
−0.244
The E° value recalculated for pH 7 is Eh(V) (see Section 15.6).
Source: W.H. Schlesinger (1997). Biochemistry: An Analysis of Global Change (2nd ed.) (San
Diego: Academic Press).
a
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Chapter 15. Oxygen and Life
environmental oxidants we are considering. The listed reduction potentials,
called Eh, are different from the E° values discussed above, because they
­apply when the pH is 7.0, which is more realistic for environmental samples
than the defined standard condition of 1M (H+). The adjustment of the potential for nonstandard concentration is discussed in Section 15.6.
Microbial populations first use the oxidant that produces the most energy
until it is depleted; only then does another agent become the dominant oxidant. Thus, the redox potential of a body of water tends to fall in a stepwise
pattern as BOD increases (Fig. 15.1).
As oxidants are consumed in the conversion of reduced carbon to CO2,
the reduction potential falls to successively lower plateaus, corresponding to
the successively lower potential redox couples O2/H2O, NO3−/ N2, MnO2/Mn2+,
Fe(OH)3/Fe2+, SO42−/HS−, and CO2/CH4. These couples do not give reversible
potentials at electrodes, but the metabolic activity of the vast array of microbes in soils and in water ensure that electron transfer does occur on a time
scale of hours or days (Table 15.4). Consequently, all redox-active materials
respond to the reduction potential established by the microbial activity.
Note, however, that, while there is a general correspondence with the Eh
values of the half-reactions, the plateau potentials in Figure 15.1 deviate substantially from the numbers listed in Table 15.4, because conditions in the
environment are far from the standard conditions that establish the Eh values.
CH2O + O2 → CO2↑ + H2O
Redox potential (mV)
500
CH2O + NO3 → N2↑ + CO2↑
CH2O + MnO2 → Mn2+ + CO2↑
300
CH2O + Fe(OH)3 → Fe2+ + CO2↑
2–
CH2O + SO4 → H2S↑ + CO2↑
100
CH2O + CH2O
or
CH3COOH → CH4↑ + CO2↑
0
–100
–300
0
4
8
12
16
mg C L–1
20
24
Figure 15.1 Sequence of redox reactions in aqueous environments. Oxygen in natural
waters at 20°C is sufficient to oxidize ~3.4 mg of organic carbon (shown here as CH2O)
per liter of water. When the rate of replenishment of O2 from the atmosphere is slower
than the rate of oxidation of CH2O, oxygen is depleted and microbes will select the next
most energetic oxidant in the sequence shown. For simplicity, only major products and
their valence states are shown. (See Table 15.5 for balanced equations.) Adapted from
W.M. Stigliani (1988). Changes in valued capacities of soils and sediments as indicators
of nonlinear and time-delayed environmental effects. Environ. Monitoring Assessment 10:
245–307.
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15.5 Natural Sequence of Biological Reductions
407
While the pH may be close to 7, the concentrations of other reactants and
products are unlikely to be 1.0 M (or 1 atm, in the case of a gas).
The biological oxidation processes supported by these environmental
oxidants are described in Table 15.5. Microbial populations first use the
Table 15.5 Redox Reactions, Products, and Consequences
Redox Reaction
Reaction Products/Consequences
1. O2 + CH2O → CO2 + H2O
The solution redox potential is highest under
aerobic conditions, when there is an
abundance of O2, and little decomposing
organic matter. The end products, CO2 and
water, are nontoxic.
2.4 NO3− + 5 CH2O + 4 H+
→ 5 CO2 + 2 N2 + 7 H2O
When molecular oxygen is depleted, available
nitrate is the most efficient oxidant.
Denitrifying bacteria consume nitrate and
release N2. Nitrous oxide (N2O) is also
released as a side product. Up to 20% of
nitrogen fertilizer applied to agricultural soils
can be lost to denitrification. Denitrifying
bacteria are active in polluted rivers and in
stratified estuaries.
3a.2 MnO2 + CH2O + 4 H+
→ 2 Mn2+ + 3 H2O + CO2
Manganese and ferric oxides are the next
most efficient oxidants, when the nitrate
concentration is low and O2 is depleted. These
minerals are also important for their capacity
to bind toxic heavy metals, organic
compounds, phosphates, and gases. When the
metal oxides are reduced, they become water
soluble, and release the bound materials.
3b.4 Fe(OH)3 + CH2O + 8 H+
→ 4 Fe2+ + 11 H2O + CO2
4a.1/ 2 SO42− + CH2O + H+
→ 1/ 2 H2S + H2O + CO2
Sulfate reduction is common in marine
sediments because of the sulfate in seawater.
In fresh water, this reaction may be important
in areas affected by sulfuric acid (acid rain)
deposition. Hydrogen sulfide (H2S) is an
extremely toxic gas, but is mostly sequestered
by precipitation of metal sulfides in soils and
sediments.
4b.MS2 + 7 O2 + H2O
→ M2+ + 2 SO42− + 2 H+
Conversion of metal sulfide (MS2) to sulfuric
acid (H2SO4) may also occur when anaerobic
sediments are exposed to the atmosphere, as
when dredge soils are raised or when
wetlands are drained, or when coal mines are
left exposed to air (acid mind drainage).
5.CH2O + CH2O → CH4 + CO2
In the absence of oxidants, partially reduced
carbon compounds can disproportionate to
CH4 and CO2. This reaction is more common
in fresh water than marine sediments because
the sulfate concentration is lower
Source: W.M. Stigliani (1988). Changes in valued capacities of soils and sediments as indicators of
nonlinear and time-delayed environmental effects. Environ. Monitoring Assessment 10: 245–307.
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Chapter 15. Oxygen and Life
o­ xidant that produces the most energy until it is depleted; only then does
­another agent become the dominant oxidant.
15.6 Concentration Dependence of the Potential: pH and Eh
What happens to the reduction potential when conditions are not standard? As
in all chemical reactions, the driving force for electrochemical processes depends on the concentrations of reactants and products. This dependence is
given by the Nernst equation
E = E° − (RT/nF ) ln Q
(15.9)
where E° is the standard potential, R is the gas constant, n is the number of
electrons transferred in the reaction, and Q is the equilibrium quotient, that is,
the concentration expression for the equilibrium constant. In the fuel cell reaction (15.2), for example, Q = 1/PO2PH22 (the water activity being defined as
unity), and n = 4. Therefore,
E = 1.24 − (RT/4F ) (−ln PO2 − 2 ln PH2)
A convenient form of the Nernst equation is
E = E° − (0.0591 V/n) log Q
(15.10)
where 0.0591 V is the value of RT/F at 25°C, multiplied by the conversion
factor from natural to base-10 logarithms (ln 10 = 2.303). For temperatures
other than 25°C, the factor 0.0591 V must be raised or lowered accordingly.
The Nernst equation applies equally to whole cell reactions or halfreactions.­Thus the potential of the hydrogen electrode [half-reaction (15.4)]
at 25°C is (after dividing through by n = 4)
E = 0 V − 0.0591 V (log PH21/2/[H+])
(15.11)
From this equation, we see that the hydrogen electrode potential becomes
more negative as (H+) diminishes. Thus H2 gas is a more powerful reductant
in alkaline solution than in acid and E falls by −0.0591 V for every unit rise in
pH. At pH 7, the hydrogen electrode potential is −0.42 V (when all other conditions are standard). This value is the Eh for the H2 reduction half-reaction.
Likewise, O2 is a less powerful oxidant in alkali than in acid, because
protons are consumed in the reduction half-reaction (15.3). The oxygen potential (again after dividing by n = 4) is
E = 1.24 V − (0.0591 V/4) log 0.0591 V (1/PO21/4[H+])
(15.12)
Again the potential drops 0.0591 V for every unit rise in pH and is 0.82 V
at pH 7. This value is the Eh for the oxygen reduction half-reaction (cf., Table
15.4).
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15.6 Concentration Dependence of the Potential: pH and Eh
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Even if no protons appear explicitly in a half-reaction, the potential may
be pH dependent because of secondary acid–base reactions. For example, the
potential of the Fe3+ reduction half-reaction
Fe3+ + e− → Fe2+
(15.13)
has no proton dependence per se, but the equilibrium quotient, [Fe2+]/[Fe3+], is
highly dependent on pH because of the acidic character of Fe3+. At quite low
pH, it forms a series of hydroxide complexes, and precipitates as the highly
insoluble Fe(OH)3 (Ksp = 10−37). In contrast, Fe2+ forms hydroxide complexes
only at high pH, and Fe(OH)2 (Ksp = 1015) is more soluble than Fe(OH)3. Consequently, the reduction potential falls with increasing pH, because [Fe3+] declines more rapidly than [Fe2+].
Worked Problem 15.3 Eh and Ksp of Fe(OH)3
Q. The Fe3+/2+ standard potential [Eq. (15.13)] is 0.77 V. From this
value and the Ksp calculate Eh for the reduction of Fe(OH)3 to Fe2+
(see Table 15.3).
A. Here Eh is the Fe3+/2+ potential at pH 7, when Fe(OH)3 is cer-
tainly precipitated. This potential can be calculated from the Nernst
equation
E = 0.77 V − 0.0591 V{log ([Fe2+]/[Fe3+])}
and [Fe3+] can be calculated from Ksp = [Fe3+][OH−]3. Substitution
gives
E = 0.77 V − 0.0591 V{log [Fe2+] − log (Ksp) + 3 log [OH−]}
At pH 7, [OH−] = 1.0 × 10−7 M
E = 0.77 V − 0.0591 V(log [Fe2+] + 37 − 21)
= −0.17 V − 0.0591 V(log [Fe2+])
The Eh value is −0.17 V, since the standard condition (except that pH
7.0) is [Fe2+] = 1 M.
Worked Problem 15.4 Effective Oxygen Potential
Q. The first plateau in Figure 15.1, corresponding to O2 reduction, is
at 0.5 V, whereas the Eh value (Table 15.5) is 0.82 V. What might account for this difference?
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Chapter 15. Oxygen and Life
A. Assuming that the environmental pH is 7, the difference must
arise from the O2 concentration dependence. The potential diminishes with decreasing O2 concentration. Recall that
E = 1.24 V − (0.0591 V/4) log (1/PO21/4[H+])
(15.14)
E = 0.82 V − (0.0591 V/4) log (1/PO21/4)
(15.15)
or, at pH 7,
If E = 0.50 V, then substituting into Eq. (15.15) gives
log (1/PO21/4) = (−log PO2)/4 = −0.32 V/(−0.0591 V/4) = 21.7
or PO2 = 10−86.8 atm. This value may seem bizarrely low, but it reflects
the fact that when microbes are actively respiring in an aqueous medium, they draw down the O2 to very low levels in their immediate
vicinity.
15.7 Electron and Proton Affinities Are Linked: pE versus pH
Most reduction reactions are accompanied by proton uptake, and oxidations
generally lead to proton release. Since adding an electron increases negative
charge while adding a proton decreases it, the coupling of electron and proton
transfers is a simple consequence of the tendency to lower the energy of the
molecule by neutralizing charge. This coupling leads to a strong dependence
on the solution pH for most half-reaction potentials.
Analagous to pH, we can define pE as −log E, with E in volts. Then, for
the hydrogen electrode [reaction (15.4)]:
pE = 0 − 1/2 log PH2 + log [H+]
(15.16)
If pH2 is maintained at 1 atm, then pE = −pH. This relation emphasizes that
hydrogen gas is far more reducing in alkaline than in acid solution. Likewise,
oxygen is harder to reduce in alkaline solution (or, conversely, O2 is more
oxidizing). From reaction (15.3):
pE = pE° + 1/4 log PO2 + log [H+] = 20.75 − pH at PO2 = 1 atm
(15.17)
Since both the hydrogen and oxygen electrodes have the same pH dependence, the difference between them is pH independent, reflecting the fact that
there is no gain or loss of protons in the overall reaction for hydrogen oxidation by oxygen [reaction (15.2)]. Thus, the potential of the hydrogen–oxygen
fuel cell is independent of the pH of the cell compartments, even though the
individual electrode potentials are strongly affected.
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21.0
17.5
System Fe–O–H
25°C, 1 atm
Fe3+
PO
14.0
2
=1
atm
10.5
pE
7.0
Fe(OH)3
Fe2+
3.5
0
–3.5
PH
2
–7.0
=1
atm
Fe(OH)2
–10.5
–14.0
0
2
4
6
8
10
12
14
pH
Figure 15.2 The pE/pH diagram for a Fe O H system.
The relationship between pE and pH is conveniently illustrated in a diagram, such as that shown for the Fe3+/2+ couple in Figure 15.2. The regions of
the diagram are labeled according to the dominant chemical species present,
and the lines show the pE/pH dependence at the edges of these stability fields.
Thus, the horizontal line at the top left of the diagram represents pE = 13.2,
the value expected for an equimolar solution of Fe3+ and Fe2+ in the absence of
hydroxide reactions. Here Fe3+ predominates above this line, while Fe2+ predominates below the line. The vertical line at pH 3.0 arises because of the
precipitation of Fe(OH)3, which happens when the Ksp is exceeded, which
depends on the pH and on [Fe3+]. For purposes of illustration, the Fe concentration was set at 10−5 M a typical environmental value, in drawing Figure
15.2.
From
Ksp = 10−38 = [Fe3+][OH−]3
(15.18)
we calculate [OH−] = (10−38/[Fe3+])1/3 = 10−11, and pH 3. Above this pH,
Fe(OH)3 precipitates, and [Fe3+] declines in conformity to the Ksp and the pH.
The effect of this decline on pE is seen in the line sloping downward from
pH 3. This line has a slope of −3.0 because of the three OH−/Fe in Fe(OH)3.
Rearranging Eq. (15.18), we have
log [Fe3+] = −38 + 3 pOH
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412
Chapter 15. Oxygen and Life
and since
pOH = 14 − pH
and
pE = 13.2 − log ([Fe2+]/[Fe3+])
(15.20)
the dependence of pE on pH is given by
pE = 13.2 − log [Fe2+] + log [Fe3+] = 22.2 − 3 pH (with [Fe2+] = 10−5 M)
Above this line, Fe2+ is oxidized and precipitates as Fe(OH)3, while below the
line Fe(OH)3 dissolves by reduction to Fe2+.
The second vertical line, at pH 9.0, arises from the precipitation of
Fe(OH)2. From
Ksp = 10−15 = [Fe2+][OH−]2
(15.21)
we calculate [OH−] = (10−15/[Fe2+])1/2 = 10−5, giving pH 9. Above this pH,
Fe(OH)2 precipitates and [Fe2+] declines. The line sloping downward above
this pH represents the phase boundary between Fe(OH)3 and Fe(OH)2. Its
slope, −1, is the difference between the two hydroxides of Fe(OH)2 and the
three hydroxides of Fe(OH)3. Rearranging Eq. (15.21), we have
log [Fe2+] = −15 + 2 pOH
(15.22)
When both hydroxides are present, Eqs. (15.19) and (15.22) can be substituted into Eq. (15.20) to obtain the dependence of pE on pH:
pE = 13.2 − log [Fe2+] + log [Fe3+] = −9.8 + pOH = 4.2 − pH
The top and bottom diagonal lines in Figure 15.2 represent the pE/pH dependence of the hydrogen and oxygen reduction reactions, Eqs. (15.3) and (15.4).
They represent the stability limits for aqueous solutions. Below the bottom
diagonal line, water is reduced to hydrogen, while above the upper diagonal
line, water is oxidized to oxygen.
Figure 15.2 is not a complete diagram of the Fe3+/2+ system because soluble complexes of the ions have been omitted from consideration. Hydroxide
complexes, already mentioned above, are always present in aqueous solutions, but they predominate only in narrow regions of pH and do not greatly
affect the appearance of the diagram. Other complexing agents can have significant effects. The naturally occurring anions chloride, carbonate, and phosphate bind Fe3+ and can lower [Fe3+], and therefore pE, as can organic constituents of soils, especially the humic acids, which can either bind the Fe3+ to
soil particles or form soluble chelates with the Fe3+. Despite these complexi-
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413
ties, Figure 15.2 presents the main features of the Fe3+/2+ system, which is
dominated by the species Fe2+ and Fe(OH)3. Over most of the available pH
range, 3–9, these are the only significant species.
15.8 Biological Oxidations
Bacteria also catalyze oxidation of reduced substances by molecular oxygen,
even though such reactions can occur spontaneously in an aerobic environment. Thus HS− oxidation to sulfate is catalyzed by sulfide oxidizers. These
bacteria manage to extract energy from the HS−/SO42− and O2/H2O redox
­couples. Another important oxidation process is nitrification, the conversion
of NH4+ to nitrate ion. Since plants take up and utilize nitrogen mainly in the
form of nitrate, this is a key reaction in nature, especially in connection
with the use of ammonium salts in fertilizers (see Section 17.2a). The
process actually occurs in two steps, ammonium to nitrite (NO2−) and nitrite
to nitrate:
NH4+ + 2 H2O → NO2− + 8 H+ + 6 e−
(15.23)
NO2− + H2O → NO3− + 2 H+ + 2 e−
(15.24)
These half-reactions are catalyzed by two separate groups of bacteria, Nitrosomonas and Nitrobacter, each utilizing the oxidizing power of O2 to extract
energy from the process. Even though oxygen is used up by bacteria, the
­energy production preserves some oxidizing power (as NO3−) for organic
­matter decomposition–respiration. The nitrate generated in this way can in
turn be used as an oxidant by other bacteria, as discussed above.
In summary, redox potential can be considered as a kind of chemical
switch in the aqueous environment, one that determines the sequence by
which oxidants are utilized by microorganisms. Changes in redox potential
can have important consequences for environmental pollution (Table 15.4).
15.9 Aerobic Earth
Oxygen was not always a constituent of the atmosphere. The leading theory
on O2 formation is that it arose from the evolution of life itself. The primitive
Earth had an atmosphere derived from outgassing of the minerals in the interior. Once the surface cooled sufficiently to condense water, and with it acidic
gases (e.g., HCl and SO2), the main atmospheric constituents would have been
N2 and CO2.
Life arose quite early in the Earth’s history; microfossils resembling
­modern cyanobacteria have been found in 3.5 billion year old rocks. How life
started is unknown, and remains one of the great scientific mystries of our
time. It is known that simple organic molecules are common in the universe,
and are present in meteorites, which would have bombarded the young Earth.
Laboratory experiments show that they could also have been formed from
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Chapter 15. Oxygen and Life
inorganic precursors when subjected to electric discharges from lightning, or
to ultraviolet (UV) irradiation. The UV flux would have been intense, since,
in the absence of an oxygen atmosphere, the Earth would have lacked an
ozone shield. Many of the organic building blocks of organisms could have
been produced in this way. Alternatively, the building blocks might have been
formed on the surfaces of sulfide minerals under the high pressures and temperatures found in hydrothermal vents on the sea floor. (These vents are found
in regions where the crustal plates are being formed through upwelling from
the Earth’s mantle.) Recent experiments show that complex organic molecules can be formed in this way. How the organic building blocks were
­assembled into the first self-replicating organisms remains an unanswered
question, although many ingenious proposals have been put forward.
The first organisms were heterotrophic, assimilating organic compounds
from their environment. Since there was no O2, they must have obtained their
energy from redox reactions other than respiration, similar to the modern anaerobic processes discussed in Section 15.8. The splitting of simple organic
molecules (e.g., acetic acid):
CH3COOH → CH4 + CO2
(15.25)
may have been the first of such processes; this reaction still provides energy
for modern acetogenic bacteria.
However, photosynthesis evolved quite early on, probably in the cyanobacteria mentioned above, which survive today as photosynthetic organisms
in the oceans. Photosynthesis made these organisms autotrophic, capable of
synthesizing their own organic molecules from CO2. They had a strong selective advantage over heterotrophs. In addition to the fossil evidence mentioned
above, carbon isotope measurements on the fossil organic carbon show photosynthesis to be at least 3.5 billion years old. The fossil carbon is found to be
depleted in the stable 13C isotope, relative to 12C, as a result of the slightly
slower diffusion of 13CO2 and its slower rate of capture by the CO2 fixing enzyme ribulose bisphosphate carboxylase.
Oxygen was a byproduct of the rise of autotrophic organisms. Because of
the reactivity of O2, it would have been a toxic byproduct; most anaerobes are
very sensitive to O2, and cannot survive in an aerobic environment. However,
O2 did not become a significant constituent of the atmosphere for a long time
after the advent of photosynthesis, because it was first consumed by oxidizable elements in the ocean and in the earth’s crust, particularly iron and sulfur.
The early ocean would have had a high concentration of Fe2+, which is abundant in silicate minerals of the mantle, and is quite soluble, in contrast to Fe3+.
Photosynthetic O2 would initially have been used up by reaction with Fe2+ to
produce precipitates of Fe(OH)3. Indeed ferric oxide begins to be seen in sedimentary rock that is ~3.5 billion years old, occurring in banded iron formations, in which Fe2O3 [the dehydrated form of Fe(OH)3] is interbedded with
siliceous sediment. These formations reach a peak occurrence in rock that is
2.5–3 billion years old.
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15.9 Aerobic Earth
415
Once the oceanic Fe2+ was used up, the accumulating O2 attacked oxidizable minerals on land, principally FeS2 (pyrite), producing Fe(OH)3 and H2SO4
[the same chemistry that still produces acid mine drainage (Section 14.5e)].
Evidence for this transition is found in the occurrence of red beds, deposits of
Fe2O3 found in geologic layers of terrestrial origin, starting ~2 billion years
ago, after the last of the banded iron formations were formed.
Finally, when the rate of O2 production exceeded its rate of consumption
by exposed oxidizable material, the O2 concentration in the atmosphere began
to rise, permitting the evolution of respiring organisms. Fossil evidence of
eukaryotic organisms has been found in rocks that are 1.3–2 billion years old.
Eukaryotes (in contrast to the more primitive prokaryotes) have mitochondria,
organelles that are specialized for respiration. Some eukaryotes can survive
on O2 at only 1% of the present concentration, suggesting that this level was
attained >1 billion years ago. Oxygen production would have accelerated with
the evolution of chloroplasts in the eukaryotes, organelles that are specialized
for photosynthesis. The rising O2 was also accompanied by the production of
stratospheric ozone, which permitted life to colonize the continents, freed
from the destructive effects of UV radiation. Fossils of multicellular organisms have been found in sedimentary rocks that are 680 million years old, but
the rise of green plants, and with them the modern O2 atmosphere, dates to
400 million years ago.
The time line for the course of O2 production is shown schematically in
Figure 15.3. The present atmospheric reservoir accounts for only ~2% of the
estimated cumulative production of O2, the rest having been used up in the
oxidation of minerals. Interestingly, the O2 concentration seems to have stayed
at ~20% of the atmospheric gases over the last 400 million years; this constancy suggests some sort of feedback control. As with any reservoir (see
Section 14.2), the amount of O2 reflects the balance between the rate of production and the rate of consumption. Over geologic time, O2 consumption
results from exposure and weathering of reduced carbon-bearing rock; this
rate is set largely by the earth’s tectonic movements. Oxygen production results from the burial of reduced carbon, whose rate depends (among other
things) on the total biomass. The biomass is limited, at least in part, by forest
fires, and it is possible that feedback control arises from the dependence of
fires on the O2 concentration. It is known that fires cannot be maintained when
the O2 concentration is <15%, while even wet organic matter burns freely at a
concentration > 25%.a
If carbon burial has balanced O2 accumulation for the last 400 million
years, what accounts for the rising O2 level starting 4 billion years ago? A
much larger carbon burial rate seems unlikely. It has recently been suggestedb
that UV photolysis of CH4 could have provided the driving force. Methane
J.E. Lovelock (1974). Gaia: A New Look at Life on Earth. (Oxford, UK: Oxford University
a
Press)
b
D.C. Catling et al. (2001). Biogenic methane, hydrogen escape, and the irreversible oxidation
of early Earth. Science 293: 839–843.
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Chapter 15. Oxygen and Life
Percent of cumulative O2 production
100
5.1 × 1022 g
80
O2 bound
as Fe2O3
(~83%)
60
40
Beginning of
atmospheric O2
20
Atmospheric
O2 at 21%
0
Occurrence of
continental
Red Beds
O2 bound
as SO42–
(~15%)
Present-day location of O2
416
Molecular
oxygen
(~2%)
Occurrence of
banded iron
formation
4.0
3.0
2.0
1.0
Time (109 years before present)
Today
Figure 15.3 Cumulative history of O2 released by photosynthesis through geologic
time. Of >5.1 × 1022 g of O2 released, ~98% is contained in seawater and sedimentary
rocks, beginning with the occurrence of banded iron formations at least 3.5 billion years
ago (bya). Although O2 was released to the atmosphere beginning ~2.0 bya, it was
consumed in terrestrial weathering processes to form red beds, so that the accumulation
of O2 to present levels in the atmosphere was delayed to 400 million years ago (mya).
[Adapted from W.H. Schlesinger (1997). Origins. Biogeochemistry: An Analysis of
Global Change, 2nd ed. (San Diego, CA: Academic Press).]
production would have been much higher when O2 levels were low; CH4 producing anaerobes would have been abundant, and the CH4 would have ­escaped
to the atmosphere without oxidation. In the absence of the ozone UV shield,
the CH4 would have been exposed to photons energetic enough to break the
C H bonds. At the top of the atmosphere, the light H atoms would have
­escaped earth’s gravitational field, and would have been lost to space. This
removal of oxidizable H atoms from the earth–atmosphere system would provide a mechanism for O2 accumulation.
15.10 Water as an Ecological Medium
15.10a The Euphotic Zone and the Biological Pump
Biological productivity depends on primary producers, organisms that fix carbon via photosynthesis, and provide the food for the animal food chain. In
water, the primary produces are cyanobacteria, phytoplankton, and algae. Because of their dependence on sunlight, they are limited to the region near the
surface, where sunlight can penetrate. This region is the euphotic zone. Its
depth depends on the clarity of the water.
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15.10 Water as an Ecological Medium
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Most biological activity takes place in the euphotic zone. The primary
producers are eaten by animals or decomposed by bacteria, in a continuing
cycle of photosynthesis and respiration. However, because of gravity, some
dead organisms fall below the euphotic zone. In the deeper layers, bacterial
decomposition continues and the waters are enriched in carbon and the other
elements of life. Because of thermal stratification, there is little physical mixing between the warmer surface layer and the cold deep layer. Consequently,
there is a biological pump, which transfers C, N, P, S, and so on, from the
surface to the deep layers and the sediments. Figure 15.4 shows the effect of
biological production on the depth profiles of nitrate and iron, as well as oxygen, in the North Pacific. Oxygen is high at the surface and diminishes sharply
over the first few hundred meters. Nitrate and iron are drawn down at the
­surface, due to uptake by organisms, but increase sharply with depth as the
organisms are decomposed; below the surface layer their concentrations remain at elevated levels.
nmol Fe kg–1
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
µmol O2 kg–1
0
50
100
150
200
250
300
0.0
0.5
1.0
Depth (km)
1.5
2.0
2.5
3.0
O2
NO3
Fe
3.5
4.0
0
10
20
30
40
50
60
70
µmol NO3 kg–1
Figure 15.4 Vertical distribution of O2, Fe, and NO3 in the central North Pacific Ocean.
[Adapted from J.H. Martin et al. (1989). VERTEX: Phytoplankton–iron studies in the
Gulf of Alaska. Deep Sea Res. 36: 649– 680.]
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Chapter 15. Oxygen and Life
In the oceans, the biological pump is responsible for increasing the carbonate concentration of the deep layers with respect to the surface layers. This
drawing down of carbonate from the surface increases the rate of transfer of
CO2 from the atmosphere. This contribution is important to the global carbon
cycle. It has been calculated that the atmospheric CO2 level would double in
the absence of the biological pump.
15.10b Eutrophication in Freshwater Lakes
Because the supply of oxygen is restricted, the species that inhabit an aquatic
ecosystem are in a dynamic balance, one that is easily disturbed by humans.
In water, the O2 concentration falls with increasing distance from the air–
water­interface. Thus, aerated soils support oxygen-utilizing microbes as well
T
Depth
N
N
N
EZ
Sediment
Summer conditions
(a)
T
Depth
EZ
N
N
N
N
N
Sediment
Winter conditions
(b)
Figure 15.5 Seasonal cycling of nutrients in lakes. EZ = thermocline and end of the
euphotic zone; stipple represents phytoplankton growth; T = temperature; N → signifies
direction of nutrient flow; enclosed arrows indicate circulation of waters. The solid line at
the right is the temperature profile of the water column.
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419
as higher life forms, while deeper in the soil, in the saturated zone where the
soil pores are filled with water, anaerobic bacteria dominate and utilize progressively lower Eh redox couples. Likewise in lakes, the sediments are generally oxygen starved and rich in anaerobic microorganisms, while in the
­water column above, the O2 concentration increases toward the surface. The
concentration of O2 at the surface is increased not only because the surface is
in contact with air, but because the surface waters support the growth of vegetation and algae, which release O2 as a product of photosynthesis.
The biological productivity of a temperate lake varies annually in a
cycle (Figs. 15.5 and 15.6). The onset of winter diminishes the solar heating
of the surface. The thermal stratification disappears and the water’s density
becomes uniform, allowing easy mixing by wind and waves, which brings
nutrient-rich waters to the surface. In winter, the nutrient supply is high, but
productivity is inhibited by low temperatures and light levels. Spring brings
sunlight and warming, leading to a bloom of phytoplankton and other water
plants. As plant growth increases, the nutrient supply diminishes and phytoplankton activity falls. Bacteria decompose the dead plant matter, gradually
replenishing the nutrient supply, and a secondary peak of phytoplankton
­activity is observed in the autumn. Because the nutrient supply is limited in
unpolluted waters, the BOD in the surface waters rarely outstrips the available
oxygen.
This natural cycle can be disrupted, however, by excessive nutrient loading from human sources (e.g., wastewaters or agricultural runoff ). The added
Phytoplankton
productivity
Phosphates and
nitrates at surface
Available sunlight
Winter
Spring
Summer
Autumn
Winter
Figure 15.6 Seasonal phytoplankton productivity as a function of sunlight and nutrient
concentration. [Adapted from W.D. Russel-Hunter (1970). Aquatic Productivity (New
York: Macmillan Publishing Co., Inc.).]
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Chapter 15. Oxygen and Life
Oligotrophic lake
Eutrophic lake
Marsh or swamp
Dry land
Figure 15.7 Eutrophication and the aging of a lake by accumulation of sediment.
nutrients can support a higher population of phytoplankton, producing “algal
blooms”. When masses of algae die off, their decomposition can deplete the
oxygen supply, killing fish and other life forms. If the oxygen supply is exhausted, the bacterial population may switch from predominantly aerobic bacteria to mainly anaerobic microorganisms that generate the noxious products
(NH3, CH4, H2S) of anaerobic metabolism. This process is called eutrophication or, more accurately, cultural eutrophication. Eutrophication is the natural
process whereby lakes are gradually filled in with solid material (Fig. 15.7).
Over time, an initially clear (oligotrophic) lake eutrophies, filling with sediment and becoming a marsh, and then dry land. This process normally proceeds over thousands of years because biological growth and decomposition
in the euphotic zone are closely balanced (the surface layers remain well oxygenated, and only a small fraction of biological production is deposited as
sediment). When this balance is upset by overfertilization of the water, the
eutrophication process accelerates greatly.
15.10c Nitrogen and Phosphorus: The Limiting Nutrients
The slow pace of natural eutrophication reflects the nutrient dynamics of an
aquatic ecosystem (Fig. 15.8). The nutrients are assimilated from the environment by the primary producers, which serve as food for secondary producers,
including fish. Dead plant and animal tissues are decomposed by bacteria,
which restore the nutrients to the water. The growth of the primary producers
is controlled by the limiting nutrient, the element that is least available in
­relation to its required abundance in the tissues. If the supply of the limiting
nutrient increases through overfertilization, the water can produce algal
blooms, but not otherwise; therefore, management of the aquatic ecosystem
requires that the supply of the limiting nutrient be restricted.
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15.10 Water as an Ecological Medium
Phosphates
(soil; detergents; sewage)
P
(usually limited)
N2 (atmosphere)
nitrates (soil; sewage)
N
(possibly limited)
421
CO2 (atmosphere)
C
(usually plentiful)
Trace elements
(Fe, Mn, Cu, etc.)
(Sufficient O2)
CO2
NO3–
PO43–
Bacterial decomposition
of plant and animal debris
Primary producers
algae and other nitrogen fixing and
photosynthetic organisms assimilate
C, N, P in the atomic ratios of 106:16:1
Growth of fish and
other secondary producers
(Sufficient O2)
Aging
process
Sedimentation
of plant and animal debris
Figure 15.8 Nutrient cycling in an aquatic ecosystem.
The major nutrient elements are C, N, and P, which are required in the
atomic ratios 106:16:1, reflecting the average composition of the molecules in
biological tissues. Numerous other elements are also required, including S, Si,
Cl, I, and many metallic elements. Because the minor elements are required
in small amounts, they can usually be supplied at adequate rates in natural
waters. On the other hand, carbon, the element required in the largest amounts,
is plentifully supplied to phytoplankton from CO2 in the atmosphere. Phytoplankton outrun the supply of CO2 only under conditions of very rapid growth,
(e.g., in some algal blooms). In these cases, the pH of the water can be driven
as high as 9 or 10 through the required shift of the carbonate equilibrium
HCO3− + H2O → OH− + CO2
(15.26)
The increase in pH can in turn alter the nature of the algal growth, selecting
for varieties that are resistant to high pH.
Normally, the limiting nutrient element is either N or P. Although nitrogen
makes up 80% of the atmosphere, it is unavailable except through the agency
of N2 fixing bacteria, living in symbiotic association with certain species of
plants. On land, these species are rare enough to make nitrogen the limiting
nutrient under most conditions. In fresh water, however, N2 fixing algal species are common, and nitrate ions are often abundant because of runoff from
the land. Consequently, nitrogen is not usually limiting, although it may be in
some regions, especially the oceans, where nitrate concentrations are low.
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Chapter 15. Oxygen and Life
Figure 15.9 Surface blooms of cyanobacteria (Microcystis aeruginosa) in lakes
Mendota and Monona, Madison, Wisconsin. Although the lakes can exhibit temporary
symptoms of nitrogen limitation during summer blooms, eutrophication of these lakes is
driven by phosphorus runoff from agricultural and urban lands. False-color LandSat
image processed to highlight surface bloom. [See color insert.]
Phosphorus remains as the element that is usually limiting to growth. In
fact, recently it was discovered through a 37-year study at the Experimental
Lakes Area in Candada that phosphorous is the main cause of eutrophication,
and if the amount of nitrogen is decreased, more algae blooms actually occur.c
In this study, the experimental lake was fertilized with constant inputs of
phosphorus and decreasing amounts of nitrogen, then during the last 16 years,
phosphorous alone was added. Nitrogen-fixing cyanobacteria were able to
provide the nitrogen inputs necessary from the atmosphere to allow biomass
production in proportion to the phosphorus added to the lake. The lake was
highly eutrophic, despite no additional inputs of nitrogen (Fig. 15.9)
Phosphorus has no atmospheric supply because there is no naturally occurring gaseous phosphorus compound. Moreover, the input of phosphorus in
runoff from unfertilized lands is usually low because phosphate ions, having
multiple negative charges, are bound strongly to mineral particles in soils. In
surface waters, most of the phosphorus is contained in the plankton biomass;
the phosphorus availability depends on recycling of the biomass by bacteria.
c
D.W. Schindler et al. (2008). Eutrophication of lakes cannot be controlled by reducing nitrogen
input: Results of a 37-year whole-ecosystem experiment. Proc. Natl. Acad. Sci. 105: 11254 –11258.
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15.10 Water as an Ecological Medium
423
Some of the phosphorus is lost to the deeper water and to the sediments
when dead organisms sink. When a lake turns over in winter, the phosphorus
in the deep waters is carried to the surface and supports the plankton bloom in
the spring. Whether this phosphorus is available to the surface waters depends
on conditions in the lake. At the bottom, phosphate ions may be adsorbed onto
particles of iron and manganese oxide. However, when the sediment becomes
anoxic (absence of oxygen), the metal ions are reduced to the divalent forms,
the oxides dissolve, and the phosphate ions are released into solution (see
notes on maganese and iron oxides in Table 15.5). Phosphate solubility is also
increased through acidification since at successively lower pH values, HPO42−,
H2PO4−, and H3PO4 are formed (Section 13.5).
Under conditions of phosphorus limitation, human inputs of phosphate
lead to enhanced biological production and the possibility of oxygen depletion. These inputs can arise from sewage, from agricultural runoff, especially
where synthetic fertilizers, which contain phosphate salts, are applied intensively, or where Concentrated Animal Feeding Operations (CAFOs, see Section 17.2c) are located (manure contains phosphates), and from polyphosphates in detergents (Section 14.5b). When phosphorus is added to lakes and
rivers where the availability of phosphorus limits biochemical productivity,
biomass production will increase in proportion to the amount of excess phosphorus added. The increased biomass raises the BOD of the water; as the
BOD increases, oxygen is depleted, leading to anaerobic conditions (anoxia).
The most notorious instance of phosphate-induced eutrophication was in Lake
Erie, which “died” in the 1960s. Excessive algal growth and decay killed most
of the fish and fouled the shoreline. A concerted effort by the United States
and Canada to reduce phosphate inputs was put into effect in the 1970s. Over
$8 billion was spent in building sewage treatment plants to remove phosphates from wastewater, and the levels of phosphate in detergents were restricted. These efforts, along with other pollution control measures, succeeded
in bringing the lake back to life. Commercial fisheries have revived, and the
beaches are once again in use.
15.10d Anoxia and Its Effects on Coastal Marine Waters
Not only is anoxia a problem in freshwater lakes, it can be a problem in the
intermediate or deep waters of an enclosed estuary, gulf, or fjord with restricted circulation between deep and surface waters. If inputs of organic carbon are high, dead biomass sinks into deeper waters where aerobic bacteria
progressively consume the oxygen; if the deep layers fail to mix with surface
layers, the oxygen is not replenished and anoxia sets in (Fig. 15.10). Because
seawater is rich in sulfate salts, the favored reaction under anaerobic conditions is sulfate reduction to hydrogen sulfide (H2S), a chemical that is extremely toxic to fish and humans. Although H2S is generally confined to the
lower layers of seawater, during storms the deeper, anoxic layers can mix with
surface layers, exposing aquatic life to the deadly gas.
Such events occurred in 1981 and 1983 in the enclosed marine areas off
the east coast of Denmark. These events killed unprecedented numbers of fish
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Chapter 15. Oxygen and Life
0%
Mild periodic
Energy to mobile predators
100%
Mild
seasonal
Energy to microbes
424
Severe
seasonal
Persistent
0%
Normoxia
Hypoxia
H2S
Anoxia
100%
Figure 15.10 In healthy waters, known as normoxia, mobile predators feed on the
organisms that live on the seabed, known as benthic organisms. As oxygen is depleted in
the water, a short increase of transfer of energy from mobile predators to benthic
organisms occurs (mild periodic), and then benthic predation takes over under hypoxic
conditions (mild seasonal persistent). Eventually, when no oxygen is left in the water
column, anoxia, microbes process all of the energy and form H2S. [Adapted from R.J.
Diaz and R. Rosenberg (2008). Spreading Dead Zones and Consequences for Marine
Ecosystems, Science 321: 926 –929.]
through suffocation or poisoning by H2S gas (Fig. 15.11). The two episodes
were triggered by sea storms, but the underlying cause was nutrient enrichment of the coastal waters. The enhanced production of biomass (and hence
organic carbon), which, upon bacterial decomposition, outstripped the oxygen
available for aerobic degradation; anaerobic conditions produced H2S, setting
the stage for disaster.
Enhanced nutrient loading affects many coastal areas, creating “dead
zones” where marine life is curtailed. Over 400 dead zones have been identified, covering >245,000 km2 around the world. The most notorious dead zone
in the United States is a 20,000-km2 segment of the Gulf of Mexico, near the
mouth of the Mississippi River (Fig. 15.12), where the O2 concentration is too
low to support aquatic life during the spring and summer. More than 40% of
US commercial fisheries are located in the Gulf of Mexico, and these have
been hard hit by the annual appearance of the dead zone. The size of the dead
zone actually varies by year or season based on the organic content of the
sediments on the shallow shelf of the Gulf of Mexico and mixing of the water
column from hurricanes and tropical storms. If the water column mixes and
the organic sediments oxidize, then there is often lower oxygen demand the
following year.
The Mississippi drains the vast mid-continent farmlands of the United
States, and discharges 1.5 million tons of dissolved nitrogen annually. Agriculture accounts for 80% of this total, 25% from animal manure, and 55%
from synthetic fertilizer. Concentrated Animal Feeding Operations (see
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15.10 Water as an Ecological Medium
†
Oxygen depletion 0.5– 4.0 mg O2/L
Generation of H2S
Fish death
North Sea
Thisted
425
Skagen
Frederikshavn
Alborg
Nordjylland
Viborg
†
Skive
†
Randers
Viborg
Holstebro
Arhus
Ringkobing
Grena
Silkeborg
Ringkobing Herning
Arhus
†
Helsingor
Hillerod
Frederiksborg
Horsens
Vejle
†
Vejle
Ribe
Frederica
Esbjerg
Kolding
Ribe
†
†
Middlefart
Sonderjylland
Fyn
Vest
Kobenhavn
Staden
Kobenhavn
Copenhagen
Roskilde
Koge
Slagelse
Odense
Haderslev
Holbaek
Kalundborg
Nyborg
Korsor
Naestved
Svendborg
Vordingborg
Nakslov
Nykobing
Storstrom
Figure 15.11 Coastal areas of eastern Denmark and southwestern Sweden affected by
oxidation depletion, fish suffocation, and generation of H2S. [Adapted from
H. Miljostyrelsen (1984). Oxygen Depletion and Fish Kill in 1981: Extent and Causes
(in Danish) (Copenhagen: Miljostyrelsen).]
Section 17.2c) are estimated to contribute 15% of the total N and P discharges
to the Gulf of Mexico.
There has been great controversy over whether N or P is the main culprit
in producing dead zones. As noted above, P is generally the limiting nutrient
in fresh water lakes, because of abundant N supplies from the surrounding
land, in the form of nitrate run-off. Nitrate levels are low in the oceans, while
the phosphate in organisms is efficiently recycled, so that N becomes limiting.
In coastal regions, there is a transition from high to low N, so that both P and
N may need to be controlled to reduce overfertilization.
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Chapter 15. Oxygen and Life
Upper Mississippi
Basin
Missouri
Basin
Arkansas
Basin
Lower
Mississippi
Basin
Gulf of Mexico
Ohio
Basin
Dead
Zone
Figure 15.12 The “dead zone” in the Gulf of Mexico due to nutrient enrichment from
land use activities in the drainage basin of the Mississippi River.
The complexities of nutrient cycling in coastal waters are illustrated by
the severely impacted Chesapeake Bay on the US Atlantic seaboard (Fig.
15.13). In winter, cold temperatures and lack of biochemical activity allow the
concentration of O2 to reach its annual maximum. Nitrogen enters in large
amounts because winter is the period of maximum freshwater flow, with accompanying transport of sediment and runoff. Sedimentation removes phosphorus from the water column, partly through settling of organic debris, and
mainly through the precipitation of manganese and iron oxides, which absorb
phosphorus efficiently and are insoluble under aerobic conditions. In the late
spring and early summer, oxygen levels decline due to increased biological
activity. Nitrogen concentrations also decline because (1) nitrogen is incorporated into biomass and sinks as the organisms die, (2) little new nitrogen is
introduced in runoff, and (3) nitrogen is depleted as increasingly anoxic conditions force a switch from oxygen to nitrate as oxidant.
The opposite situation prevails for phosphorus. Under anaerobic conditions, P is liberated from the sediments, in large part due to the reduction of
manganese and ferric oxides to Mn2+ and Fe2+. In the 2+ valence states, the
metals are soluble and release the bound phosphorus formerly adsorbed to the
insoluble oxides of the metals. The P is readily mixed with the surface layers
given the mechanical turbulence of estuarine environments. Thus, as conditions cycle from aerobic to anaerobic and back, the P is continuously recycled
between the surface waters and the sediments. During anaerobic periods,
phosphates are released to the water column to be taken up by microorganisms; during aerobic periods, phosphates are returned to the sediments. The
amount of phosphate trapped in this cycle is vast, much greater than the annual quantities entering the estuaries from sewage effluents or other sources;
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15.10 Water as an Ecological Medium
427
Nitrogen
Late spring
Oxygen concentration (mg L–1)
10
Sediments remove nitrogen from water
during late spring/early summer.
8
6
4
2
0
J
F
M A M
J J A
Month
S
O
N D
Nitrogen
Phosphorus
Late winter
Riverine inputs of nitrogen (as nitrate) are especially
high in late winter/early spring; phosphorus moves
into sediments.
Summer
Phosphorus
Sediments supply phosphorus to the water in
summer as oxygen concentration decreases in the
water overlying the sediments.
Figure 15.13 Oxygen concentration in water overlying the sediments with major
seasonal net fluxes of nitrogen and phosphorus (insets) in the Patuxent River at the
estuary of Chesapeake Bay. [Adapted from C.F. D’Elia (1987). Too much of a good
thing: Nutrient enrichments of the Chesapeake Bay. Environment 29(2): 6 –11, 30 –33.]
it represents the cumulative inputs of many years. Thus, even though Maryland and Virginia banned detergents with phosphates in the 1980s, phytoplankton productivity is still excessive, and nitrogen is likely the limiting
­nutrient.
The Chesapeake Bay receives some of the highest atmospheric NOx emissions in the world, mainly due to the density of traffic in the adjacent areas.
Part of the strategy for cleaning up the Chesapeake Bay might include reducing NOx from vehicle exhausts, demonstrating once again the link between the
atmosphere and the hydrosphere.
15.10e Wetlands as Chemical Sinks
Wetlands are anoxic and have large amounts of organic carbon. They create a
natural buffer zone for nearby fresh or marine waters by trapping nitrates. The
nitrates enter the wetlands in runoff, but are utilized by bacteria to oxidize
stored carbon via the reduction of nitrate to N2 or N2O, which are vented to the
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Chapter 15. Oxygen and Life
NO3–(1,2)
SO42–(2)
N2, N2O
River or lake
S–2(2)
S–2(3)
Wetlands as a sink for nitrate and sulfate
(a)
SO42–(2)
NO3–(1,2)
NO3–
SO42–
River or lake
SO42–
S2–(3)
Dry lands as a transporter of nitrate and sulfate
(b)
Figure 15.14 (a) Ability of wetlands to buffer against nitrate and sulfate inputs to water
bodies; (b) under conditions where wetlands become dry, none of the protective reducing
reactions occur. In addition, accumulated sulfides may oxidize to sulfate as sulfuric acid,
and leach into adjacent rivers or lakes. (1. Runoff of nitrogenous fertilizer. 2. Input from
acid deposition. 3. Sulfide minerals from former marine sediments.) [Adapted from W.M.
Stigliani (1988). Changes in valued capacities of soils and sediments as indicators of
nonlinear and time-delayed environmental effects. Environ. Monitoring Assessment 10:
245–307.]
atmosphere (Fig. 15.14). By depleting the nitrates before they can enter the
estuary, the surrounding wetlands limit the excessive growth of biomass and
subsequent anoxic conditions in the estuary. Destruction of wetlands has
greatly contributed to overfertilization from runoff, and their restoration
would ameliorate the problem.
If the original wetlands are of marine origin, they are likely to contain
high concentrations of sulfur in the form of reduced sulfide minerals (e.g.,
pyrite). Under the redox/pH conditions prevalent in wetlands, these sulfides
are highly insoluble and immobilized (Fig. 15.14a). Draining the wetlands
(Fig. 15.14b) exposes these compounds to oxidizing conditions, producing a
situation similar to acid mine drainage (Section 14.5e). The episodes of fishkills along the Danish coast depicted in Figure 15.11 might have been avoided
if the original coastal wetlands had not been drained.
Another example of this phenomenon occurred in a coastal area of
­Sweden near the Gulf of Bothnia, where wetlands were drained in the early
1900s for use as agricultural lands. As shown in Figure 15.15, draining the
wetland shifted the Eh/pH conditions diagonally to the upper left, from the
values typical of waterlogged soils to conditions close to those of acid mine
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15.10 Water as an Ecological Medium
0.8
Ac
id
0.6
Ox
mi
ne
ze
dm
dra
ge
in-
0.2
eta
ina
Ra
0.4
Eh (volts)
idi
fed
ld
ep
429
os
its
str
ea
ms
gw
G
Wa ater roun
dw
ter
ate
lo
r
Magged
rin so
e s ils
ed
im
en
ts
Bo
0.0
–0.2
– 0.4
–0.6
–0.8
2
4
6
8
10
12
pH
Figure 15.15 A plot of Eh/pH as a function of different aquatic environments. Oval
enclosed by dashed line indicates region of highest solubility of heavy metals. [Adapted
from W. Salomons (1995). Long-term strategies for handling contaminated sites and
large-scale areas. In: Biogeodynamics of Pollutants in Soils and Sediments, W. Salomons
and W.M. Stigliani, Eds. (Berlin: Springer-Verlag).]
drainage. The draining exposed sulfides to the atmosphere, and their oxidation
to sulfuric acid acidified the soil and nearby lakes. The pH in one of these
lakes, Lake Blamissusjon, dropped from 5.5 or higher in the last century to a
current value of 3. Even though agricultural activities ceased in the 1960s, the
lake has not recovered; it is widely known as the most acidic lake in Sweden.
Wetlands store more carbon (~3000 g C m−2 year−1) than does reforested
agricultural land (~100 g C m−2 year−1).d Because dead plants are covered by
water their access to O2 is limited, and decomposition is slow. Organic peat
soils formed from this process have been found that are 60 ft, deep and
7000 –10,000 years old. Freshwater marshes produce CH4, cancelling the
greenhouse gas diminution from the CO2 capture. However, CH4 production
is low in saltwater marshes, because sulfate is available as an anaerobic oxidant (Section 15.5). Thus CO2 absorption by salt water marshes contributes
significantly to greenhouse amelioration.
15.10f Redox Effects on Metals Pollution
Changes in the redox potential can have important consequences for environmental pollution, especially with respect to metal ions (e.g., Cd, Pb, Ni, and
As). In general, the solubility of heavy metals is highest in oxidizing and
acidic environments (Fig. 15.15). However, at neutral to alkaline pH in
­oxidizing environments, these metals adsorb onto the surface of insoluble
Fe(OH)3 and MnO2 particles, especially when phosphate is present to act as a
J. Pelley (2008). Can wetland restoration cool the planet? Environ. Sci. Technol. 8994.
d
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Chapter 15. Oxygen and Life
bridging ion. When the redox potential becomes reducing as a result of microbial action, Fe(OH)3 and MnO2 in soils and sediments are reduced and solu­
bilized. The adsorbed metal ions likewise become solubilized and move into
groundwater [or into the water column of lakes when there is Fe(OH)3 or
MnO2 in the sediment]. Conversely, if sulfate is reduced microbially to HS−
metal ions are immobilized as insoluble sulfides. But as we have seen, if
­sulfide-rich sediments are exposed to air through drainage or dredging operations, then HS− is oxidized back to sulfate, and the heavy metal ions are released.
Important instances of biological redox mediation of metal pollution involve As (Section 19.5c) and Hg (Section 19.5a). Release of arsenate from
Fe(OH)3 laden sediment in the presence of organic matter is the likely mechanism for As contamination of well water in Bangladesh and India. And the
environmental route to Hg toxicity involves sulfate reducing bacteria that live
in anaerobic sediments and methylate mercury, producing the highly toxic
(CH3)Hg+.
15.10g Fertilizing the Ocean with Iron
Although N and P are the limiting aquatic nutrients near land, it has become
evident that in large areas of open ocean, it is actually Fe that limits biological
production. Among the “trace metals” essential for life, Fe is required in
­largest amounts. Iron is utilized in many enzymes involved in electron transport, and in processing O2 and N2, as well as their reduction and (for N2) oxidation products. Thus most organisms require a steady supply of Fe. Since Fe
is abundant in the earth’s crust, Fe limitation is not a problem for land plants,
or for phytoplankton growing near land. However, the concentration of Fe in
the ocean is extremely low (Table 14.1), because of the low solubility of
Fe(OH)3 (Section 15.9) in the alkaline (pH 8) seawater.
In much of the oceans, the settling of dust from the land provides phytoplankton with sufficient Fe for growth. Prevailing winds blow sands from the
Sahara and Gobi deserts far out over the Atlantic and Pacific oceans. Recent
satellite measurements show a fairly good correlation between patterns of
dust in the air and phytoplankton growth in the oceans below. However, there
are large areas that are relatively dust-free, especially in the equatorial Pacific
Ocean, and the waters ringing Antarctica at >60° south latitude, called the
Southern Ocean. These areas have less phytoplankton than could be supported
by the available N and P. It has been known for some time that adding Fe to
samples of these waters stimulates phytoplankton growth in the laboratory,
and a series of field experiments in the 1990s showed that spreading Fe over
areas of nutrient-rich ocean produced phytoplankton blooms.
Iron limitation on biological productivity is an important ingredient in
the carbon cycle, because phytoplankton take up CO2 and transport some of
it to the deep ocean when they die. This reaction is the mechanism of the
­bio­logical pump for CO2, discussed in Section 15.10a. In iron-limited areas,
adding Fe to the oceans could increase the speed of the biological pump,
drawing down the atmospheric CO2. Indeed it has been suggested that Fe
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Problems
431
supplementation could offer a “geoengineering” solution to the problem of
rising atmospheric CO2. However, this solution has been set aside for several
reasons:
1. The remedy would be very expensive, because the Fe stimulation of
phytoplankton blooms is a transient effect. The blooms quickly fade as
the excess Fe precipitates out of the photic zone. (The duration depends
somewhat on the form of the added Fe. Ferrous salts are soluble, but
rapidly oxidize to insoluble Fe(OH)3. Ferric chelates are longer lived,
but the chelating agents (see Section 19.5d.iii, Section 14.5b, and Fig.
14.12) would add to the expense). Consequently, Fe would have to be
added continuously to have a permanent effect.
2. Modeling indicates that the maximum effect on the atmospheric CO2
concentration would be a ~60 ppm lowering, making a relatively small
difference in the rising CO2 level.
3. There could be unforeseeable consequences to the biology of the oceans
from such an intervention.
4. There would have to be international agreement on ocean alteration,
particularly in the region of Antarctica, which is protected by
international law.
The evidence that Fe can fertilize the oceans, and that dust is an important
source of Fe, raises the possibility that changes in global dustiness may have
contributed to the temperature changes that produced the ice ages. Data from
ice cores and from deep sea sediments indicate that there was much more Fe
in ocean water during the ice ages. Thus the biological pump would have been
stimulated; the ~60 ppm lowering in the CO2 level that might have been available from this mechanism corresponds approximately to the CO2 lowering
that is also detected in ice cores (see Section 6.12). The increased Fe might
have resulted from dust due to drying of the continents and expansion of deserts. However, as is usual in reconstructing the past, it is difficult to decide
which factor is cause and which is effect.
Problems
15.1. Calculate the equilibrium partial pressure of
oxygen in a water sample at pH 7.0, which contains equal concentrations of NH4+ and NO3−. E° =
0.89 V for the half-reaction of nitrite to ammonia,
NO2 + 8 H + 6 e → NH4 + 2 H2O
−
+
−
+
and E° = 1.24 V for the half-reaction involving O2
reduction
4 H+ + O2(g) + 4 e− → 2 H2O
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15.2. Copper reacts with dilute nitric acid to form
copper(II) nitrate and nitric oxide gas. Write a balanced equation for this reaction.
15.3. Bromide (Br−) ions and permanganate ions
(MnO4−), react in basic solution to form solid
manganese(IV) oxide (MnO2), and bromated ions
(BrO3−). Balance the equation for this r­ eaction.
15.4. From standard free energies, one can esti-
mate the standard potential of a carbohydrate fuel
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14 June 2011 10:14 AM
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Chapter 15. Oxygen and Life
cell as 1.24 V. Calculate the standard potential for
the CO2/CH2O electrode. Calculate Eh for this
electrode.
15.5. Figure 15.1 indicates that if no other oxi-
dant is available, CH2O is able to oxidize itself
(via microbial biochemistry) to CO2, producing
CH4 in the process. From the electrode half reaction in Problem 15.4 and the data in Table 15.4 and
Figure 15.1, calculate how much free energy per
mole of carbohydrate is available from this reaction. Compare this with the free energy available
from CH2O oxidation by O2.
15.6. Also, using the answer to Problem 15.4 cal-
culate the potential of a solution that is saturated
with CO2 from the atmosphere (~400 ppm), and
contains 0.4 × 10−3 M CH2O. What is the equilibrium partial pressure of CH4 over this solution?
15.7. What is the pE value of an acid mine water
sample having [Fe3+] = 8.0 × 10−3 M and [Fe2+] =
4.0 × 10−4 M?
15.8. Calculate the ratio of NH4+ to NO3− at a pH
of 6.5 for anaerobic water that has a pE of 4.
15.9. (a) What class of molecules is responsible
for most of the reducing power in aqueous
­environments?
(b) What parameter is a measure of reducing
power?
umn of the lake below the euphotic zone during
the summer when there is no circulation with the
upper layer? The bacterial decomposition reaction
is
(CH2O)n + n O2 → n CO2 + n H2O
The solubility of oxygen in pure water saturated with air at 20°C is 8.9 mg L−1; 1 m3 = 1000 L.
15.12. Assume that algae need C, N, and P in the
atomic ratios 106:16:1. What is the limiting nutrient in a lake that contains the following concentrations: total C = 20 mg L−1, total N = 0.80 mg L−1,
and total P = 0.16 mg L−1? If it is known that half
the phosphorus in the lake originates from the use
of phosphate detergents, will banning phosphate
builders slow down eutrophication?
15.13. Name the six most important oxidants in
the aquatic environment, and how the redox potential regulates their reactivity.
15.14. (a) If a lake contains high concentrations
of dissolved Mn2+ and Fe2+, what would be the
concentration of dissolved NO3− and why?
(b) What environmental effect may accompany reduction of MnO2 and Fe(OH)3?
15.15. In anaerobic marine environments, what
toxic gas can be generated and by which reaction
(name reactants and products)?
hundred kg of n-propanol
(CH3CH2CH2OH) are accidentally discharged into
a body of water containing 108 L of H2O. By how
much is the BOD (in mg L−1) of this water increased? Assume the following reaction:
15.16. Explain the “phosphate trap” in the estuary of Chesapeake Bay. Why was a local ban on
phosphorus in detergents not particularly helpful
in mitigating eutrophication in the estuary?
C3H8O + 9/2O2 → 3 CO2 + 4 H2O
lands with high concentrations of organic carbon
can serve as natural buffers against sulfates and
nitrogen oxides (give reactions).
(b) When other oxidants are absent from
such wetlands, which redox reaction is likely
to predominate, and which products will be
­emitted?
15.10. Five
15.11. A lake with a cross-sectional area of 1 km2
and a depth of 50 m has a euphotic zone that extends 15 m below the surface. What is the maximum weight of the biomass (in g of C) that can be
decomposed by aerobic bacteria in the water col-
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15.17. (a) Explain why anaerobic freshwater wet-
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15.18. An estuarine creek in New Jersey contains
large amounts of mercury bound as sulfide (with
K = 10−52) under the prevailing environmental conditions (pH 6.8; Eh = −230 mV). Environmental
scientists have been asked to assess the potential
impacts of the polluted sediments. They conclude
Additional Reading
433
that the mercury poses no danger in its current
state. However, they caution against any action
that would expose it to air and increase its redox
potential. Explain why the scientists come to this
conclusion?
Additional Reading
Metal Cycling in Water
J. Hamilton-Taylor, E.J. Smith, W. Davison, and M. Sugiyama (2005).
Resolving and Modeling the Effects of Fe and Mn Redox Cycling on
Trace Metal Behavior in a Seasonally Anoxic Lake. Geochim.
Cosmochim. Acta 69: 1947–1960.
J.H. Kang, Y.G. Lee, K.Y. Lee, S.M. Cha, K.H. Cho, Y.S. Lee, S.J. Ki, I.H.
Yoon, K.W. Kim, and J.H. Kim (2009). Factors Affecting Metal Exchange
Between Sediment and Water in an Estuarine Reservoir: A Spatial and
Seasonal Observation. J. Environ. Monitoring 11: 2058–2067.
Eutrophication
Y.G. Lee, J.H. Kang, S.J. Ki, S.M. Cha, K.H. Cho, Y.S. Lee, Y. Park, S.W.
Lee, and J.H. Kim (2010). Factors Dominating Stratification Cycle and
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