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Chapter 20: Problem 39

Iron(II) can be oxidized to iron(III) by permanganate ion in acidic solution. The permanganate ion is reduced to manganese(II) ion. (a) Write the oxidation half-reaction, the reduction half-reaction, and the overall redox equation. (b) Calculate \(E^{\circ}\) for the reaction. (c) Calculate the percentage of Fe in an ore if a \(0.3500-\mathrm{g}\) sample is dissolved and the \(\mathrm{Fe}^{2+}\) formed requires for titration \(55.63 \mathrm{~mL}\) of a \(0.0200 \mathrm{M}\) solution of \(\mathrm{KMnO}_{4}\)

Short Answer

Expert verified
Answer: Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ + e⁻ Reduction Half-Reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Overall Redox Equation: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O E° = +2.28 V Percentage of Fe in the ore ≈ 88.74%

Step by step solution

01

Oxidation Half-Reaction

Fe^{2+} \rightarrow Fe^{3+} + e^-
02

Reduction Half-Reaction

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O Now, we will balance the two half-reactions by multiplying them with suitable numbers, and then we will add them to get the overall redox equation.
03

Balancing half-reactions and writing overall redox equation

5 \times (Fe^{2+} \rightarrow Fe^{3+} + e^-) \\ \\ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \\ Now, we can add the two balanced half-reactions: \\ 5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O #b) Calculation of E° for the given reaction#
04

Reduction and Oxidation potentials

For MnO_4^-: E°_{reduction} = +1.51 V \\ For Fe^{2+}: E°_{oxidation} = -0.77 V Now, we will calculate the E° for the given reaction by using the Nernst equation. The E° for a redox reaction is given by: E° = E°_{cathode} - E°_{anode}
05

E° Calculation

E° = E°_{reduction} - E°_{oxidation} \\ E° = (+1.51 V) - (-0.77 V) \\ E° = +2.28 V #c) Calculation of percentage of Fe in the ore#
06

Moles of KMnO4

We know that the \(\mathrm{Fe}^{2+}\) formed requires for titration \(55.63 \mathrm{~mL}\) of a \(0.0200 \mathrm{M}\) solution of \(\mathrm{KMnO}_{4}\). \\ Moles of KMnO4 = Molarity × Volume(L) \\ Moles of KMnO4 = 0.0200 mol/L × 55.63 mL × (1 L/1000 mL) \\ Moles of KMnO4 = 1.1126 × 10^{-3} mol
07

Moles of Fe

From the balanced equation, 1 mol of MnO4^- requires 5 mol of Fe^2+ for complete reduction. \\ Moles of Fe = 5 × Moles of KMnO4 \\ Moles of Fe = 5 × 1.1126 × 10^{-3} mol \\ Moles of Fe = 5.563 × 10^{-3} mol
08

Mass of Fe in the sample

We can now find the mass of Fe in the sample by multiplying the moles of Fe by its molar mass. \\ Mass of Fe = Moles of Fe × Molar mass of Fe \\ Mass of Fe = 5.563 × 10^{-3} mol × 55.845 g/mol \\ Mass of Fe = 0.3106 g
09

Percentage of Fe in the ore

Finally, we will calculate the percentage of Fe in the ore. \\ Percentage of Fe = (Mass of Fe in the sample / Mass of the sample) × 100 \\ Percentage of Fe = (0.3106 g / 0.3500 g) × 100 \\ Percentage of Fe ≈ 88.74 %

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