Choosing an oxidant for iron(II) in chloride-containing solution
A technician must oxidise all of the iron(II) in 2.0 dm³ of a solution that is 0.050 mol dm⁻³ in Fe²⁺, so that iron(III) hydroxide can be precipitated and sold as a pigment. The solution was made from iron(II) chloride and is therefore also 0.10 mol dm⁻³ in chloride ion; it is 1.0 mol dm⁻³ in sulfuric acid. Three oxidants are on the shelf: solid potassium dichromate(VI), potassium manganate(VII) solution, and 30 % w/w hydrogen peroxide (density 1.11 g cm⁻³). Standard reduction potentials (V): Fe³⁺/Fe²⁺ +0.77; Cr₂O₇²⁻/Cr³⁺ +1.33; MnO₄⁻/Mn²⁺ +1.51; MnO₂/Mn²⁺ +1.23 (in acid); Cl₂/Cl⁻ +1.36; H₂O₂/H₂O +1.78; O₂/H₂O₂ +0.68. The technician must balance pigment purity, cost and waste disposal.
1. Write balanced equations for the oxidation of Fe²⁺ by each of the three oxidants and calculate E°cell for each. What does this tell you about choosing between them?
Read the worked answer
Dichromate(VI) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (+1.33 V) 6Fe²⁺ → 6Fe³⁺ + 6e⁻ Overall: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O E°cell = 1.33 − 0.77 = +0.56 V; ΔG° = −6F(0.56) ≈ −324 kJ mol⁻¹
Manganate(VII) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (+1.51 V) Overall: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O E°cell = 1.51 − 0.77 = +0.74 V; ΔG° ≈ −357 kJ mol⁻¹
Hydrogen peroxide H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (+1.78 V) Overall: H₂O₂ + 2H⁺ + 2Fe²⁺ → 2Fe³⁺ + 2H₂O E°cell = 1.78 − 0.77 = +1.01 V; ΔG° ≈ −195 kJ mol⁻¹
All three have E°cell > 0 and large negative ΔG°, so all are thermodynamically capable of oxidising Fe²⁺ essentially completely. Thermodynamics cannot discriminate between them - the choice must be made on side reactions, contamination, cost and disposal, which is exactly the judgement the technician has to make.
2. Using the quoted potentials, explain why potassium manganate(VII) is a poor choice for this particular solution, and what the practical consequences would be for the pigment.
Read the worked answer
The solution contains 0.10 mol dm⁻³ Cl⁻, and chloride can be oxidised:
2Cl⁻ → Cl₂ + 2e⁻ (Cl₂/Cl⁻ = +1.36 V)
With MnO₄⁻/Mn²⁺ at +1.51 V, E°cell = +0.15 V > 0, so the side reaction is thermodynamically allowed. It is slow at room temperature but becomes significant in hot, strongly acidic solution, and especially near the end-point, when [Fe²⁺] has fallen so low that the potential of the mixture is set by the MnO₄⁻/Mn²⁺ couple and rises towards +1.51 V.
Consequences
- Chlorine is evolved - toxic, corrosive, and a fume-hood problem.
- Permanganate is consumed beyond the 5:1 stoichiometry, so the end-point (persistent pink) no longer corresponds to complete oxidation and the reagent cost rises.
- In less acidic regions, or where Mn²⁺ accumulates, MnO₄⁻ can be reduced only as far as brown MnO₂ (MnO₂/Mn²⁺ = +1.23 V). Colloidal MnO₂ carried down with the hydroxide gives a brown-black, off-specification pigment, and Mn²⁺ left in the liquor is an effluent problem.
Judgement: permanganate gives an uncontrollable stoichiometry and an unacceptable contamination risk here. Notice that dichromate (1.33 V) is below the Cl₂/Cl⁻ potential, so it cannot oxidise chloride - that is the chemically decisive comparison.
3. Recommend which oxidant the technician should use and justify your choice quantitatively, including the mass or volume of reagent needed, and comment on how the later precipitation step affects the decision.
Read the worked answer
n(Fe²⁺) = 2.0 × 0.050 = 0.100 mol
- Dichromate: 0.100/6 = 0.0167 mol; m = 0.0167 × 294.2 = 4.9 g
- Permanganate: 0.100/5 = 0.0200 mol of MnO₄⁻
- Hydrogen peroxide: 0.100/2 = 0.0500 mol; m(H₂O₂) = 0.0500 × 34.0 = 1.70 g; mass of 30 % solution = 1.70/0.30 = 5.67 g, i.e. ≈ 5.1 cm³
Recommendation: hydrogen peroxide, for purity reasons. It is the strongest oxidant of the three (E°cell = +1.01 V), so oxidation is essentially complete, and its only reduction product is water - no metal ion is introduced into a pigment that will be sold on colour and composition. A small excess is harmless: it decomposes to O₂ and water, and can be destroyed by warming the solution to about 50 °C before precipitation. The stoichiometric amount is, however, hard to judge, so a slight excess is normal practice.
Caveats to state: H₂O₂ can oxidise chloride in principle (1.78 − 1.36 = +0.42 V), but the reaction is slow at room temperature without a catalyst, so the solution should be kept cool and stirred while the peroxide is added slowly, and no transition-metal catalyst should be present.
If dichromate were used instead: Cr³⁺ remains in solution. Fe(OH)₃ precipitates around pH 3–4 whereas Cr(OH)₃ needs about pH 5–6, so careful pH control could still give a tolerable pigment, but Cr(VI) is carcinogenic and the spent liquor would need reduction and controlled disposal - a real cost penalty.
Permanganate is rejected because of chloride oxidation and MnO₂ inclusion.