Electric Work

• Energy drives all changes including chemical reactions. In a redox reaction, the energy released in a reaction due to movement of charged particles give rise to a potential difference. The maximum potential differenc is called the electromotive force, (EMF), E and the maximum electric work W is the product of charge q in Coulomb (C), and the potential ΔE in Volt (= J / C) or EMF.

W = q Δ E

• Note that the EMF ΔE is determined by the nature of the reactants and electrolytes, not by the size of the cell or amounts of material in it. The amount of reactants is proportional to the charge and available energy of the galvanic cell.

Gibb's Free Energy

• The Gibb's free energy DG is the negative value of maximum electric work,

ΔG = - W      = - q ΔE

• A redox reaction equation represents definite amounts of reactants in the formation of also definite amounts of products. The number (n) of electrons in such a reaction equation, is related to the amount of charge trnasferred when the reaction is completed. Since each mole of electron has a charge of 96485 C (known as the Faraday's constant, F),

q = n F

and, ΔG = - n F DE

At standard conditions,

DG° = - n F DE°

Nerst Equation

• The general Nernst equation correlates the Gibb's Free Energy DG and the EMF of a chemical system known as the galvanic cell. For the reaction

a A + b B = c C + d D [C]c [D]d Q = --------- [A]a [B]b

ΔG = ΔG° + R T ln Q and

ΔG = - n F ΔE. Therefore

- n F DE = - n F DE° + R T ln Q

where R, T, Q and F are the gas constant (8.314 J mol-1 K-1), temperature (in K), reaction quotient, and Faraday constant (96485 C) respectively. Thus, we have

Nerst Equation

R T [C]c [D]d ΔE = ΔE° - ----- ln ---------

n F [A]a [B]b

This is known as the Nernst equation. The equation allows us to calculate the cell potential of any galvanic cell for any concentrations. Some examples are given in the next section to illustrate its application. It is interesting to note the relationship between equilibrium and the Gibb's free energy at this point. When a system is at equilibrium, ΔE = 0, and Therefore, we have,

R T [C]c [D]d ΔE° = ----- ln ---------

n F [A]a [B]b

Nerst Equation

• At any specific temperature, the Nernst equation derived above can be reduced into a simple form. For example, at the standard condition of 298 K (25°), the Nernst equation becomes

0.059 [C]c [D]d ΔE = ΔE° - ----- log --------- n [A]a [B]b

Nerst Equation

For the cell

Zn | Zn2+ || H+ | H2 | Pt

we have a net chemical reaction of

Zn(s) + 2 H+ = Zn2+ + H2(g)

and the standard cell potential ΔE° = 0.763.

If the concentrations of the ions are not 1.0 M, and the H2 pressure is not 1.0 atm, then the cell potential ΔE may be calculated using the Nernst equation:

0.0592 V P(H2) [Zn2+] ΔE = ΔE° - ------- log ------------ n [H+]2

Eh – pH or Pourbaix Diagrams

Eh-pH or Pourbaix Diagrams are plots of Eh versus pH showing regions or fields where dissolved species and precipitates are stable. They can be used to quickly determine the equilibrium stability fields for aqueous species.

The effective overall boundaries of the diagram are determined by the stability field for water. One boundary is determined by the stability of water with respect to reduction to H2 at 1 bar:

2 e- + 2 H2O (l) ------> H2 (g, 1 bar) + 2 OH- (aq)

Writing the usual expression for the Gibb’s free energy change:

G = Go + R T ln [ a H2 (g) a2 OH- (aq) / a2 H2O (l) ]

We also have:

G = - n F

where F, Faraday’s constant, is the charge on a mole of electrons and is equal to 96,570 C, and n is the number of moles of electrons transferred in the oxidation or reduction.

Using the fact that the activities of molecular hydrogen at 1 bar and liquid water can be set equal to 1 gives:

- n F = Go + 2.303 R T log a2 OH- (aq)

or:

= - Go / (n F) - (2.303 R T / (n F) log a2 OH- (aq)

At 25 oC:

2.303 R T / F = 2.303 (8.314 J/mol K) (298.15 K) / (96,486.8 C)

= 0.05917 V

Thus:

= - Go / (n F) - ( 0.05917 V / n ) log a2 OH- (aq)

= - Go / (2 F) - 2 ( 0.05917 V / 2 ) log a OH- (aq)

= - Go / (2 F) - ( 0.05917 V ) log a OH- (aq)

= - Go / (2 F) + ( 0.05917 V ) pOH

= - Go / (2 F) + ( 0.05917 V ) ( 14.0 – pH )

Go can be calculated from standard tabulated Gibb’s free energies of formation at 298.15 K:

Go = 2 Go m,

f, OH

- (aq)

- 2 Go m,

f, H2O

(l)

= 2 ( - 157.300 kJ ) - 2 ( - 237.2 kJ )

= + 159.8 kJ

= - Go / (2 F) + ( 0.05912 V ) ( 14.0 – pH )

= - (+ 159.8 10+3 J) / (2 (96,570 C) ) + 0.8283 V

- ( 0.05911 V ) pH

= ~ 0 V - ( 0.05911 V ) pH

This equation can be plotted on an Eh-pH diagram to give one of the stablility boundaries for water:

Below this line water is unstable with respect to reductive decomposition to H2 (g) at 1 bar of pressure.

How would this line change, if the H2 (g) pressure were set at 10 bar or 0.1 bar?

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

2 e - + 2 H2O (l) ----> H2 (g, 1 bar) + 2 OH - (aq)

With this background the stability boundary for the oxidative decompostion of water to O2 (g) at 1 bar is relatively easily calculated:

H2O (l) ------> ½ O2 (g, 1 bar) + 2 H+ (aq) + 2 e-

Note for consistency in constructing the diagram this decomposition must be written as a reduction:

½ O2 (g, 1 bar) + 2 H+ (aq) + 2 e- ------> 1 H2O (l)

= - Go / (n F) - ( 0.05911 V / n ) log [ a2 H2O (l) / a O2 (g) a2H+ (aq) ]

= - Go / (2 F) - ( 0.05911 V / 2 ) log a-2H+ (aq)

= - Go / (2 F) - ( 0.05911 V ) pH

Calculating Go in this case gives:

Go = + 1 Go m,

f, H2O

(l) = + 1 ( - 237.2 kJ )

= - 237.2 kJ

gives:

= - Go / (2 F) - ( 0.05917 V ) pH

= - (- 237.2 10+3 J / (2 (( 96,486.8 C )) - ( 0.05911 V ) pH

= + 1.23 V - ( 0.05911 V ) pH

This equation can be plotted on an Eh-pH diagram to give the other stablility boundary for water:

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

) H2O (l) ----> 1/2 O2 (g, 1 bar) + 2 H + (aq) + 2 e -

Above this line water is unstable with respect to oxidative decomposition to O2 (g) at 1 bar of pressure.

The regions within the stability field for water where Fe3+ and Fe2+ are stable can be determined by considering the reduction of iron (III) to iron (II):

Fe3+ (aq) ------> Fe2+ (aq) + e –

Go = Gom,

f, Fe2+

- Go m,

f, Fe3+

= ( - 78.87 kJ ) - ( - 4.6 kJ )

= - 74.27 kJ

= - Go / (n F) - ( 0.05911 V / n ) log [ a Fe2+ / a

Fe3+ ]

= - ( -74.27 10+3 J / (1 (( 96,570 C ))

- ( 0.05911 V / 1 ) log [ a Fe2+ / a Fe3+ ]

= + 0.769 V - ( 0.05911 V ) log [ a Fe2+ / a Fe3+ ]

To further develop this equation requires making a choice about the activities of Fe3+ and Fe2+. The usual choice is to take the activities of dissolved species as 10-6, although other choices can be made.

= + 0.769 V - ( 0.05911 V ) log [10-6 / 10-6 ]

= + 0.769 V

Notice that is independent of pH, because the reduction we were considering did not involve, H+ (aq) or OH- (aq) and will plot as a straight line parallel to the pH axis on the Eh-pH diagram.

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

Fe3+ (aq) ----> Fe

2+ (aq) + e -

Fe3+ (aq) stable

Fe2+ (aq) stable

Note that the more oxidized species, in this case Fe3+, is stable at the higher Eh.

Now consider the relative stability of Fe3+ (aq) and FeO(OH) (s), the mineral goethite:

FeO(OH) (s) + 3 H+ (aq) ------> Fe3+ (aq) + 2 H2O (l)

Note that in this equilibrium no species is being oxidized or reduced.

Go = Gom,

f, Fe3+

+ 2 Gom,

f, H2O (l) - Go

m, f, FeO(OH)

(s)

= ( - 4.6 kJ ) + 2 ( - 237.2 kJ ) - ( - 488.57 kJ )

= + 9.6 kJ

Why was Gom,

f, H+

not included in this calculation?

(n = 0 ) = - Go / F - ( 0.05911 V ) log [ a Fe3+ / a FeO(OH) (s) ]

- ( 0.05911 V ) 3 pH

0 = - ( + 9.6 10+3 J ) / ( 96,570 C )

- ( 0.05911 V ) log [ 10-6 / 1 ]

- ( 0.05911 V ) 3 pH

= ( - 9.94 10-5 V ) - 6 (- 0.05911 V ) - ( 0.1773 V ) pH

= + 0.255 V - ( 0.1773 V ) pH

pH = 1.44

Notice that when the equilibrium did not involve and oxidation or

reduction, pH is independent of and will plot as a straight line

parallel to the axis on the Eh-pH diagram.

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

Fe3+

Fe2+

FeO(OH) (s)

pH = 1.44

Now consider the relative stability of Fe2+ (aq) and FeO(OH) (s):

FeO(OH) (s) + 3 H+ (aq) + e- ------> Fe2+ (aq) + 2 H2O (l)

Go = Gom,

f, Fe2+

+ 2 Gom,

f, H2O (l) - Go

m, f,

FeO(OH) (s)

= ( - 78.87 kJ ) + 2 ( - 237.2 kJ ) - ( - 488.57 kJ )

= - 64.7 kJ

= - Go / (n F) - ( 0.05911 V / n ) log [ a Fe2+ ]

- ( 0.05911 V ) 3 pH

= - ( - 64.7 10+3 J / (1 (( 96,570 C ))

- ( 0.05911 V / 1 ) log [ 10-6 ]

- ( 0.05911 V ) 3 pH

= + 0.670 V + 0.355 V - ( 0.1773 V ) pH

= + 1.025 V - ( 0.1773 V ) pH

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh (V

)

Fe3+

Fe2+

FeO(OH) (s)

FeO(OH) (s) + 3 H+ (aq) + e- --> Fe2+ (aq) + 2 H2O (l)

Any oxidation or reduction equilibrium that involves H+ (aq) or OH- (aq) will plot as a sloped line on these Eh-pH diagrams.

Now consider the relative stability of FeO(OH) (s) and Fe3O4 (s), magnetite:

3 FeO(OH) (s) + H+ (aq) + e- ------> Fe3O4 (s) + 2 H2O (l)

Go = Go m,

f, Fe3O4 (s) + 2 Go

m, f, H3O (l) - 3 Go

m, f,

FeO(OH) (s)

= ( - 1015 kJ ) + 2 ( - 237.2 kJ ) - 3 ( - 488.57 kJ )

= - 24.13 kJ

= - Go / (n F) - ( 0.05911 V ) 1 pH

= - ( - 24.13 10+3 J / (1 (( 96,570 C ))

- ( 0.05911 V ) 1 pH

= + 0.250 V - ( 0.05911 V ) pH

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh (V

)

Fe3+

Fe2+

FeO(OH) (s)

3 FeO(OH) (s) + H+ (aq) + e- --> Fe3O4 (s) + 2 H2O (l)

Fe3O4 (s)

Now consider the relative stability of Fe2+ (aq) and Fe3O4 (s), magnetite:

Fe3O4 (s) + 8 H+ (aq) + 2 e- ------> 3 Fe2+ (aq) + 4 H2O (l)

Go = 3 Gom,

f, Fe2+

+ 4 Gom,

f, H2O (l) - Go

m, f,

Fe3O4 (s)

= 3 ( - 78.87 kJ ) + 4 ( - 237.2 kJ ) - ( - 1015 kJ )

= - 170 kJ

= - Go / (n F) - ( 0.05911 V / n ) log [ a3 Fe2+ ]

- (( 0.05911 V ) 8 / 2 ) pH

= - ( - 170 10+3 J / (2 ( 96,570 C )))

- 3 ( 0.05911 V / 2 ) log [ 10-6 ]

- (( 0.05911 V ) 8 / 2 ) pH

= + 0.880 V + 0.532 V - ( 0.1773 V ) pH

= + 1.025 V - ( 0.2364 V ) pH

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

Fe3+

Fe2+

FeO(OH) (s)

Fe3O4 (s) + 8 H+ (aq) + 2 e- --> 3 Fe2+ (aq) + 4 H2O (l)

Fe3O4 (s)

Note that Fe2+ (aq) - Fe3O4 (s) line meets the intersection of the FeO(OH) (s) - Fe3O4 (s) and Fe2+ (aq) - FeO(OH) (s) lines, as, of course, it should. These expectations can be used to reduce the effort required to construct these diagrams.

Based on the above diagram what iron species and what iron equilibria would you expect to dominate natural waters containing iron?

How does the Eh-pH diagram change if activities of the aquesous ions involved in the various equilibria are changed by a factor of 10 in either direction from the 10-6 values that have been used so far?

Using the result for the Fe2+ (aq) and Fe3O4 (s) equilibria previously derived for aFe2+ = 10-6:

= + 0.880 V - 3 ( 0.05911 V / 2 ) log [ 10-6 ] - ( 0.1773 V ) pH

equations can be quickly derived for:

aFe2+ = 10-5:

= + 0.880 V - 3 ( 0.05911 V / 2 ) log [ 10-5 ] - ( 0.1773 V ) pH

= 1.323 V - ( 0.1773 V ) pH

and for aFe2+ = 10-7:

= + 0.880 V - 3 ( 0.05911 V / 2 ) log [ 10-7 ] - ( 0.1773 V ) pH

= 1.500 V - ( 0.1773 V ) pH

The positions of the other lines of equilibria can relatively quickly established based on the intersections of the above lines with lines that don’t shift and the fact the changing these activities does not affect the slope of the lines.

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

Fe3+

Fe2+

FeO(OH) (s)

Fe3O4 (s)

a = 10-7

a = 10-5

a = 10-6

Note that where the aqueous ionic species is in the products of the equilibria used to derive these lines, decreasing that ionic species activity increases it’s stability field.

Where does elemental iron occur in this Eh-pH diagram?

Consider the reduction of Fe2+ (aq) to Fe (s):

Fe2+ (aq) + 2 e – ------> Fe (s)

Go = Gom,

f, Fe (s)

- Go m,

f, Fe2+

= ( 0 kJ ) - ( - 78.87 kJ )

= + 78.87 kJ

= - Go / (n F) - ( 0.05911 V / n ) log [ 1 / a

Fe2+ ]

= - ( +78.87 10+3 J / (2 (( 96,570 C ))

- ( 0.05911 V / 2 ) log [ 1 / a Fe2+ ]

= - 0.4084 V - ( 0.05911 V / 2 ) log [ 1 / 10-6 ]

= - 0.4084 V - 0.1773 V

= - 0.5857 V

Also consider the reduction of magnetite to elemental iron:

Fe3O4 (s) + 8 H+ (aq) + 8 e- ------> 3 Fe (s) + 4 H2O (l)

Go = 4 Gom,

f, H2O (l) - Go

m, f, Fe3O4 (s)

= 4 ( - 237.2 kJ ) - ( - 1015 kJ )

= + 66.2 kJ

= - Go / (n F - (( 0.05911 V ) 8 / 8 ) pH

= - ( + 66.2 10+3 J / (8 ( 96,570 C )))

- ( 0.05911 V ) pH

= - 0.086 V - ( 0.05911 V ) pH

Plotting these lines on the Eh-pH diagram gives:

Would you expect to find elemental iron in a wet oxygen rich environment?

Under what conditions of Eh and pH is siderite, FeCO3 (s), stable? To answer this question assume that PCO2 (g) is 10-3.47 bar, roughly the global average value for the partial pressure of carbon dioxide.

Eh - pH or Pourbaix Diagram

-1.10

-0.60

-0.10

0.40

0.90

1.40

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

pH

Eh

(V

)

Fe3+

Fe2+

FeO(OH) (s)

Fe3O4 (s)

Fe (s)

The iron – oxygen diagram that we have developed is a strong function of the species considered and would look differently, if other species were taken into account, e.g. FeOH2+ (aq), Fe(OH)2

+, etc. For example, if hematite, Fe2O3 (s), had been considered, it would occupy the entire stability field now occupied by geothite, FeO(OH) (s) and since hematite is thermodynamically more stable, it would be the species represented on the diagram. Geothite is a precursor to hematite and kinetically it takes several months to convert geothite to hematite and so the diagram developed in these notes would apply to the short time formation of iron precipitates.

An Atlas of Eh pH Diagrams, Intercomparison of Thermodynamic Databases, Geological Survey of Japan, Open File Report No. 419, National Institute of Advanced Industrial Science and TechnologyResearch Center for Deep Geological Environments Naoto TAKENO, May 2005 is available on the web at:

http://www.gsj.jp/GDB/openfile/files/no0419/openfile419e.pdf

Pages 101 to 103 of this reference cover the iron oxygen system.

There are a number of commercial software packages that will generate Eh pH diagrams, notably STABCAL:

http://www.mtech.edu/hhuang/predominance.htm

developed by Hsin-Hsiung Huang of the Metallurgical Engineering Department at Montana Tech.

These comments just point out that these diagrams should be used with caution.