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Iron(II) hydroxide is a greenish precipitate that is formed from iron(II) ion by the addition of a base. This precipitate gradually turns to the yellowish- brown iron(III) hydroxide from oxidation by \(\mathrm{O}_{2}\) in the air. Write a balanced equation for this oxidation by \(\mathrm{O}_{2}\).

Short Answer

Expert verified
The balanced equation is: \( 4\text{Fe(OH)}_2 + \text{O}_2 \rightarrow 4\text{Fe(OH)}_3 \).

Step by step solution

01

Identify Reactants and Products

First, identify the reactants and products involved in the oxidation process. The reactants here are iron(II) hydroxide, denoted as \( \text{Fe(OH)}_2 \), and oxygen, \( \text{O}_2 \). The product is iron(III) hydroxide, \( \text{Fe(OH)}_3 \). The process involves iron being oxidized from the +2 to the +3 oxidation state.
02

Write Unbalanced Equation

Next, write the unbalanced chemical equation using the reactants and products identified. It should look something like: \[ \text{Fe(OH)}_2 + \text{O}_2 \rightarrow \text{Fe(OH)}_3 \]
03

Balance Iron Atoms

Ensure that the number of iron atoms is the same on both sides of the equation. Currently, one iron atom is present on each side, so iron is balanced: \[ \text{Fe(OH)}_2 + \text{O}_2 \rightarrow \text{Fe(OH)}_3 \]
04

Balance Oxygen Atoms from Hydroxides

Balance the number of hydroxide groups. On the left, two hydroxides are present, and three are needed on the right. Multiply \( \text{Fe(OH)}_2 \) by 2 and \( \text{Fe(OH)}_3 \) by 2: \[ 2\text{Fe(OH)}_2 + \text{O}_2 \rightarrow 2\text{Fe(OH)}_3 \]
05

Balance Oxygen Atoms from O2

Now the total oxygen atoms on the left are four from hydroxide and two more from \( \text{O}_2 \), equaling six oxygens. On the right, there are six oxygens from \( 2\text{Fe(OH)}_3 \) (each having three hydroxides). The equation is now balanced regarding oxygen.
06

Verify Balancing of All Elements

Finally, verify that all elements are balanced. Iron: 2 on each side; Oxygen: 4 from hydroxide + 2 from \( \text{O}_2 \) on the left and 6 from hydroxide on the right; Hydrogen, 6 from hydroxide on both sides. The balanced equation is: \[ 4\text{Fe(OH)}_2 + \text{O}_2 \rightarrow 4\text{Fe(OH)}_3 \]

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Key Concepts

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

Oxidation
Understanding the concept of oxidation is crucial when talking about chemical reactions, especially those involving metal ions like iron. In simple terms, oxidation involves the loss of electrons by an atom or ion.
For iron, oxidation can be observed when iron(II) ions, which are stable at a +2 oxidation state, undergo a chemical change and become iron(III) ions, which are more stable at a +3 oxidation state. Thus, during this process, iron is oxidized.
This transformation is essential to many natural and industrial processes, including rusting, where iron reacts slowly with oxygen in the air. In our example of iron(II) hydroxide converting to iron(III) hydroxide, exposure to oxygen leads to the change, with oxygen being the oxidizing agent.
Key points about oxidation include:
  • Involves loss of electrons
  • Commonly associated with the gain of oxygen atoms
  • Typically results in a higher oxidation state of the element
Balancing Equations
Balancing chemical equations ensures that the number of atoms for each element involved in a reaction is the same on both sides. This represents the Law of Conservation of Mass, implying that matter is neither created nor destroyed in a chemical reaction.
When balancing, both the number of atoms and the charge should be equal on both sides of the equation. For example, if you're given a chemical process with elements like iron, oxygen, and hydrogen, you need to carefully adjust coefficients to match the total number of atoms for each element involved.
In our specific case, where iron(II) hydroxide is oxidized to iron(III) hydroxide by oxygen, balancing is done stepwise:
  • First, balance the metals (iron atoms in this case)
  • Next, balance the non-metals (oxygen and hydrogen atoms)
  • Consistently check and adjust through trial and error as needed
This ensures a correctly balanced chemical equation, such as: \[ 4\text{Fe(OH)}_2 + \text{O}_2 \rightarrow 4\text{Fe(OH)}_3 \] which is the final product in this oxidation reaction.
Iron Hydroxide
Iron hydroxide is a versatile compound commonly encountered in chemical reactions, especially involving transition metals. It exists primarily in two forms: iron(II) hydroxide and iron(III) hydroxide. These compounds include iron ions combined with hydroxide ions.
Iron(II) hydroxide is typically formed by the addition of a base to iron(II) ions. It appears as a greenish precipitate. Over time, due to oxidation, it changes into iron(III) hydroxide, which is a yellowish-brown substance, characteristic of iron oxidation products like rust.
Observations you might make with iron hydroxide:
  • Changes color as it oxidizes - from green to yellowish-brown
  • Found in various natural settings such as bogs or industrial waste systems
  • Plays a role in processes like corrosion, water purification, and pigment production
Understanding the behavior of iron hydroxide during reactions and its transformations can help in grasping broader concepts of chemistry, such as electron transfer processes and compound stability.

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Most popular questions from this chapter

A voltaic cell is constructed from the following halfcells: a magnesium electrode in magnesium sulfate solution and a nickel electrode in nickel sulfate solution. The half-reactions are $$ \begin{aligned} &\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{e}^{-} \\ &\mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \end{aligned} $$ Sketch the cell, labeling the anode and cathode (and the electrode reactions), and show the direction of electron flow and the movement of cations.

Explain the electrochemistry of rusting.

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The composition of the hull of a submarine is mostly iron. Pieces of zinc, called "zincs," are placed in contact with the hull throughout the inside of the submarine. Why is this done?

A constant current of \(1.25\) amp is passed through an electrolytic cell containing a \(0.050 M\) solution of \(\mathrm{CuSO}_{4}\) and a copper anode and a platinum cathode until \(2.10 \mathrm{~g}\) of copper is deposited. a. How long does the current flow to obtain this deposit? b. What mass of silver would be deposited in a similar cell containing \(0.10 \mathrm{MAg}^{+}\) if the same amount of current were used?

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