Problem 39 - Chemistry: Principles and Reactions

Question

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 \(\mathrm{KMnO}_{4}\)

Step-by-Step Solution

Answer

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 1: Oxidation Half-Reaction

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

Step 2: 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.

Step 3: 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#

Step 4: 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}

Step 5: 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#

Step 6: 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

Step 7: 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

Step 8: 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

Step 9: 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 %

Key Concepts

Oxidation and Reduction Half-Reactions

To understand oxidation and reduction half-reactions, we need to zoom in on the redox processes. These reactions signify a chemical change where electrons are transferred between two substances. Understanding these half-reactions is crucial because they reveal how electrons move around in redox reactions.

Oxidation is the process where a substance loses electrons. For example, in the oxidation of iron(II) ions to iron(III) ions, the reaction can be shown as:

• Oxidation half-reaction: \( ext{Fe}^{2+} \rightarrow ext{Fe}^{3+} + ext{e}^- \)

Reduction, on the other hand, is when a substance gains electrons. In our example, permanganate ions are reduced to manganese(II) ions with this reaction:

• Reduction half-reaction: \( ext{MnO}_4^- + 8 ext{H}^+ + 5 ext{e}^- \rightarrow ext{Mn}^{2+} + 4 ext{H}_2 ext{O} \)

Combining the balanced half-reactions provides the overall redox equation, telling us the complete story of how electrons are exchanged and the products that emerge. In this case:

• Overall redox equation: \( 5 ext{Fe}^{2+} + ext{MnO}_4^- + 8 ext{H}^+ \rightarrow 5 ext{Fe}^{3+} + ext{Mn}^{2+} + 4 ext{H}_2 ext{O} \)

Electrode Potentials

Electrode potentials, often depicted by \( E^{\circ} \), are a measure of a chemical species' tendancy to gain or lose electrons under standard conditions. These potentials are crucial in determining the direction and feasibility of redox reactions.

Every half-reaction has an associated reduction potential, showing the likelihood of gaining electrons. For instance, the reduction potential for the permanganate ion to manganese(II) ion is \( +1.51 \, \text{V} \), showing a strong tendency to be reduced.

The oxidation potential is the opposite, often calculated by reversing the sign of the reduction potential. The oxidation of iron(II) to iron(III) carries a potential of \( -0.77 \, \text{V} \).

To find the overall cell potential, you calculate:

• \( E^{\circ}_{\text{reaction}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \)

Using this formula, we get \( 2.28 \, \text{V} \), indicating a strong driving force for the reaction to proceed spontaneously forward under standard conditions.

Stoichiometry in Titrations

Stoichiometry in titrations plays a vital role when determining the amount of substance present in a solution. It involves using balanced chemical equations to relate volumes and concentrations to find quantitative information about a reaction.

In this exercise, the titration of \( \text{Fe}^{2+} \) with \( \text{KMnO}_4 \) involves the delicate balance of equations where stoichiometry confirms the amount of iron in the solution. Given that 1 mole of \( \text{MnO}_4^- \) requires 5 moles of \( \text{Fe}^{2+} \) for complete reaction, the calculation proceeds based on the volume and molarity of the \( \text{KMnO}_4 \) solution.

We determine the moles of \( \text{KMnO}_4 \) using:

• Moles = Molarity \( \times \) Volume

Once determined, these moles are multiplied by the stoichiometric coefficient (5 in this case) to find moles of iron. Mass of iron can then be calculated, and ultimately, the percentage of iron in the ore is computed by:

• \( \text{Percentage of Fe} = \frac{\text{Mass of Fe in sample}}{\text{Total sample mass}} \times 100 \)

Through this method, we discover that the ore is composed of approximately \( 88.74\% \) iron. This concept clarifies the tight relationship between moles, mass, and percentages in chemical solutions.