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Chapter 17: Problem 17

Write a half-reaction for the oxidation of magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)\) to hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) in acidic groundwater.

Short Answer

Expert verified
Answer: The balanced half-reaction is 2(Fe3O4) + H+ + 2e- -> 3(Fe2O3).

Step by step solution

01

Determine the oxidation states of iron in magnetite and hematite

For magnetite (Fe3O4), let x represent the oxidation state of iron. With four oxygen atoms with an oxidation state of -2 each, we have: 3x + 4(-2) = 0 => 3x = 8 => x = 8/3 So, the iron atoms in magnetite have an average oxidation state of +8/3. In hematite (Fe2O3), again let x represent the oxidation state of iron, we have: 2x + 3(-2) = 0 => 2x = 6 => x = 3 The iron atoms in hematite have an oxidation state of +3.
02

Balance the iron atoms

To balance the iron atoms in the half-reaction, we need to have the same number of iron atoms on both sides. In magnetite, there are three Fe atoms per one formula unit, and in hematite, there are two Fe atoms per one formula units. To balance the iron atoms, we need to multiply hematite by 3 and magnetite by 2, like so: 2(Fe3O4) -> 3(Fe2O3)
03

Balance the oxygen atoms and add protons (H+)

Having balanced the iron atoms, we now need to balance the oxygen atoms. There are eight oxygen atoms in 2(Fe3O4) and nine oxygen atoms in 3(Fe2O3). To balance the oxygen atoms, we can add one proton (H+): 2(Fe3O4) + H+ -> 3(Fe2O3)
04

Add electrons (e-) to balance charge

The oxidation states of iron change from +8/3 in magnetite to +3 in hematite. The total change in oxidation state for 2(Fe3O4) to 3(Fe2O3) is: 2 * 3 * (8/3) - 3 * 2 * 3 = 16 - 18 = -2 The change in oxidation state is -2, so we need to add two electrons (e-) to balance the charge: 2(Fe3O4) + H+ + 2e- -> 3(Fe2O3)
05

Write the final balanced half-reaction

The balanced half-reaction for the oxidation of magnetite to hematite in acidic groundwater is: 2(Fe3O4) + H+ + 2e- -> 3(Fe2O3)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Reaction Balancing
Balancing chemical reactions involves ensuring that the number of atoms for each element is the same on both sides of the equation. In a half-reaction, we focus only on the ions and electrons participating in the redox process.
For instance, when magnetite (\(\text{Fe}_3\text{O}_4\)) is oxidized to hematite (\(\text{Fe}_2\text{O}_3\)), it happens in acidic conditions.
Hence, protons (\(\text{H}^+\)) are involved in maintaining balance. To begin, identify the chemical species involved. Make sure you understand how many of each atom participates.
  • Balance the atoms of interest (Fe and O in this example).
  • Add relevant quantities of (\(\text{H}^+\)) or (\(\text{OH}^-\)) if under acidic or basic conditions.
  • Lastly, ensure the charge is balanced by adding electrons to one side of the reaction.
Oxidation States
Oxidation states help us track electron transfers in redox reactions. They denote the hypothetical charge an atom has if all bonds were ionic.
For magnetite (\(\text{Fe}_3\text{O}_4\)), finding the oxidation state involves solving a simple algebraic equation. Here, three iron atoms share an average oxidation state of (\(+8/3\)), factoring in the four oxygen atoms each with -2.
In hematite (\(\text{Fe}_2\text{O}_3\)), the iron atoms hold a (\(+3\)) oxidation state. Understanding these numbers lets us figure out how electrons are being transferred and what species get oxidized or reduced.
  • Calculate oxidation states for each element using known values for others, like oxygen usually being -2.
  • Keep track of changes in oxidation states to determine electron transfer.
  • Pay attention to what element is losing or gaining electrons.
Acidic Conditions
Reactions in acidic conditions involve additional hydrogen ions (\(\text{H}^+\)) besides water and electrons in balancing redox reactions. This environment affects which species can exist and participate in the reaction.
For the conversion of magnetite to hematite, adding (\(\text{H}^+\)) is critical to conserving atoms in the equation.
This leads to the successful balance of oxygen and hydrogen atoms, negating the need for further elements to maintain equilibrium.
  • Use (\(\text{H}^+\)) to balance additional O.
  • Introduce water molecules where appropriate for balancing hydrogen appropriately.
  • Remember, acidic conditions utilize (\(\text{H}^+\)), whereas basic conditions would typically use (\(\text{OH}^-\)).
Electron Transfer
Electron transfer is the cornerstone of oxidation-reduction (redox) reactions. Identifying which species gains or loses electrons is essential. This drives the classification into oxidation and reduction processes.
Here, iron goes from (\(+8/3\)) in magnetite to (\(+3\)) in hematite, indicating a reduction.
Electrons removed in oxidation are used to balance charges across the equation. In this example, 2 electrons are included on the reactant side.
  • Identify the oxidizing agent (losing electrons) and reducing agent (gaining electrons).
  • Balance the overall equation by adding required electrons.
  • The aim is to ensure the net charge is the same before and after the reaction.

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Most popular questions from this chapter

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