Redox andRedox and

pE pE –– pH DiagramspH Diagrams

OCN 623 – Chemical Oceanography

Reading: Libes, Chapter 7

Equations are

written as

REDUCTIONS!!

This lecture will go briefly over the principles involved in

deriving and using these equations. These principles

are all based on electrochemical cells.

Redox Equations Worth KnowingRedox Equations Worth Knowing

nRT

GpE

KpE

EhRT

FpE

reduc

oxidpEpE

reduc

oxid

nEE

nFEG

n

nb

a

b

a

3.2

log

3.2

log

log0592.0

1

1

∆−=•

=•

=•

+=•

+=•

−=∆•

ο

ο

ο

Electrochemical CellsElectrochemical Cells

Consider the following simple electrochemical cell

operating at 25°C, used to measure the electrochemical

potential of this reaction: Cu2+(aq) → Cu0(s)

ions

We arbitrarily assign a potential of 0 to

the reaction in the left cell:

2H+(aq) + 2e- → H2(g) E°= 0.000 V

Then the potential for the reaction in

the right cell is:

Cu2+(aq) + 2e- → Cu0(s) E°= 0.337 V

(always write as a reduction)

The standard potentials for all redox reactions are similarly

determined against the standard hydrogen electrode:

= E° = E°H

An electrochemical cell is capable of doing work – by

driving electrons across a potential difference.

This can be measured as a change in free energy:

∆Gr° = -nFE°

where

n = number of moles of electrons (equivalents) involved in

the reaction

F = Faraday constant = 23.1 kcal V-1 equiv-1

E° = the cell potential (V) at standard state

For the general case: ∆Gr = -nFE

Nernst EquationNernst Equation

We know from a previous class:

y

x

reactants

products RTG∆G

}{

}{ln+°∆=

Substituting ∆G = -nFE, we get the Nernst Equation:

Or:

yreactants

xproducts

nF

RTEE

}{

}{ln −°=

xproducts

yreactants

nF

RTEE

}{

}{ln +°=

reduced species

oxidized species

At 298°K:

xproducts

yreactants

nEE

}{

}{log

0592.0 +°=

Important points:

• Geochemists usually use the symbol EH instead of E

(indicating the hydrogen scale is being used)

• The Nernst Equation relates the EH of a cell to the

standard EH and to the activities of reactants and

products under given conditions

• When at standard state (all activities = 1), EH = E°

• We can use EH as an indicator of the state of natural

waters:

+−+ ↔+ 23 VeV

What species of V dominates in seawater with

measured Eh = 0.729 V?

1. From Table 7.1: Ehº= -0.26

2. Plug into equation:

+=+

+

2

3

log1

059.0

V

VEhEh ο

16

2

3

2

3

2

3

2

3

106

log8.16

log059.0989.0

log059.026.0729.0

x=

=

=

+−=

+

+

+

+

+

+

+

+

V

V

V

V

V

V

V

V

∴ V3+ dominates

Example #1: Use Eh values to Example #1: Use Eh values to

Calculate Vanadium SpeciationCalculate Vanadium Speciation

{ }{ } { }

{ }{ }{ }

{ }{ }{ }b

a

b

a

n

na

b

reduc

oxidK

en

reduc

oxidK

e

eoxid

reducK

loglog1

log

:logtheTake

1

+=

=

=

−

−

−

Definition of pEDefinition of pE

a(oxidized-species) + ne- → b(reduced-species)

From equilibrium theory:

{ } { }

{ }{ }

{ }{ }b

a

b

a

reduc

oxid

npEpE

KpE

reduc

oxidK

npE

npEee

n

n

log1

log:but

loglog1

bydivideandlog1

log:but

1

1

+=∴

=

+=∴

≡−= −−

ο

ο

Example #2: Calculation of pEExample #2: Calculation of pE

Assume:

• pE in a given environment is controlled by this reaction:

Fe3+ + e- → Fe2+

(∴n = 1)

• {Fe3+} = 10-5

• {Fe2+} = 10-3

( )

{ } 11

3

5

10

0.1110

10log0.13

0.13log

log1

−−

−

−

=

=

+=∴

==

+=

e

pE

n

KpE

reduc

oxid

npEpE

ο

ο

Table 7.1

Example #3: Effect of Atmospheric OExample #3: Effect of Atmospheric O22

Assume: Natural water at pH 7.5 in equilibrium with the atmosphere

Conclusion: This environment has lower electron activity than

Example #2, and is thus more oxidizing

{ }

( ) ( ) ( )

( )

{ } { }{ }

{ }

{ } 08.13

5.7

2

1

2

4

1

2

22

2

1008.13

10log

Libes7.3,Table75.20

21.021.0

1

2

1

4

1

2

−−

−++

−+

=∴=

=+=

=

↔++

=∴=

epE

H

OH

HOpEpE

pE

OHeHO

OatmP

n

laqg

O

ο

ο

1

Redox Reactions Have Characteristic pRedox Reactions Have Characteristic pE ValuesE Values

HERT

FpE

3.2=

nRT

GpE

3.2

∆−=

Garrels & Christ (1965)

¼ O2 + H+ + e- ← ½ H2O

H2O + e- → ½ H2 + OH-

Thus, if an environment is characterized by a Thus, if an environment is characterized by a

certain redox reaction, it has a characteristic pEcertain redox reaction, it has a characteristic pE

pEpE--pH DiagramspH Diagrams

pE-pH stability field diagrams show in a comprehensive

way how protons (pH) and electrons (pE)

simultaneously shift equilibria of reactions under various

conditions

These diagrams also indicate which species predominate

under any given condition of pE and pH

Two equations are used to produce the diagrams:

{ }{ }r

o

reduc

oxid

npEpE log

1+°=

Kn

pE log1

=°

Oxidizing limit of diagrams: ¼ O2 + H+ + e- → ½ H2O

{ } { }{ }

pHpE

npEpE

−=

+°=+

75.20

OH

HOlog

1

21

2

41

2

Reducing limit of diagrams: H2O + e- → ½ H2 + OH-

OH- + H+ → H2O

H+ + e- → ½ H2

{ }{ }

pHpE

npEpE

−=

+°=+

21

2H

Hlog

1

pE° = +20.75 (Table 7.3)

n = 1

pH = -log{H+}

Set limit: {O2} = 1

pE° = 0.0 (Table 7.3)

n = 1

pH = -log{H+}

Set limit: {H2} = 1

Oxidizing limit of diagrams:

O2 + H+ + e- → ½ H2O

pHpE −= 75.20

Reducing limit of diagrams:

H+ + e- → ½ H2

pHpE −=

• Phase-boundary lines on a

pE-pH diagram indicate stability

field boundaries – defined as

lines where activities of both

adjacent dominant species are

equal.

• Lines are defined by reactions

between adjacent dominant

species

• Reactions must have known

log K or pE° values.

Construction of pEConstruction of pE--pH DiagramspH Diagrams

{ }{ }{ }

{ }{ }

{ } { } 0.2,

log0.2log

2.0logKHSOHSO6.

4

2

4

4

2

4

4

2

4

2

44

==

−=−=

=

−=+↔

−−

−

−

−

+−

+−−

pHHSOSOWhen

pHHSO

SOK

HSO

HSOK

Acid-base reactions with no pE dependency

{ }{ }{ }{ }

{ }{ }

{ } { } pHpEHSSOWhen

pHHS

SOpE

Kn

pE

OHHS

HSOpEpE

8

9

8

34,

8

9log

8

1

8

34

8

34log

1

log8

1

34.0KlogO4HHS8e9HSO5.

2

4

2

4

4

2

92

4

2

2

4

−==

−+=

==

+=

=+↔++

−−

−

−

−

+−

−−+−

ο

ο

=1

Redox reactions of dissolved species

Redox reactions of dissolved and solid species

( )

{ } { }( ){ }{ }

{ }

{ } pHpESOWhen

pHSOpE

OHS

HSOpE

s

6

870.5,10

6

8log

6

1

6

2.36

log6

1

6

2.36

36.2KlogO4HS6e8HSO1.

22

4

2

4

4

2

82

4

2s

2

4

−==

−+=

+=

=+↔++

−−

−

+−

−+−

=1

Activity of dissolved

species must be given

SI

Homework Homework –– Due Thurs, Jan. 24, 2013Due Thurs, Jan. 24, 2013

1) Write a balanced equation for the nitrification of NH4+ to

NO3- (oxidation by O2).

2) Calculate the Eh at 25°C for this reaction in offshore

Hawaiian seawater at 100 m depth. Assume [NH4+] = 1 µM,

[NO3- ] = 150 nM, and the partial pressure of O2 = 0.21 atm;

assume the activity coefficients for NH4+ and NO3

- are 0.7 .

(Hint: calculate ∆G° and then Eh.)

3) In which direction would the reaction proceed in an

environment with an Eh of -200 mV? Why?