Environmental Biogeochemistry of Trace Metals (CWR 6252)

Background on pE-pH Diagrams andChemical Elements' Aqueous Speciation

(From Vanloon and Duffy, Oxford University Press, 2000)

Both the phosphorus and cadmium distribution diagrams share the obvious limitation thatthey describe behaviour of a particular chemical in terms of only a single environmentalvariable-pH or chloride concentration, respectively. Clearly. in complex real systems.there are many variables operating simultaneously. A further step towards accuratelydescribing natural systems in graphical form is then to create a two-variable diagram. Insuch diagrams, the dominant species are plotted in two dimensions as a function of twoindependent variables. There are obvious merits in creating diagrams that take intoaccount 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 aflat surface.

Avery widely used type oftwo-variable diagram for describing chemical behaviour in thehydrosphere is the pH/pH diagram. also called a Pourbaix diagram. We shall discuss how toconstruct and interpret such plots, but before doing this it is necessary to introduce theconcept of pE.

Analogous to pH. the measure of acidity in aqueous solutions. pH is defined as thenegative 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 solutionwhich implies that reducing conditions obtain. Conversely. a large positive value of pEimplies low electron activity in solu~on and oxidizing conditions. In practice pE values inwater 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 ofpH are done.To show how pE is calculated and measured it is helpful to consider some examples.

- 1-

(10.35)

(10.34)

(10.36)

(10.37)

(10.3B)

(10.39)

(10041)

(10.42)

(10.43)

(10045)

(10.44) I

= -nFE"

E'logK"q = 0.0591

E'PE-pE'--­

- - 0.0591

E E'l apeHp =--+ og--

0.0591 apeH

!J.G' = -2.303RT log K"q

1 Keq x apeH-=a e- a pe2+

Consider the simple half reaction

Fe3+(aq) +e- ;= Fe2+(aq)

Using the definition ofpE and tal<ing logs ofboth sides ofeqn 10.36, we have

Since

In this case, n = 1, so

TWO·VARIABUO DIAGRAMS: plO/pll OIACiRAMS

and

and

Under standard conditions. apeH = ape2+ = 1:

For non-standard conditions:

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 particularenvironmental system can be calculated.

1[:

ii'(n has the.usual electrochemical me~g-=~.e. the number ofelec~ons transferred irl the !I:

·--·--half.reaC't1on);-at-29B..j(·(R·=·B03'!4JK~-mol-and"F=-964B5·G-mol-);.we,have----------.-----;j!

nFE' nE" / illlogK,q = 2.303,RT 0.0591 (10040) v' :11

-2-

_ 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, asan approximation. concentrations) that obtain under any conditions. not just those atequilibrium. TI,e general form ofeqn 10.45 is then

1pE = pE" - -logQ.

n

10.2.1 Methods of calculating pEa

(10.48)

Values of p~ for a number of half reactions of environmental interest are given inAppendix 11.When additionalvalues 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.

nEo10gKeq =0.0591 =npE" (10.49)

pE"_ logK,q .- (10.50)

n

This relation is applicable when an E" value is not available. but where the appropriateequilibrium 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

The reaction is the sum of

Fe(OH),;= Fe'+(aq) + 30H" (aq)

Fe'+ (aq) + e" ;= Fez+(aq)

3H,O+(aq) + 30H" (aq) ;= 6HzO

(10.51)

(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.

1Kc =--= 1042

(Kw)'

and log Kc = +42.0. For the original. overall reaction

logKm = logK, + 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 10.38and 10.49:

t>G' = -2.303RTnpB"

-t>G''--'-'- .pB" = 2-=:acc03°C'R"'Tc-"11 _

Consider the redox reaction

SO;'(aq) + 10H,O+(aq) + 8e- ;= H,S (aq) + 14H,O

(10.52)

(1053)

(10.54)

Using thermochemical tables and noting that ~G~ for the aqueous electron is 0, and forthe hydronium ion. 6.Gt has the same value as for water:

t>G' = -27.86 + 14 x (-237.18) - 10 x (-237.18) - (-744.60)

= -231.98kJ

pB" = -(-231.98) kJ x 1000Jkr '2.303 x 8.314Jmol 'K 1 x 298.2K x 8 mol

= 5.08

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 wastesTraditional leather tanning processes involve treating the hides with an aqueous solutionof 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 canoxidize Cr'+ to Cr,O~-. We will calculate the extent of oxidation for a situation whereoxygen 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 valueto calculate the concentrations ofCr3+ and Cr20~- assuming that the chromium species arealso in equilibrium with the system.

The relevant reaction for atmospheric O2 in equilibrium with water (often called awell aerated system) is

Using eqn 10.48:

0, (g) + 4H,O+ (aq) +4e- ;=' 6H,O

BO = 1.23 Vo 1.23 V

pB = 0.0591 V 20.8

1 1pB = pE" - -log •

n Po,/po x (aH,O+)1 1

= 20.8 - -log------;----;=:.4 0.209 x (10 6.5)'

= 14.1

(10.55)

___. .__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 theatmosphere. Po, = 21200 Pa; po = 101325 Pa.

For the Cr system:

Cr,O;- (aq) + 14H,O+ (aq) + 6e- ;=' 2Cr'+ (aq) + 17H,O (10.56)

E" = 1.36V and pE" = 23.0

1 [C,.'+]'pB=pBo --log, 1.

6 [Cr,O? ](aH,O+)

Since the chromium and oxygen systems are in equilibrium. pB is the same for both:

1 [Cr'+]'14.1 = 23.0 - -6 log , 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,.'+]'= 7.8 --log[ , ]

6 Cr,O?

[Cr'+]'log [ 'I -37.8

Cr,O?

[C,.'+]'1.6 x 10-38

[Cr,O; J

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 ofleathertanning involves a two-bath process inwhich 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-themateriaL The hides are ~~n transferred into ano~ertanI<:containingreducing agents suchas sucrose in order to reduce tl,e exc~ss chromium (VI) to chromiUm (ill). The excess is ~endischarged in tile waste water.

The leatiler industry is important tilroughout tile globe, witillndia tile world's majorproducer of leatiler goods. In that country tilere are large industria) establishmentsmanufacturing shoes, luggage, garments, and so on, but much oftile industry is small-scaleand scattered in numerous villages in rural areas of the C01,1D.tIy. The l10rthern part ofTamil Nadu state in Soutillndia is ~~ centre of tile leatiler industry.

The quantity of chromium (ill) wasted in effluents of a small-scale chrome tannery inIndia is estimated to be approxhnately 0.41>g per 100 kg ofraw hides. This is about onehalf oftilat added in tile first batil. Chromium (Ill) has low toxicity, but in tile hexavalentform it is botil toxic and carcinogenic to humans. It is also toxic tp plants and large areasof tile leatller-producing areas ofTamil Nadu are devoid ofvegetation. The consequencesto humans in that region aTe not documented. In any case, the release of such largeamounts of chromium with its-subsequent conversion to the thermodyn<;l.mically stableCr20~- species"is a major environmental iss:ue.

---10.2.3 pE/pH diagramsWe are now in a position to construct a pE/pH diagram. The diagram will talce tile fprm ofa two-dimensional plot ofpE (ordinate, y axis) versus pH (abscissa, x axis) in which areas intile 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 theseboundaries.

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 andtile otiler is a gas, tile boundary condition is set as P" = 101325 Pa = 1 atm. This is tileminimum 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 anexample, this type ofboundary condition is used in tile case of tile evolution ofhydrogengas 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 areinvolved, orwhere tilere is a soluble species reacting to form one that is insoluble. Examplesof these two cases are

and

Sn4+ (aq) + 2e- "" Sn2+ (aq)

Mn02 .+ 4H3 0+ (aq) +2e- "" Mn2+ (aq) + 6H20

(10.58)

(10.59)

In tilese cases, we arbitrarily define a concentration (as an estimate ofactivity) belowwhicha particular species is considered to be soluble. Any con,entration may be chosen but it is

- b-

------ - ----- -- -~------ --- -------

Equation 10.48 for this reaction is written as

1 ,___________ pE = pE" - 2.!Qgi&,/.1'" x...(!!oH:.)_).. -.(.lO.61.) _

then

I

i

(10.60)

(10.62)

(1Q.63)

E" = -0.828 V

pE" = -14.0

2H,O + 2e- "" H, (g) + 20I1 (aq)

2H,O+ (aq) + 2e- "" H, (g) + 2H,O

DISTRIBUTION OF SPEaESIN AQUATIC SVSTEMS

common to use values between 10-5 and 10-2 mol L-'1. Ofcourse. inconstructing a diagramone should consiste:n.tly use the same concentration for all boundaries.

We will illustrate how boundary conditions are chosen and used in the construction ofa pElpH diagram by considering the aqueous sulfur system where the species of interestate SO~- (aq). HSO' (aq). S (s). HS- (aq). and H2S (aq). Before beginning calculationsinvolving these species. however, there are some :greliminary calculations common to alldiagrams pertaining to the hydrosphere.

As we had noted in Chapter 10. water itself has a limited range of pE and pH valueswithin which it is stable. Under highly reducing conditions (low pEl. water is reduced:

For this reaction

E" = 1.229V

pE" = EO/0.0591 = 20.80

(note again that here and in subsequent calculations, we will use concentrations ratherthan activities except for hydroxyl and hydronium ions).

For the boundary involving a gas. we choose the condition that

pH+pOH= 14

pE=-pH

PH, = 1'" = 101325 Pa

pE = -14.0 -log(aow)

=-14.0+pOH

Since

This line then defines the boundary for water stability with respect to reduction and isshown as the lower line on Fig. 10.4. Where the pE value is less than the pH value. water isunstable. It is possible also to calculate the s;une line by ta1cing the reduction reaction to be

Considering the other extreme. highly oxidizing conditions. water is unstable withrespect to O2 evolution and the reaction is written as

For this reaction

\

TWO-VARIABLE DIAC;RAMS: pE/pH DIAC;BAM~

(10.64)

642-12

Once again, the boundarycondition requires that the pressure ofthe gas equal atmosphericpressure

-p

,pE 4

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 thestability region for water on a pElpH diagram. Above the top boundary, water is oxidizedwith the evolntion ofoxygej}, while below the lower boundary it is reduced and releaseshydrogen. .

The production of oxygen from water ~ an oxidation Iea~on. It is important to' note.however, thatwe use the MAC couvention ofdefining E" andE, and pE" andpE, in terms ofthe reverse, reduction process:

10.2.4 The sulfur systemNow we can deal with the sulfur species. To create the pElpH diagram, we will be definingboundaries between species and superimposing these on the water stability diagram. Wemay consider the varions pairs of species in any order but it is helpful to have an idea ofwhat to expect before beginning calculations-for example we expect that HSO' will beimportant at low pH and SO~- at high pH. SimiJarlySO~- would exist under oxidizing, highpE conditions, while HS- is a reduced, low pE form. For all soluble species, we will choose10-2 mol L_1 as our arbitrary definition for the boundary.

I=Ig. '~;4 Region {If stability anc! stability bQundari,es ,foq(Vater, On a pflpH diagram. " " ' ,

-----------i::-~:f:.:_,t-::·~-.~:=::;:,~-==j;,:::~';~=,·-L~~..~~Z~.2,.=,_;;~;::,:;;:::~~t.=:::..~C~=,-~:_=..'7:j~_,::::,~).:::_;;,

DISTRIBUTION OF SPECIES IN AQUATIC SYSTEMS

The so~-IHSO' boundaryThe equation describing this boundary requires hydronium ion, but there is no oxidation orreduction involved:

(10.65)

To make things easier, we use H+ as an abbreviation for H30+, the hydronium ion, but ofcourse the result would be the same if the latter species were used·in the equations andcalculations.

!:>.GO = !:>.G~ (HSO.) - AG~ (SO~-) - AG~ (W)

= -755.99 - (-744.60) - 0

= -11.39kJ = -11390J

----------------_.

_AGOlog 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-zM, and

1K=-

ali'

logK = pH = 1.995

Therefore, the boundary between HSO. and SO~- is a vertical line at pH 1.995. Below thisvalue, 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)+4HzO

AGO = AG~ (S) +4AG~ (HzO) - AG~ (HSO.) -7AG~ (H+) - 6AG~ (e-)

= 0 +4(-237.18) - (-755.99) - 0 - 0

= -192.73 kJ = -192 730J

(10.66)

EO _ -AGo +192730P - 2.303 nRT 2.303 x 6 x 8.314 x 298.2

PE=p~-~log 16 [HSO.] (aH' )7

5.626

._- ---- ---._----------.. -- .,---

At the boundary, [lISO"] = 10-2 molL-1:

7 1 1 1pE = 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.

+2041202.303 x 6 x 8.314 x 298.2

The SO~- fS" bOllndary

SO~- (aq) + 8H+ (aq) + 6e- "" So (s) + 4HzO

!::..Go = !::..G~ (S) +4!::..G~ (H20) - !::..G~ (SO~-) - 8!::..G~ (H+) - 6!::..G~ (e-)

= 0 + 4(-237.18) - (-744.60) - 0 - 0

= -204.12Ig = -204120J

EO _ -- ~GoP - 2.303nRT

~ 5.958

PE=pE"-~log 16 [SO~-](aH+)8

- 10-

.._-- -- --- ----------_._--- ---

(10.67)

• DISTRIBUTION OF SPECIES IN AQUA11C SYSTEMS

This is line 'd' on Fig. 10.5; above it is the domain of So, below it H,S.

(10.68)

(10.69)

./-..27..860

+231980

2.303 x 2 x 8.314 x 298.2

2.303 x 8 x 8.314 x 298.2

2.303nRT

-l>.Go

2.303nRT

= 2.400

pE"

l>.GO = l>.G' (H,S) -l>.G' (S) - 2l>.G' (H+) - 2l>.G' (e-)

= -27.86 - 0 - 0 - 0

= -27.86kJ = -27860J

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

S (s) + 2H+ (aq) + 2e- ;= H,S (aq)

At the boundary, [SO~-] = 10-' mol L-1:

8 1 1 1pE = 5.958 - -log- - -log--

6 aH+ 6 10-2

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.

= 5.079

E = E" _ ~lo [H,S]P P 8 g[SO~-](aH+)10

The 5·/H.5 boundary

The 50~-/H25 boundary

SO~- (aq) + lOW (aq) + 8e- ;= H,S (aq) + 4H,O

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

/ {-

TWO·VARIABLE OIAC;BAM$: pElpH DIAC;BA/vIS

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.

As for the HSO./SO~- boundary. this is not a redox reaction and therefore the line will bevertical with the protonated species on the left.

tlGo = tlGf (HS-) + tlGf (H+) - tlGf (H2S)

= 12.08 + 0 - (-27.86)

= 39.94kJ = 39940J

(10.70)

(10.71)

+1920402.303 x 8 x 8.314 x 298.2

H2S (aq) .= HS- (aq) +W (aq)

K

logK

-tlGopE" = =--2.-=-30=-3:-nR-=-T

=4.204.

E= E" _ !10 [HS-]P p 8 g [SO~-](aH+)9

K=aH+log K = log[H+j = -pH = -6.995

pH = 6.995

tlGo = tlGf (HS-) HtlGf (H20) - tlGf (SO~-) - 9tlGf (H+) - 8tlGf (e-)

= 12.08 +4(-237.18) - (~744.60) - 0 - 0

= -192.04kJ = -192040J.

The H.S/HS- boundary

This is line 'f' on Fig. 10.5; to the left is the domain ofH2S and to the right HS-.

The SO~-/HS- boundary

so~- (aq) + 9H+ (aq) +8e- .= HS- (aq) +4H20

At the boundary, [H2S] = [SO~-] = 10-2molL-':

10 1pE = 5.079 - -log--a aH+

= 5.079 - 1.25 pH

- tlGo -39 9402.303 nRT = =-2-=.3-=-03=-X-":8::.3~1:":4~X-2=-9'"'8c-=-.2

= -6.995

[HS-]aH+[H2S]

At the boundary, [HS-j = [H2S] = 10-2 molL-':

-/2-

';-"

21', DISTRIBUTION OF SPECIES IN AQUATIC SYSTEMS

At the boundary, [HS-] = [50:-1 = lO-zmolL-1:

9 1pE = 4.202 - -10g-

B aH+

= 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 ofparticular environmental situations follow. For mine wastes, at a pH of2.5 and exposed to the atmosphere, so that they are well aerated (pE", 15), corresponding topoint Xon Fig. 10.5, the most inlportant sulfur species in solution would be sulfate. Wherethe originally mined material was sulfide ore, such as from the copper-nJckel Ore$ ofSudbury, Ontario, Canada, sulfur might initially go into solution as sulfide but wouldeventually (the kinetics are relatively fast) be oxidized to sulfate. For a swamp or paddy (rice)field where soil containinga high content oforganic matter (OM) is submerged, the OM actsas a reducing agent and creates low pE conditions. For example, such a soil might havepH = 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 thesediment. Organic matter in soil typically contains about 2% sulfur.

Additional important pEjpH diagrams will be given in the problems section of thischapter 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 forcommonly 'encountered aqueous, soil, and sedimentenvironments.

MEASUREM!'NTS OF pE 219

~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 laboratoryor 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.

In principle it sllould be simple to measure pE in a real environment In practice, non~

equilibrium conditions make stable ~and meaningfp.l readings exceedingly difficult. Therequirements for me~surement are an indicator electrode which is iliert and develops apotential 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 isused so that the indicated pote!ltial may be measured against a known value. As with allpotential measurements it is essential to' ensure· that negligible c~rrentbe drawn duringthe 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 foundto be +713 mVvs. the saturated calomel electrode (SCE). The value of E is recalculatedvs. the normal hydrogen electrode (NHE), using the known potential, +0.242 V ofthe SCE:

Evs. NHE = 0.713 +0.242V= 0.955V

-14-

220 DISTRIBUTION OF SPEOES IN AQUATIC SYSTEMS

The value ofpE is then readily calculated using eqn 10.44:

0.955pE = 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 thatare part ofseveral different redox couples and that are themselves not in equilibrium. Theresult will be an unstable. drifting potential. For this and other reasons it is frequently notpossible 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 theoxidizing or reducing categories. As the template indicates, an intermediate redox status israre 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 formof the species that is present. In the aqueous environment; the species distributiondepends on a number of factors including solution pH. pE. and the nature and availability ofcomplexing ligands.

2 In order to determine the species distribution of a particular substance. it is usual to makean assumption that all forms are in equilibrium with the surroundings. Furthermore. it isfrequently assumed that concentration and activity of species are the same. Accuratecalculations, however. require activity estimations based on detailed knowledge of thesolution composition and ionic strength. The two assumptions and others mean thatcalculations in actuai complex situations are only approximate and can sometimes beerroneus.

3 Single variable distribution diagrams show how the concentratipns pf different specieschange as a function of the one defined variable.

4 Two variable diagrams indicate regions on a two-dimensionai plpt where individual speciespredominate. 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 natureof 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 andApplications ofAquatic Chemistry, John Wiley &Sons, New York; 1993.