The Fe3+/Fe2+ and the H3AsO4/H3AsO3 systems are important redox systems in analytical chemistry, bec — Analytical Chemistry Chemistry Question
Redox Systems: Fe3+/Fe2+ and H3AsO4/H3AsO3
The Fe3+/Fe2+ and the H3AsO4/H3AsO3 systems are important redox systems in analytical chemistry, because their electrochemical equilibrium can be shifted by complex formation or by varying the pH.
Given standard redox potentials:
Fe2+/ Fe Eº1 = –0.440 V
Fe3+/ Fe Eº2 = –0.036 V
[Fe(CN)6]3–/ [Fe(CN)6]4– Eº4 = +0.356 V
H3AsO4 / H3AsO3 Eº5 = +0.560 V
I2 / 2 I– Eº6 = +0.540 V
Calculate the standard redox potential, Eº3, of the reaction Fe3+ + e– → Fe2+.
Model Answer
Eº3 = (3 Eº2 – 2 Eº1) = 0.772 V
The standard redox potential of the Fe3+/ Fe2+ system in 1 mol dm–3 HCl is 0.710 V.
Give an estimate for the stability constant of the complex [FeCl]2+.
Model Answer
E3 = Eº3 + 0.059 lg([Fe3+] / [Fe2+]) = 0.710 V
[Fe3+] / [Fe2+] = 0.0890
A very rough estimate:
[FeCl2+] / ([Fe3+][Cl–]) = 0.911 / 0.089 = 115
Both Fe3+ and Fe2+ ions form a very stable complex with CN– ions.
Calculate the ratio of the cumulative stability constants for the formation of [Fe(CN)6]3– and [Fe(CN)6]4– ions.
Model Answer
Eº4 = 0.356 V
[Fe(CN)6]3– / [Fe3+][CN–]6 = β6(FeIII)
[Fe(CN)6]4– / [Fe2+][CN–]6 = β6(FeII)
Eº4 = Eº3 + 0.059 lg(β6(FeII) / β6(FeIII))
β6(FeII) / β6(FeIII) = 10^(–7.05) = 8.90 × 10^–8
H3AsO4 and K4Fe(CN)6 are dissolved in water in a stoichiometric ratio. What will the [H3AsO4]/[H3AsO3] ratio be at equilibrium if pH = 2.00 is maintained?
Model Answer
H3AsO4 + 2 H+ + 2 e– = H3AsO3 + H2O
Eº5´ = Eº5 + (0.059/2) lg [H3O+]2 = Eº5 – 0.059 pH = 0.442 V
E = (2 × 0.442 + 0.356) / 3 = 0.413 V
0.413 = 0.442 + (0.059 / 2) × lg([H3AsO4] / [H3AsO3])
[H3AsO4] / [H3AsO3] = 0.107
Are the following equilibrium concentrations possible in an aqueous solution? If yes, calculate the pH of the solution.
[H3AsO4] = [H3AsO3] = [I3–] = [I–] = 0.100 mol dm–3.
Model Answer
E6 = Eº6 + 0.059/2 × lg([I3–] / [I–]3) = 0.540 + (0.059/2)×2 = 0.599 V
E5 = 0.599 V = 0.560 – 0.059 pH
pH = – 0.66