Redox student part4 pourbaix diagrams fe

Pourbaix diagrams
Plots of E vs pH
We will, as an example, derive the Pourbaix diagram for iron
Two Latimer diagrams pertain
In acid ([H+] = 1 M):
Fe3+

0.77 V

-0.44 V

Fe(OH)2

Fe

In alkali ([OH-] = 1 M)
-0.56 V

Fe3+

-0.887 V

Fe(OH)2

Fe

Pourbaix diagrams:
•correlate Latimer diagrams at pH 0 and pH 14
•take into account speciation or oxidation state of the element

Fe3+

0.77 V

-0.44 V

Fe(OH)2

The half reaction
Fe3+ + e → Fe2+

Eo = 0.77 V

does not involve a proton so Eo is independent of pH

Fe


Fe3+ + e → Fe2+


Fe3+ will precipitate out of solution as pH is increased. We can calculate the pH
at which this will occur from the KSP for Fe(OH)3.
Fe(OH)3(s) Ý Fe3+ + 3OH–

KSP = 4.11 x 10-37 M4

At what pH will [Fe3+] = 1.00 M?
KSP = 4.11 x 10-37 M4 = [Fe3+][OH–]3

[OH–] = (4.11 x 10-37/1)0.333
= 7.43 x 10-13 M
So [H+] = 10-14/7.43 x 10-13 = 1.35 x 10-2 M
hence pH = 1.87


Fe(OH)3 Ý Fe3+ + 3OH-

Vertical lines in a Pourbaix
diagram indicate where two
species of an element in the
same oxidation state are in
equilibrium


To calculate the
Fe(OH)3|Fe2+
line...


Fe3+ + e → Fe2+

Eo = 0.77 V

3OH- + 3H+ → 3H2O

∆Go = -74.3 kJ mol-1
-239.7 kJ mol-1

Fe(OH)3 → Fe3+ + 3OHFe(OH)3 + 3H+ + e → Fe2+ + 3H2O

207.6 kJ mol-1
∆Go = -nFEo
-106.4
kJ mol-1
= -1 x 96485
x 0.77
∆Go = -RT ln KSP
= -8.315 x 298 x ln (4.11 x 10-37)

∆Go = -nFEo
Eo = -∆Go /nF
= 106400/1 x 96485
= 1.10 V


Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O

Eo = 1.10 V

E = Eo – RT/nF ln Q
E = 1.10 – 3 x 0.0592 x pH

This must cross the Fe3+/Fe(OH)3 line when

0.77 = 1.10 – 3(0.0592)pH
or pH = 1.87
which confirms the result we got from the KSP calclation

Fe3+

0.77 V

-0.44 V

Fe(OH)2

Fe


1.1

Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O


1.1

Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O


From the KSP for Fe(OH)2
Fe(OH)2 Ý Fe2+ + 2OH–

KSP = 1.61 x 10-15 M3

At what pH will [Fe2+] = 1.00 M?
KSP = 1.61 x 10-15 M3 = [Fe2+][OH–]2
[OH–] = (1.61 x 10-15/1)0.5
= 4.01 x 10-8 M
So [H+] = 10-14/4.01 x 10-8 = 2.49 x 10-7 M

hence pH = 6.61


1.1

Fe(OH)2 Ý Fe2+ + 2OH-


The half reaction
Fe2+ + 2e → Fe

Eo = -0.44 V

does not involve a proton so Eo is independent of pH
1.1

Fe2+ + 2e → Fe


An expression for the potential for the Fe(OH)3|Fe(OH)2 couple can be derived
from the following data
Fe(OH)3 + 3H+ + e → Fe2+ + 2H2O

Eo 1.10 V

∆Go -106.4 kJ mol-1

3H2O → 3H+ + 3OH-

239.7 kJ mol-1

Fe2+ + 2OH- → Fe(OH)2

-84.4 kJ mol-1

Fe(OH)3 + e → Fe(OH)2 + OH-

Eo -0.51 V

∆Go 48.9 kJ mol-1

E = Eo – RT/nF ln Q
E = -0.51 + 0.0592 x pOH
E = -0.51 + 0.0592 x (14 – pH)
E = 0.316 – 0.0592 x pH


1.1

0.316

Fe(OH)3 + e → Fe(OH)2 + OH-


...and finally the value of Fe(OH)2|Fe couple can be found by
similar considerations, and the Nernst equation applied.
E = -0.060 – 0.0592 x pH

Overlaying Pourbaix diagrams
The feasibility of a reaction can be predicted by overlaying the relevant
Pourbaix diagrams


stability field for As(V)

stability field for As(III)


At pH < 5.5 and at
pH > 9, Fe3+ has the
potential to oxidise
As3+ to As5+

5.5

9


For example

0.65
0.45

Fe(OH)3 + e + 3H+ → Fe2+ + 3H2O
E = 0.65
2
As3+ → As5+ + 2e
E = -0.45

As3+ + 2Fe(OH)3 + 6H+ → 2Fe2+ + 6H2O + As5+ E = 0.20 V


For 5.5 < pH < 9
As5+ will oxidise Fe2+
to Fe3+


The effect of complex formation on Eo values
The Eo value of a metal ion is very dependent on the ligands of the ion

Example, for the Fe3+|Fe2+ couple
Ligand
phenanthroline
H2O
CN-

Eo /V
1.14
0.77
0.36


Ligand
phenanthroline
H2O
CN-

N

N

N
N

N

N

N
Fe

Fe
N

N
N

π back bonding from
metal to phen ligand
stabilises Fe(II)

Eo /V
1.14
0.77
0.36