The Fe 3+/Fe 2+ and the H3AsO4/H3AsO3 systems are important redox systems in analytical chemistry, b — Analytical Chemistry Chemistry Question
Theoretical Problem 12
The Fe 3+/Fe 2+ 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.
Calculate the standard redox potential, Eº3, of the reaction Fe 3+ + e – → Fe 2+.
Model Answer
Eº3 = (3 Eº2 – 2 Eº1) = 0.772 V
The standard redox potential of the Fe 3+ / Fe 2+ 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([Fe 3+] / [Fe 2+]) = 0.710 V
[Fe 3+] / [Fe 2+]) = 0.0890
A very rough estimate:
Kst = [FeCl 2+] / ([Fe 3+][Cl –]) = 0.911 / (0.089×0.089) = 115
Both Fe 3+ and Fe 2+ 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
βIII6 = [Fe(CN)6 3–] / ([Fe 3+][CN –]6)
βII6 = [Fe(CN)6 4–] / ([Fe 2+][CN –]6)
Eº4 = Eº3 + 0.059 lg(βIII6 / βII6)
βIII6 / βII6 = 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.
Fe 2+ / Fe Eº1 = –0.440 V
Fe 3+ / 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
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