The Geochemistry of Rocks and Natural Waters

Course no. 210301

Introduction to

Electrochemistry

A. Koschinsky

Electrochemistry - Redox reactionsElectrochemistry translates the chemical energy of a reduction–oxidation reaction into

electrical energy. Reduction–oxidation reactions involve the transfer of electrons. Electrical energy comes from the movement of electrons through a wire.

When the substances involved in oxidation and reduction half–reactions are physically separated, it is called an electrochemical cell. Each half reaction occurs on the surface of an electrode. Each electrode is immersed in a solution containing ions needed for the half–reaction. The solutions are connected by a salt bridge so that ions can move between solutions. The force moving electrons through the wire is called the electromotive force, electrochemical potential, or voltage (E). The unit for this force is volts (V).

.

Since electrons are negative, a negative charge might build up with the reduction and a positive charge with the oxidation. The salt bridge moves ions into those solutions to maintain electric neutrality without mixing of solutions

A- --> A + e- B + e- --> B-

Electrochemistry - Redox reactionsSince electrochemical potential (E) measures the driving force of a reaction, it is similar in

concept to Gibbs free energy (G). The relationship is

G = –nFE        

where n is the number of electrons transferred in the redox reaction and F is Faraday's constant, 9.65 x 104 C/mol e–. Thus a positive potential represents a spontaneous

reaction and a negative potential denotes a non-spontaneous reaction.

Like free energy, electrochemical potential is related to the equilibrium constant (K), where

Using the actual values for the constants at 25 °C and rearranging, the equation becomes

E0 = E at standard conditions

Electrochemistry - Redox reactionsCell potentials (E) depend primarily on the identity of the reaction. Since cells separate the

half–reactions, it is often convenient to talk about the potential of the half–reaction or the electrode potential. Only potential differences can be measured, but no single half-cell reaction. So that electrode potentials could be tabulated, one half–cell was defined as having an electrode potential of exactly 0 V. This half–cell is called the standard hydrogen electrode (SHE). It uses an inert platinum electrode with 1 M H+ and 1 atm H2 for the following reaction

2 H+ + 2 e– <--> H2

Tables list potentials for reduction half–reactions at standard state (1 M, 25 °C and 105 Pa) with SHE as the oxidation half–reaction. These are called standard reduction potentials. To obtain standard oxidation potentials, the opposite sign is used. These tabulated values can be used to determine the standard cell potential for any electrochemical cell. The standard cell potential is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E°cell = E°c – E°a 

Electrochemistry - Redox reactionsTo correct for non-standard conditions, the Nernst equation is used to determine the cell

potential (Ecell).

 

where R is the gas constant, T is temperature, n is the moles of electrons in the reaction, F is Faraday's constant, and Q is the reactant quotient, where all solutions are in units of molarity and concentration of gases are expressed as the partial pressure in atmospheres.

An alternative representation of redox state: porpeA "p" unit is a log10-based quantity. We can report the activity of any chemical species in p units, such as pH.

The p is the -log10 of the electron activity: plog ae-

The p is related to the reduction potential by the FRT factor of the Nernst equationp= E / 0.059

Electrochemistry - Redox reactions

The Nernst equation may be recast using p formalism:

pp01/n log Q

Example:

n

Electrochemistry - Redox reactions

Half cell potential is given by:

E = E0 - (RT/nF) ln{[Red]/[Ox]}

where E0 is the thermodynamic driving force for the reaction when the other 1⁄2 cell reaction is SHE (standard hydrogen electrode, E0 = 0 V).

NOTE: All of these E0 values are based on writing the equation as a reduction (this is also known as the standard reduction potential):

Cl2 + 2e- ---> 2Cl- E0 = 1.36 V Strong tendency to occur as a reduction (i.e. as written)

Na+ + e- ---> Na E0 = -2.71 V Strong tendency to occur as an oxidation (i.e. opposite direction as written)

Electrochemistry - Redox reactionsExamples: E = E0 - (RT/nF) ln{[Red]/[Ox]}

1. Calculate the half cell potential for a Zn electrode in 10-2 M Zn(NO3)2

reaction: Zn2+ + 2e- ---> ZnE = E0 - (0.059/n) log 1/[Zn2+] E = -0.76 - 0.059/2 log {1/0.01} = -0.82 V

The reaction proceeds as an oxidation, therefore Zn is the anode. Any half-cell potential calculated using the Nernst Equation is an E for a hypothetical cell which has the half cell of interest as the cathode and the SHE as the anode. However, the negative sign on our example indicates that if assembled, the Zn electrode would be the anode.

2. Consider half-cell SHE, but p(H2) and [H+] are different: [p{H2} = 0.5 atm and [H+] = 0.01 M]

reaction: 2H+ + 2e- ---> H2

E = E0 - 0.0059/2 log {pH2/[H+]2} = 0 - 0.059/2 log [0.5/(0.01)2]E = -0.1093 V

Electrochemistry - Redox reactionsExercise 13

What is the p (or redox potential E) of the following solution:

Water (pH = 7) in equilibrium with atmospheric oxygen (pO2 = 0.2);

K = 10 20.75

K = [H2O]1/2 / [H+] [e-] pO21/4 = 10 20.75

log K = 20.75 = 0 - log H+ - log e- - 1/4 log pO2 = p + pH - 1/4 log pO2

p= 20.75 - 7 - 1/4(-0.7) = 13.58

E = 0.80 V

1/4 O2(g) + H+ + e- = 1/2 H2O

Electrochemistry - Redox reactionsExercise 14

What is the p (or redox potential E) of the following solution:

Lake water (pH = 7) in equilibrium with MnO2 (manganate) and 10-5 M Mn2+

MnO2 + 4 H+ + 2 e- = Mn2+ + 2 H2O

G0 = -228.0 + 2 x (-237,18) - (-453.1) = -249.26 kJ mol-1

Log K = -G0 / 2.3 RT = 249.26 / 5.7066 = 43.6

Log K = 43.6 = log[Mn2+] + 4 pH + 2 p

p = 1/2 (43.6 + 5 - 28) = 10.3

E = 0.61 V

Eh-pH diagramsLearn to construct and use Eh-pH (or pe-pH) diagrams

• Diagrams that display relationships between oxidized and reduced species and phases.

• They are a type of activity-activity diagram!

• Useful to depict general relationships, but difficulties of using field-measured pe (Eh) values should be kept in mind.

• Constructed by writing half reactions representing the boundaries between species/phases.

pH0 2 4 6 8 10 12 14

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atm

Fe(OH)3(s)

ΣF106molL1

F2+

F3+

pe-pH diagram for the Fe - O2 - H20 system

UPPER STABILITY LIMIT OF WATER (pe-pH)

The following half reaction defines the conditions under which water is oxidized to oxygen:

1/2O2(g) + 2e- + 2H+ H2OThe equilibrium constant for this reaction is given

by

2221

2

1+−

HO aap

K

+− −−−HeO aapK log2log2loglog

221

pHpepK O 22loglog22

1 ++−

Solving for pe we get

This equation contains three variables, so it cannot be plotted on a two-dimensional diagram without making some assumption about pO2. We assume that pO2 = 1 atm. This results in

We next calculate log K usingGr° = -237.1 kJ mol-1

pHpKpe O −+2

loglog 41

21

pHKpe − log21

53.41)K)(298.15molKJ314.8(303.2

molJ100,237log 11

1

−−

−

K

pHpe − 77.20

LOWER STABILITY LIMIT OF WATER (pe-pH)

At some low pe, water will be reduced to hydrogen by the reaction

H+ + e- 1/2H2(g)

We set pH2 = 1 atm. Also, Gr° = 0, so log K = 0.

+−

He

H

aap

K2

1

2

pHpepK H ++−2

loglog 21

pHpe −

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Water stable

A pe-pH diagram showing the stability limits of water. At conditions above the top dashed line, water is oxidized to O2; at conditions below the bottom dashed line, water is reduced to H2. No natural water can persist outside these stability limits for any length of time.

UPPER STABILITY LIMIT OF WATER (Eh-pH)

To determine the upper limit on an Eh-pH diagram, we start with the same reaction

1/2O2(g) + 2e- + 2H+ H2Obut now we employ the Nernst equation

20

21

2

1log0592.0+

−HO apn

EEh

20

21

2

1log2

0592.0+

−HO ap

EEh

As for the pe-pH diagram, we assume that pO2 = 1 atm. This results in

This yields a line with slope of -0.0592.

221

2log0296.023.1 ++

HO apEh

pHpEh O 0592.0log0148.023.12−+

volts23.1)42.96)(2()1.237(0

0 −−

ℑ−

nGE r

pHEh 0592.023.1 −

LOWER STABILITY LIMIT OF WATER (Eh-pH)

Starting withH+ + e- 1/2H2(g)

we write the Nernst equation

We set pH2 = 1 atm. Also, Gr° = 0, so E0 = 0. Thus, we have pHEh 0592.0−

+

−H

H

ap

EEh2

1

2log1

0592.00

pH0 2 4 6 8 10 12 14

Eh (volts)

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Eh-pH diagram showing the stability limits of water at 25°C and 1 bar. Note the similarity to the pe-pH diagram.

pH0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Range of Eh-pH conditions in natural environments based on data of Baas-Becking et al. (1960) Jour. Geol. 68: 243-284.

Fe-O2-H2O SYSTEM

pH

Fe(OH)(s)2

Fe3+Fe(OH)(s)3

Fe2+A preliminary mapping of the

species and phases in pe-pH space.

Fe(OH)3/Fe(OH)2 BOUNDARY

First we write a reaction with one phase on each side, and using only H2O, H+ and e- to balance, as necessary

Fe(OH)3(s) + e- + H+ Fe(OH)2(s) + H2O(l)Next we write the mass-action expression for the

reaction

Taking the logarithms of both sides and rearranging we get

+−

Heaa

K 1

pHpeaaKHe

+−− +− logloglog

And then

Next, we calculate rG° and log K.Gr° = Gf°Fe(OH)2 + Gf°H2O - Gf°Fe(OH)3Gr° = (-486.5) + (-237.1) - (-696.5)

Gr° = -27.1 kJ mol-1

So now we have

This is a line with slope -1 and intercept 4.75.

pHKpe −log

75.4)K)(298.15molKJ314.8(303.2

molJ100,27log 11

1

−−

−

K

pHpe − 75.4

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O Fe(OH)

2 (s)

Fe(OH)

3 (s)

pe-pH diagram showing the first Fe boundary. This boundary will surely intersect another boundary and be truncated, but at this point we don’t know where this intersection will occur. So for now, the boundary is shown stretching across the entire Eh-pH diagram.

Fe(OH)2/Fe2+ BOUNDARYAgain we write a balanced reaction

Fe(OH)2(s) + 2H+ Fe2+ + 2H2O(l)

No electrons are required to balance this reaction. The mass-action expression is:

2

2

+

+

H

F

aa

K

pHaKFe

2loglog 2 +− +

+− 2loglog 21

21

FeaKpH

Gr° = Gf°Fe2+ + 2Gf°H2O - Gf°Fe(OH)2

Gr° = (-90.0) + 2(-237.1) - (-486.5)Gr° = -77.7 kJ mol-1

To plot this boundary, we need to assume a value for ΣFe a Fe2+ m Fe2+. This choice is arbitrary - here we choose ΣFe =10-6 mol L-1. Now we have

61.13)K)(298.15molKJ314.8(303.2

molJ700,77log 11

1

−−

−

K

++ −− 22 log81.6log)61.13( 21

21

21

FeFeaapH

81.9)6(81.6 21 −−pH

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Fe(OH)

2 (s)

Fe(OH)

3 (s)

ΣF106molL1

This diagram illustrates the plotting of the second boundary required for this diagram. Note that the portion of the Fe(OH)3(s) /Fe(OH)2(s) boundary from about pH 10 to pH 0 was erased as it is metastable. Also, the portion of the Fe2+ /Fe(OH)2 boundary at high pe is also metastable and has been erased. The next boundary to be calculated is the Fe(OH)3(s) /Fe2+ boundary.

Fe2+

Fe(OH)3/Fe2+ BOUNDARYAgain we write a balanced reaction

Fe(OH)3(s) + 3H+ + e- Fe2+ + 3H2O(l)The mass-action expression is:

3

2

+−

+

H

F

aaa

K

pHpeaKFe

3loglog 2 ++ +

pHaKpeFe

3loglog 2 −− +

Gr° = Gf°Fe2+ + 3Gf°H2O - Gf°Fe(OH)3

Gr° = (-90.0) + 3(-237.1) - (-696.5)Gr° = -104.8 kJ mol-1

To plot this boundary, we again need to assume a value for ΣFe a Fe2+ m Fe2+. We must now stick with the choice made earlier, i.e., ΣFe =10-6 mol L-1. Now we have

36.18)K)(298.15molKJ314.8(303.2

molJ800,104log 11

1

−−

−

K

pHapeFe

3log36.18 221 −− +

pHpHpe 336.243)6(36.18 −−−−

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Fe(OH)

2 (s)

Fe(OH)

3 (s)

ΣF106molL1

F2+

The third boundary is now plotted on the diagram. This boundary will probably intersect the Fe2+/Fe3+ boundary, but at this point, we do not yet know where the intersection will be. Thus, the line is shown extending throughout the diagram.

Fe3+/Fe2+ BOUNDARYWe write

Fe3+ + e- Fe2+ Note that this boundary will be pH-independent.

Gr° = Gf°Fe2+ - Gf°Fe3+

Gr° = (-90.0) - (-16.7) = -73.3 kJ mol-1

+−

+

3

2

Fee

Fe

aaa

K

8.12p

84.12)K)(298.15molKJ314.8(303.2

molJ300,73log 11

1

−−

−

K

13

2

+

+

F

F

aa

Kpe log

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Fe(OH)

2 (s)

Fe(OH)3(s)

ΣF106molL1

F2+

F3+

The Fe2+/Fe3+ boundary now truncates the Fe2+/Fe(OH)3 boundary as shown. There remains just one boundary to calculate - the Fe(OH)3(s) /Fe3+ boundary. Because the reaction for the Fe2+/Fe3+ boundary does not include any protons, this boundary is horizontal, i.e., pH-independent.

Fe(OH)3/Fe3+ BOUNDARY

Fe(OH)3(s) + 3H+ Fe3+ + 3H2O(l)

Gr° = Gf°Fe3+ + 3Gf°H2O - Gf°Fe(OH)3

Gr° = (-16.7) + 3(-237.1) - (-696.5) = -31.5 kJ mol-1

3

3

+

+

H

F

aa

K pHaKFe

3loglog 3 + +

+− 3loglog 31

31

FeaKpH

52.5)K)(298.15molKJ314.8(303.2

molJ500,31log 11

1

−−

−

K

84.3)6()52.5( 31

31 −−pH

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Fe(OH)

2 (s)

Fe(OH)3(s)

ΣF106molL1

F2+

F3+

Final pe-pH diagram for the Fe-O2-H2O system. Note that the solubility of iron phases is greater when the dissolved iron species is the reduced Fe2+. In other words, Fe is more soluble under reducing conditions. Because most natural waters have pH values in the range 5.5-8.5, they will not contain much iron unless redox conditions are relatively reducing.

pH

0 2 4 6 8 10 12 14

pe

-12

-8

-4

0

4

8

12

16

20 T = 25oCpH2

= 1 atm

pO2 = 1 atmO

2H2 O

H2

H2 O

Fe(OH)

2 (s)

Fe(OH)3(s)

ΣF106molL1

F2+

F3+

Eh-pe and Eh-pH diagrams for the Fe-O2-H2O system at two different Fe concentrations

pe-pH diagram for the Fe-O- H-C-H2O system

pe-pH diagram for the Fe-O2-C-H2O system involving other Fe mineralsHematite = Fe3+

2O3 (Fe2O3)

Magnetite = Fe2+Fe3+2O4 (Fe3O4)

Siderite = FeCO3

Pyrite = FeS2

Another type of stability (activity-activity) diagram: Complexation of Hg in solution depending on the concentrations of bromide and iodide as ligands