Redox student part4 pourbaix diagrams fe

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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+] = M): Fe3+ 0.77 V -0.44 V Fe(OH)2 Fe In alkali ([OH-] = M) -0.56 V Fe3+ -0.887 V Fe(OH)2 Fe Pourbaix diagrams: •correlate Latimer diagrams at pH 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 – 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)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 For example 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 < 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 Ligand phenanthroline H2O CN- CN - Eo /V 1.14 0.77 0.36 - NC CN - Fe - NC CN CN - - Negatively charged ligands favour the higher positive charge of Fe(III) NH H 3N Co3+|Co2+ NH Co H 3N 0.11 V NH NH3 is a better σ donor ligand than H2O and so stablises Co(III) NH OH H 2O OH Co H 2O OH OH 1.84 V [...]...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... 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... 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... 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... 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 Ligand phenanthroline H2O CN- CN - Eo /V 1.14 0.77 0.36 - NC CN - Fe - NC CN CN - - Negatively charged ligands favour the higher positive charge of Fe( III) NH 3 H 3N Co3+|Co2+ NH 3 Co H 3N 0.11 V NH 3 NH3 is a better σ donor ligand
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