Transcript
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)
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