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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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Most popular questions from this chapter

Write a balanced equation to show (a) the reaction of chromate ion with strong acid. (b) the oxidation of water to oxygen gas by permanganate ion in basic solution. (c) the reduction half-reaction of chromate ion to chromium(III) hydroxide in basic solution.

A solution of potassium dichromate is made basic with sodium hydroxide; the color changes from red to yellow. Addition of silver nitrate to the yellow solution gives a precipitate. This precipitate dissolves in concentrated ammonia but re-forms when nitric acid is added. Write balanced net ionic equations for all the reactions in this sequence.

Rust, which you can take to be \(\mathrm{Fe}(\mathrm{OH})_{3}\), can be dissolved by treating it with oxalic acid. An acid-base reaction occurs, and a complex ion is formed. (a) Write a balanced equation for the reaction. (b) What volume of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) would be required to remove a rust stain weighing \(1.0 \mathrm{~g}\) ?

Write a balanced equation and give the names of the products for the reaction of (a) magnesium with chlorine. (b) barium peroxide with water. (c) lithium with sulfur. (d) sodium with water.

A \(0.500-\mathrm{g}\) sample of zinc-copper alloy was treated with dilute hydrochloric acid. The hydrogen gas evolved was collected by water displacement at \(27^{\circ} \mathrm{C}\) and a total pressure of \(755 \mathrm{~mm} \mathrm{Hg}\). The volume of the water displaced by the gas is \(105.7 \mathrm{~mL}\). What is the percent composition, by mass, of the alloy? (Vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) at \(27^{\circ} \mathrm{C}\) is \(26.74 \mathrm{~mm} \mathrm{Hg}\).) Assume only the zinc reacts.
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