Environmental Biogeochemistry of Trace Metals (CWR 6252)
Background on pE-pH Diagrams and
Chemical Elements' Aqueous Speciation
(From Vanloon and Duffy, Oxford University Press, 2000)
Both the phosphorus and cadmium distribution diagrams share the obvious limitation that
they describe behaviour of a particular chemical in terms of only a single environmental
variable-pH or chloride concentration, respectively. Clearly. in complex real systems.
there are many variables operating simultaneously. A further step towards accurately
describing natural systems in graphical form is then to create a two-variable diagram. In
such diagrams, the dominant species are plotted in two dimensions as a function of two
independent variables. There are obvious merits in creating diagrams that take into
account the simultaneous involvement of two factors. but we shall see that some information (a detailed description of concentrations) is lost in order to depict the results on a
flat surface.
Avery widely used type oftwo-variable diagram for describing chemical behaviour in the
hydrosphere is the pH/pH diagram. also called a Pourbaix diagram. We shall discuss how to
construct and interpret such plots, but before doing this it is necessary to introduce the
concept of pE.
Analogous to pH. the measure of acidity in aqueous solutions. pH is defined as the
negative logarithm of the electron activity:2
pH = -log a,
(10.33)
A large negative value of pE indicates a large value for the electron activity in solution
which implies that reducing conditions obtain. Conversely. a large positive value of pE
implies low electron activity in solu~on and oxidizing conditions. In practice pE values in
water range from approximately -12 to 25. We Shall see the reason for this range shortly.
While the definition of pE is simple and understandable to chemists. direct measurements ofelectron activity are not easily made in the way that measurements of pH are done.
To show how pE is calculated and measured it is helpful to consider some examples.
- 1-
TWO·VARIABUO DIAGRAMS: plO/pll OIACiRAMS
Consider the simple half reaction
Fe3 +(aq) + e- ;= Fe2+(aq)
(10.34)
(10.35)
1
-=
Keq x
a e-
apeH
a pe 2+
(10.36)
Using the definition ofpE and tal<ing logs of both sides of eqn 10.36, we have
(10.37)
Since
!J.G' = -2.303RT log K"q
(10.3B)
1[:
= -nFE"
(n has the.usual electrochemical
(10.39)
ii'
me~g-=~.e. the number of elec~ons transferred irl the
!I:
·--·--half.reaC't1on);-at-29B..j(·(R·=·B03'!4JK~-mol-and"F=-964B5·G-mol-);.we,have----------.-----;j!
nE"
0.0591
nFE'
logK,q = 2.303,RT
/
(10040)
v'
In this case, n = 1, so
E'
logK"q = 0.0591
(10041)
E'l
(10.42)
and
apeH
pE =--+ og-0.0591
apeH
Under standard conditions. apeH = ape2+ = 1:
(10.43)
and
E'
'--PE- - p E
- 0.0591
(10.44)
For non-standard conditions:
(10045)
When the standard pE' value and the actnal activities (usually approximated by con·
centrations) of Fe3 + and Fe2 + are substitnted into this equation, the pE of a particular
environmental system can be calculated.
-2-
I
il
:11
_ DISTRIBUTION OF SPECIES IN AQUATIC SYSTEMS
In the general case, for a reaction
(10.46)
where A and Bare the oxidized and reduced forms ofa redox couple. the reaction quotient
(Q) is defined as
(10.47)
The reaction quotient takes the form afan equilibrium constant. but uses activities (or, as
an approximation. concentrations) that obtain under any conditions. not just those at
equilibrium. TI,e general form of eqn 10.45 is then
1
(10.48)
pE = pE" - -logQ.
n
10.2.1
Methods of calculating pEa
Values of p~ for a number of half reactions of environmental interest are given in
Appendix 11. When additional values are required. there are several methods that make use
-----
--at:other"""'adily_obtainable.infonuation..J:be.flr.stmethorlinv.olv£s.llsing.e.qnJ.O.~.w.he.re_a
tabulated E" value for a half reaction is available. the pE" is readily calculated. For
Fe'+(aq) + e" ;= Fe'+(aq)
(10.34)
E" = +0.771 V
Therefore
0.771 V
-pE" = + 0.0591 V
13.0.
Being a ratio of two potentials. the pE" value is a dimensionless number.
A second method for calculating pEa makes use ofthe relations in eqns 10040 and 10.44.
nEo
10gKeq = 0.0591 = npE"
(10.49)
_ log K,q .
pE" n
(10.50)
This relation is applicable when an E" value is not available. but where the appropriate
equilibrium constant is known.
In some cases, several reactions may be combined to produce an overall half reaction.
'i- Consider the half reaction for which no tabulated E" value is easily found:
Fe(OH), + 3H,O+(aq) + e" ;= Fe'+(aq) + 6HzO
(10.51)
The reaction is the sum of
Fe(OH),;= Fe'+(aq) + 30H" (aq)
Fe'+ (aq) + e" ;= Fez+(aq)
3H,O+(aq) + 30H" (aq) ;= 6HzO
(10.51a)
(10.51b)
(10.51c)
.__.
.
_
TWO-VARIABLE DIAGRAMS: pE/pH DIAGRAMS
For reaction 10.51a. K,=K,p=9.1 x 10-3 • and log Ka =-38.0. For reaction 10.51b. p~ =
logKb = 0.771/0.0591 and log Kb = +13.0. For reaction 10.51c.
1
Kc = - - = 1042
(Kw )'
and log Kc = +42.0. For the original. overall reaction
log Km = log K, + log Kb + log Kc
= -38.0 + 13.0 + 42.0
= +17.0
Using equation 10.50 (n = 1):
P~verall = +17.0.
There is a third method for calculating pEJ values and this requires combining eqns 10.38
and 10.49:
t>G' = -2.303RTnpB"
-t>G'
.pB" = 2-=:acc03°C'R"'Tc-"11
'--'-'-
(10.52)
_
(1053)
Consider the redox reaction
SO;'(aq) + 10H,O+(aq) + 8e- ;= H,S (aq) + 14H,O
(10.54)
Using thermochemical tables and noting that ~G~ for the aqueous electron is 0, and for
the hydronium ion. 6.G has the same value as for water:
t
t>G' = -27.86 + 14 x (-237.18) - 10 x (-237.18) - (-744.60)
= -231.98kJ
r'
-(-231.98) kJ x 1000J k
2.303 x 8.314Jmol 'K 1 x 298.2K x 8 mol
= 5.08
pB" =
This final method for calculating pB" values is perhaps the most generally useful but.
depending on circumstances. anyone of the three methods may be employed. Once pB"
values are available, it is possible to determine pE for particular non-standard environ~
mental conditions. A two'part example follows.
10.2.2
Chromium in tannery wastes
Traditional leather tanning processes involve treating the hides with an aqueous solution
of chromium. Suppose the waste water from a tannery contains 26mgL- 1 chromium,
originally in the Cr3 + state. As the effluent flows downstream, the dissolved oxygen can
oxidize Cr'+ to Cr,O~-. We will calculate the extent of oxidation for a situation where
oxygen in the stream water is in equilibrium with atmospheric oxygen. and has a pH of
-- 4-
_ DISTRIBUTION OF SPECIES IN AQUAllC SYSTEMS
6.5 (aH,O+ = 10-6.5). The first step is to calculate the value ofpB. We can then use this value
to calculate the concentrations ofCr3 + and Cr20~- assuming that the chromium species are
also in equilibrium with the system.
The relevant reaction for atmospheric O2 in equilibrium with water (often called a
well aerated system) is
0, (g)
+ 4H,O+
(aq)
+ 4e-
;='
6H,O
(10.55)
BO = 1.23 V
o
1.23 V
pB = 0.0591 V
20.8
Using eqn 10.48:
1
1
pB = pE" - -log
•
n
Po,/po x (aH,O+)
1
1
= 20.8 - -log ------;----;=:.
4
0.209 x (10 6.5)'
= 14.1
___.
.__Note_that_in..these.calClllations,..the..pressure..is.giY.en..as.a..ra.tio_oJ'.2g,jE:-whkhjs...n.ume"'D-'--_ _
ically identical to pressure in atmospheres. For oxygen. which makes up 20.9% of the
atmosphere. Po, = 21200 Pa; po = 101325 Pa.
For the Cr system:
Cr,O;- (aq)
+ 14H,O+
E"
(aq) + 6e-
;='
2Cr'+ (aq) + 17H,O
(10.56)
= 1.36V and pE" = 23.0
1
[C,.'+]'
o
pB=pB - - l o g ,
[Cr,O? ](aH,O+)
6
1.
Since the chromium and oxygen systems are in equilibrium. pB is the same for both:
1
14.1 = 23.0 - -6 log
[Cr'+]'
,
1.
[Cr,O? 1(10- 6 .5 )
1
1
1
[C,.'+]'
= 23.0 - -log
-log~--=+-,
6
(10- 6 •5 )14 6
[Cr,O;]
1
[C,.'+]'
, ]
Cr,O?
= 7.8 --log[
6
[Cr'+]'
log [
'I
Cr,O?
[C,.'+]'
[Cr,O; J
-37.8
1.6 x 10-38
This very small ratio indicates that virtually all of the Cr'+ would be oxidized to Cr,O;and this, in fact, has serious environmental consequences. A widely used method ofleather
tanning involves a two-bath process in which the hides are soaked for several hours in a tank
5-
containing chromic acid in order to eff~,t cross-linking between proline and hy4roxyproline residues· in protein molecules in the'animal skin·as·-a-wayoftonghening-the
materiaL The hides are ~~n transferred into ano~ertanI<:containingreducing agents such
as sucrose in order to reduce tl,e exc~ss chromium (VI) to chromiUm (ill). The excess is ~en
discharged in tile waste water.
The leatiler industry is important tilroughout tile globe, witillndia tile world's major
producer of leatiler goods. In that country tilere are large industria) establishments
manufacturing shoes, luggage, garments, and so on, but much oftile industry is small-scale
and scattered in numerous villages in rural areas of the C01,1D.tIy. The l10rthern part of
Tamil Nadu state in Soutillndia is ~~ centre of tile leatiler industry.
The quantity of chromium (ill) wasted in effluents of a small-scale chrome tannery in
India is estimated to be approxhnately 0.41>g per 100 kg ofraw hides. This is about one
half oftilat added in tile first batil. Chromium (Ill) has low toxicity, but in tile hexavalent
form it is botil toxic and carcinogenic to humans. It is also toxic tp plants and large areas
of tile leatller-producing areas ofTamil Nadu are devoid ofvegetation. The consequences
to humans in that region aTe not documented. In any case, the release of such large
amounts of chromium with its-subsequent conversion to the thermodyn<;l.mically stable
Cr20~- species"is a major environmental iss:ue.
---
10.2.3
pE/pH diagrams
We are now in a position to construct a pE/pH diagram. The diagram will talce tile fprm of
a two-dimensional plot ofpE (ordinate, y axis) versus pH (abscissa, x axis) in which areas in
tile diagram define tile regions where particular species are dominant. In saying a particular species dominates. we must define conditions at the boundary between domains.
This requires tilat we calculate and draw lines on tile diagram corresponding to these
boundaries.
An important issue, tilen, is hoW to define tile boundary conditions in different situations. For the case ofa reaction between two species where one is dissolved in the water and
tile otiler is a gas, tile boundary condition is set as P" = 101325 Pa = 1 atm. This is tile
minimum pressure required for gas evolution from the aqueous solution. In other words.
when Pg " is greater tilan 101325 Pa, tile gas phase itself is said to predominate. As an
example, this type of boundary condition is used in tile case of tile evolution ofhydrogen
gas from an acidic solution:
2H,O+ (aq)
+ 2e- "" H2 (g) + 2H20
(10.57)
Anotiler type of condition is required for tile situation where ouly soluble species are
involved, or where tilere is a soluble species reacting to form one that is insoluble. Examples
of these two cases are
Sn4 + (aq)
+ 2e- "" Sn 2+ (aq)
(10.58)
+ 2e-
(10.59)
and
Mn02 .+ 4H3 0+ (aq)
"" Mn2+ (aq) + 6H2 0
In tilese cases, we arbitrarily define a concentration (as an estimate ofactivity) belowwhich
a particular species is considered to be soluble. Any con,entration may be chosen but it is
- b------
- ----- --
-~------
--- - - - - - - -
DISTRIBUTION OF SPEaESIN AQUATIC SVSTEMS
common to use values between 10-5 and 10- 2 mol L-'1. Ofcourse. in constructing a diagram
one should consiste:n.tly use the same concentration for all boundaries.
We will illustrate how boundary conditions are chosen and used in the construction of
a pElpH diagram by considering the aqueous sulfur system where the species of interest
ate SO~- (aq). HSO' (aq). S (s). HS- (aq). and H2 S (aq). Before beginning calculations
involving these species. however, there are some :greliminary calculations common to all
\
diagrams pertaining to the hydrosphere.
As we had noted in Chapter 10. water itself has a limited range of pE and pH values
within which it is stable. Under highly reducing conditions (low pEl. water is reduced:
2H,O + 2e- "" H, (g) + 20I1 (aq)
(10.60)
For this reaction
E" = -0.828 V
pE" = -14.0
Equation 10.48 for this reaction is written as
_ _ _ _ _ _ _ _ _ _ _ pE = pE" -
1
,
2.!Qgi&,/.1'"
x...(!!oH:.)_)..
-.(.lO.61.)
_
(note again that here and in subsequent calculations, we will use concentrations rather
than activities except for hydroxyl and hydronium ions).
For the boundary involving a gas. we choose the condition that
PH,
= 1'" = 101325 Pa
pE = -14.0 -log(aow)
=-14.0+pOH
Since
pH+pOH= 14
then
pE=-pH
This line then defines the boundary for water stability with respect to reduction and is
shown as the lower line on Fig. 10.4. Where the pE value is less than the pH value. water is
unstable. It is possible also to calculate the s;une line by ta1cing the reduction reaction to be
2H,O+ (aq) + 2e- "" H, (g) + 2H,O
(10.62)
Considering the other extreme. highly oxidizing conditions. water is unstable with
respect to O2 evolution and the reaction is written as
(1Q.63)
For this reaction
E" = 1.229V
pE" = EO/0.0591 = 20.80
I
i
TWO-VARIABLE DIAC;RAMS: pE/pH DIAC;BAM~
,pE
4
-12
4
2
6
I=Ig. '~;4 Region {If stability anc! stability bQundari,es ,foq(Vater
"
"
' ,
-----------i::-~:f:.:_,t-::·~-.~:=::;:,~-==j;,:::~';~=,·-L~~..~~Z~.2,.=,_;;~;::,:;;:::~~t.=:::..~C~=,-~:_=..'7:j~_,::::,~).:::_;;,
, On a pflpH diagram.
The production of oxygen from water ~ an oxidation Iea~on. It is important to' note.
however, that we use the MAC couvention ofdefining E" andE, and pE" andpE, in terms of
the reverse, reduction process:
(10.64)
Once again, the boundary condition requires that the pressure ofthe gas equal atmospheric
pressure
Po, = po = 101325 Pa
pE = 20.BO -log(1/Pa,o+)
= 20.80 -pH
This, then, defines the upper line on Fig. 10.4 and the region between the two lines is the
stability region for water on a pElpH diagram. Above the top boundary, water is oxidized
with the evolntion ofoxygej}, while below the lower boundary it is reduced and releases
hydrogen.
.
10.2.4
The sulfur system
Now we can deal with the sulfur species. To create the pElpH diagram, we will be defining
boundaries between species and superimposing these on the water stability diagram. We
may consider the varions pairs of species in any order but it is helpful to have an idea of
what to expect before beginning calculations-for example we expect that HSO' will be
important at low pH and SO~- at high pH. SimiJarlySO~- would exist under oxidizing, high
pE conditions, while HS- is a reduced, low pE form. For all soluble species, we will choose
10-2 mol L_1 as our arbitrary definition for the boundary.
-p
DISTRIBUTION OF SPECIES IN AQUATIC SYSTEMS
The so~- IHSO' boundary
The equation describing this boundary requires hydronium ion, but there is no oxidation or
reduction involved:
(10.65)
To make things easier, we use H+ as an abbreviation for H3 0+, the hydronium ion, but of
course the result would be the same if the latter species were used·in the equations and
calculations.
!:>.GO = !:>.G~ (HSO.) - AG~ (SO~-) - AG~ (W)
= -755.99 - (-744.60) - 0
= -11.39kJ = -11390J
----------------_.
_AGO
log K = 2.303 RT
= _ _-'+.::1:.:1.::3"'90'--__
_ _ _--"2..303...&,8...;lJ.:>-4.ax.,2"'9:Q.8.""2
= 1.995
.
.
K= [HSO.]
[SO~ ]aH'
At the boundary, [HSO'] = [SO~-] = 10- z M, and
1
K=ali'
logK = pH = 1.995
Therefore, the boundary between HSO. and SO~- is a vertical line at pH 1.995. Below this
value, HS04" is the dominant form; above it SO~- is most important. See llne 'a' on Fig. 10.5.
The HSO. ISO boundary
HSO' (aq)+7W (aq)+6e-'= SO (s)+4H zO
(10.66)
AGO = AG~ (S) +4AG~ (HzO) - AG~ (HSO.) -7AG~ (H+) - 6AG~ (e-)
= 0 +4(-237.18) - (-755.99) - 0 - 0
= -192.73 kJ = -192 730J
EO _ -AGo
P - 2.303 nRT
+192730
2.303 x 6 x 8.314 x 298.2
5.626
1
PE=p~-~log
6
[HSO.] (aH' )7
._-
----
---._----------..
--
.,---
At the boundary, [lISO"] = 10-2 molL-1:
7
1
1
1
pE = 5.626 -'-log- - -log-6
aH+
6
10...,.2
7
= 5.626 -i5PH - 0.333
= 5.293 -1.167pH
This is line 'b' on Fig. 10.5; above it is the domain ofHSO"; below it is elemental sulfur, So.
The SO~- fS" bOllndary
SO~- (aq) + 8H+ (aq) + 6e- "" So (s) + 4HzO
!::..Go = !::..G~ (S) +4!::..G~ (H2 0) - !::..G~ (SO~-) - 8!::..G~ (H+) - 6!::..G~ (e-)
= 0 + 4(-237.18) - (-744.60) - 0 - 0
= -204.12Ig = -204120J
EO _ -- ~Go
P - 2.303nRT
~
+204120
2.303 x 6 x 8.314 x 298.2
5.958
1
PE=pE"-~log
[SO~-](aH+)8
6
- 10.._ - -
-- --- - - - - - - - - - - _ . _ - - - - - -
(10.67)
•
DISTRIBUTION OF SPECIES IN AQUA11C SYSTEMS
At the boundary, [SO~-] = 10-' mol L-1:
8
1
1
1
pE = 5.958 - -log- - -log--2
6
aH+
6
10-
8
= 5.958 - i5PH - 0.333
= 5.625 - 1.333 pH
This is line 'c' on Fig. 10.5; above it is the domain of SO~-, below it So.
The 5·/H.5 boundary
S (s) + 2H+ (aq) + 2e- ;= H,S (aq)
(10.68)
l>.GO = l>.G' (H,S) -l>.G' (S) - 2l>.G' (H+) - 2l>.G' (e-)
= -27.86 - 0 - 0 - 0
= -27.86kJ = -27860J
2.303nRT
./-..27..860
2.303 x 2 x 8.314 x 298.2
= 2.400
1
[H,S]
pE = pE" - -log--,
2
(aH+)
At the boundary, [H,S] = 10-' mol L-1:
pE = 2.440 + 1- pH
=3.440-pH
This is line 'd' on Fig. 10.5; above it is the domain of So, below it H,S.
The 50~-/H25 boundary
SO~-
(aq) + lOW (aq) + 8e- ;= H,S (aq) + 4H,O
l>.Go = l>.G' (H,S) +4l>.G' (H,O) -l>.G' (SO~-) -lMG, (H+) - 8l>.G' (e-)
= -27.86 + 4(-237.18) - (-744.60) - 0 - 0
= -231.98kJ = -231980J
-l>.Go
2.303nRT
pE"
+231980
2.303 x 8 x 8.314 x 298.2
= 5.079
E=
P
E" _ ~lo
[H,S]
P
8 g[SO~-](aH+)10
/ {-
(10.69)
TWO·VARIABLE OIAC;BAM$: pElpH DIAC;BA/vIS
At the boundary, [H2S] = [SO~-] = 10-2 molL-':
10
1
pE = 5.079 - -log--
a
aH+
= 5.079 - 1.25 pH
This is line 'e' on Fig, 10.5 (a very small segment, difficult to distinguish from line 'd'): above
itis the domain of SO~-. below it H2S.
The H.S/HS- boundary
H2S (aq) .= HS- (aq)
+W
(aq)
(10.70)
As for the HSO. /SO~- boundary. this is not a redox reaction and therefore the line will be
vertical with the protonated species on the left.
tlGo = tlGf (HS-)
+ tlGf (H+) -
tlGf (H2S)
= 12.08 + 0 - (-27.86)
= 39.94kJ = 39940J
K
-39 940
- tlGo
logK
2.303 nRT =
=-2-=.3-=-03=-X-":8::.3~1:":4~X-2=-9'"'8c-=-.2
= -6.995
[HS-]aH+
[H2 S]
At the boundary, [HS-j = [H2S] = 10-2 molL-':
K=aH+
log K = log[H+j = -pH = -6.995
pH = 6.995
This is line 'f' on Fig. 10.5; to the left is the domain ofH2 S and to the right HS-.
The SO~-/HS- boundary
so~-
(aq) + 9H+ (aq) + 8e- .= HS- (aq) + 4H20
tlGo = tlGf (HS-) HtlGf (H20) - tlGf (SO~-) - 9tlGf (H+) - 8tlGf (e-)
= 12.08 + 4(-237.18) - (~744.60) - 0 - 0
= -192.04kJ = -192040J.
-tlGo
pE" = =--2.-=-30=-3:-nR-=-T
=4.204.
+192040
2.303 x 8 x 8.314 x 298.2
E = E" _ !10
P
p
[HS-]
8 g [SO~-](aH+)9
-/2-
(10.71)
21',
DISTRIBUTION OF SPECIES IN AQUATIC SYSTEMS
At the boundary, [HS-] = [50:-1 = lO- z molL-1:
9
1
B
aH+
pE = 4.202 - -10g= 4.204 -1.125 pH
This is line 'g' on Fig. 10.5; above it is the domain of SO:-, below it HS-.
The completed diagram is shown in Fig. 10.5. To interpret the plot, it is helpful to make
use ofa template (Fig. 10.6) which contains the HzO stability boundaries and within these,
approximate pEjpH regions for a variety of environments.
Two examples of particular environmental situations follow. For mine wastes, at a pH of
2.5 and exposed to the atmosphere, so that they are well aerated (pE", 15), corresponding to
point Xon Fig. 10.5, the most inlportant sulfur species in solution would be sulfate. Where
the originally mined material was sulfide ore, such as from the copper-nJckel Ore$ of
Sudbury, Ontario, Canada, sulfur might initially go into solution as sulfide but would
eventually (the kinetics are relatively fast) be oxidized to sulfate. For a swamp or paddy (rice)
field where soil containing a high content oforganic matter (OM) is submerged, the OM acts
as a reducing agent and creates low pE conditions. For example, such a soil might have
pH = 6 and pE = -3, corresponding to pointY in the figure. This is near several boundaries,
butit would nothe sUlJ'rising to detect the presence ofHzS in the interstitial water of the
sediment. Organic matter in soil typically contains about 2% sulfur.
Additional important pEjpH diagrams will be given in the problems section of this
chapter and others are discussed at later points in the book.
Fig. 10.6 Telnplate for use with pE/p!i plots. The.
i.ndicated ar\,as show typical PI' and pH values for
commonly 'encountered aqueous, soil, and sediment
environments.
';-"
MEASUREM!'NTS OF pE 219
be simple to measure pE in a real environment In practice, non~
equilibrium conditions make stable ~and meaningfp.l readings exceedingly difficult. The
In principle it sllould
requirements for
me~surement
are an indicator electrode which is iliert and develops a
potential in response to the ratio of redox couples whicll are in true equilibrium. A platinum electrode of small size can serve this purpose. In ao.dition, a reference electrode is
used so that the indicated pote!ltial may be measured against a known value. As with all
potential measurements it is essential to' ensure· that negligible c~rrent be drawn during
the measurement prol:ess. This is readily ensured by using an electronic voltmeter.
Figure 10.7 shows an apparatus suitable for determining pE.
Suppose the potential in soil pore water is measured in the field and the value is found
to be +713 mVvs. the saturated calomel electrode (SCE). The value of E is recalculated
vs. the normal hydrogen electrode (NHE), using the known potential, +0.242 V ofthe SCE:
Evs. NHE = 0.713 +0.242V
= 0.955V
~S~tJ'~~a:t~d:';':' '
;-calflrn~I'"
j'efl9:r~npe:
,electrode
.
,':',' '. ': _--_.',.-\---,-__.,'i':. _:".,".,:',r-' :'~,-,',-_- -__
_" -",' .",'
','J
Ag. 10.7 Apparatus for measuring pE' in:" ,
water, soil. or sediment, in, the laboratory
or in the ','
field. The indicator
electrode
is .
...
",
I, often a platinum di~c or wire; any appropriate reference electrode can be used. ~ ,
,p =the potential measuring device. usually'
'an electronic voltmeter.
.,,'
,'
','
...
-14-
'
"
. . . _,
220
DISTRIBUTION OF SPEOES IN AQUATIC SYSTEMS
The value ofpE is then readily calculated using eqn 10.44:
0.955
pE = 0.0591
= 16.23
A pH of16.23 indicates an oxidizing regime.
As has been suggested. frequently non-equilibrium conditions obtain in water. soil. and
sediments. Furthermore, a typical environmental sample will comprise many species that
are part ofseveral different redox couples and that are themselves not in equilibrium. The
result will be an unstable. drifting potential. For this and other reasons it is frequently not
possible to make accurate and precise pE readings as can be done for pH. It is, however,
usually possible to obtain an approximate value and define a regime as generally in the
oxidizing or reducing categories. As the template indicates, an intermediate redox status is
rare and usually transient as the system moves to a more stable higher or lower pE value.
1
The environmental behaviour of an element or compound depends on the particular form
of the species that is present. In the aqueous environment; the species distribution
depends on a number of factors including solution pH. pE. and the nature and availability of
complexing ligands.
2
In order to determine the species distribution of a particular substance. it is usual to make
an assumption that all forms are in equilibrium with the surroundings. Furthermore. it is
frequently assumed that concentration and activity of species are the same. Accurate
calculations, however. require activity estimations based on detailed knowledge of the
solution composition and ionic strength. The two assumptions and others mean that
calculations in actuai complex situations are only approximate and can sometimes be
erroneus.
3
Single variable distribution diagrams show how the concentratipns pf different species
change as a function of the one defined variable.
4
Two variable diagrams indicate regions on a two-dimensionai plpt where individual species
predominate. No detailed information about concentrations within the domains is given.
s The pH and pE of the aqueous environment are two key properties which define the nature
of chemical species. The pE is a measure of redox status and is high when the water is well
. aerated and low where oxygen is excluded or has been consumed.
1
Brookins, D. G., Eh-pH Diagrams for Geochemistry. Springer, Berlin; 1987.
2
Morel, F. M. M. and J. G. Hering, Principles and Applications ofAquatic Chemistry, John Wiley &
Sons, New York; 1993.