Redox Student POURBAIX

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+ Fe(OH)2 Fe0.77 V -0.44 V

In alkali ([OH-] = 1 M)

Fe3+ Fe(OH)2 Fe-0.56 V -0.887 V

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

The half reaction

Fe3+ + e → Fe2+ Eo = 0.77 V

does not involve a proton so Eo is independent of pH

Fe3+ Fe(OH)2 Fe0.77 V -0.44 V

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)3Ý Fe3+ + 3OH-

Vertical lines in a Pourbaixdiagram 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 ∆Go = -74.3 kJ mol-1

3OH- + 3H+→ 3H2O -239.7 kJ mol-1

∆Go = -nFEo

= -1 x 96485 x 0.77

Fe(OH)3 → Fe3+ + 3OH- 207.6 kJ mol-1

∆Go = -RT ln KSP

= -8.315 x 298 x ln (4.11 x 10-37)

Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O -106.4 kJ mol-1

= 106400/1 x 96485= 1.10 V

∆Go = -nFEo

Eo = -∆Go /nF

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)pHor pH = 1.87

which confirms the result we got from the KSP calclation

Fe3+ Fe(OH)2 Fe0.77 V -0.44 V

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)2Ý 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)

95.5

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

For example

2

0.65

0.45

Fe(OH)3 + e + 3H+→ Fe2+ + 3H2O E = 0.65

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 Eo /Vphenanthroline 1.14H2O 0.77CN- 0.36

N N

Fe

N

N N

N

N

N

N N

Fe

π back bonding frommetal to phen ligandstabilises Fe(II)

Ligand Eo /V

phenanthroline 1.14

H2O 0.77

CN- 0.36

Ligand Eo /V

phenanthroline 1.14

H2O 0.77

CN- 0.36

Fe-NC

-NC CN-

CN-

CN-

CN-

Negatively charged ligands favour the higher positive charge of Fe(III)

Co

H3N

H3N NH3

NH3

NH3

NH3

Co

H2O

H2O OH2

OH2

OH2

OH2

Co3+|Co2+

0.11 V

1.84 V

NH3 is a better σdonor ligand than H2O and so stablisesCo(III)