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Chapter 20: Problem 87

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH})\), iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4} .\) (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?(\mathbf{b})\) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is -2 .

Short Answer

Expert verified
The oxidation number of Fe in iron oxyhydroxide (FeO(OH)) is +1. In magnetite (Fe3O4), assuming a unique oxidation number for Fe, it is approximately 2.67. However, there are actually two different oxidation numbers for Fe in magnetite: two Fe atoms have an oxidation number of +3 and one Fe atom has an oxidation number of +2.

Step by step solution

01

Determine the oxidation number of Fe in iron oxyhydroxide

Let's denote the oxidation number of Fe as x. In this compound, FeO(OH), the oxidation numbers of oxygen (-2) and hydrogen (+1) are already provided. Since the compound is neutral, the sum of the oxidation numbers should be equal to zero: x + (-2) + (+1) = 0 Now, we just need to solve the equation for x.
02

Solve for x in iron oxyhydroxide

By solving the equation, we find the oxidation number for Fe in FeO(OH): x + (-2) + (+1) = 0 x - 1 = 0 x = +1 Therefore, the oxidation number of Fe in iron oxyhydroxide is +1.
03

Determine the unique oxidation number of Fe in magnetite, assuming oxygen's oxidation number is -2

In magnetite, Fe3O4, there are three Fe atoms and four oxygen atoms. Let's denote the oxidation number of Fe as x. Since we are given the oxidation number of oxygen as -2, we can write an equation with the overall charge of the compound being zero: 3x + 4(-2) = 0 Now, let's solve the equation to find x.
04

Solve for x in magnetite

By solving the equation, we find the unique oxidation number for Fe in magnetite: 3x + 4(-2) = 0 3x - 8 = 0 3x = 8 x ≈ 2.67 Therefore, when we assume a unique oxidation number for Fe in magnetite, it is approximately 2.67.
05

Determine the two actual oxidation numbers of Fe and their relative stoichiometry in magnetite

Since the unique oxidation number of Fe in magnetite (2.67) cannot be correct, as oxidation numbers must be integers, there must be two different oxidation numbers for Fe in Fe3O4. Let's denote these oxidation numbers as x (for m Fe atoms) and y (for n Fe atoms). The overall charge of the compound is still zero: mx + ny + 4(-2) = 0 We also know the total number of Fe atoms in the compound is 3: m + n = 3 Now we need to suggest values for m, n, x, and y that satisfy both equations, keeping in mind that Fe can have oxidation numbers of +2 or +3.
06

Propose the correct oxidation numbers of Fe and the stoichiometry

An acceptable proposal can be that there are two Fe atoms with an oxidation number of +3 and one Fe atom with an oxidation number of +2. Therefore: m = 2, n = 1, x = +3, y = +2 These values satisfy both equations: 2(3) + 1(2) + 4(-2) = 6 + 2 - 8 = 0 2 + 1 = 3 So, in magnetite Fe3O4, there are two Fe atoms with an oxidation number of +3 and one Fe atom with an oxidation number of +2.

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Key Concepts

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

Iron Corrosion
Understanding iron corrosion is essential as it is a common issue impacting structures and machinery. Iron corrosion occurs when iron reacts with oxygen and moisture, forming various iron oxides and hydroxides.
These products are often seen as rust.
Here are a few of the important resultant compounds:
  • Rust (\( \mathrm{Fe}_{2}\mathrm{O}_{3} \)): A reddish-brown compound that forms when iron reacts with oxygen and water over time.
  • Iron Oxyhydroxide (\( \mathrm{Fe}(\mathrm{O})(\mathrm{OH}) \)): This yellowish-brown compound represents another stage of corrosion, involving water and oxygen in its formation.
  • Magnetite (\( \mathrm{Fe}_{3}\mathrm{O}_{4} \)): A black compound that forms under particular conditions, when less oxygen is available.
These compounds form due to reactions in the environment around the iron, influenced by factors like humidity and exposure time.
Corrosion can significantly impair the strength and function of iron products, making understanding it vital.
Chemical Compounds
Chemical compounds are substances made up of different elements bonded together in specific proportions.
In the case of iron corrodes, the compounds formed play key roles in daily corrosion issues.Iron corrodes primarily into two types of oxides:
  • Iron(III) oxide (\( \mathrm{Fe}_{2}\mathrm{O}_{3} \)): Commonly termed rust, this compound forms when iron oxidizes in the presence of water and oxygen.
  • Magnetite (\( \mathrm{Fe}_{3}\mathrm{O}_{4} \)): This mixed oxidation state compound forms under specific conditions, like limited oxygen.
Each of these compounds exhibits unique physical and chemical properties, owed to their distinct iron and oxygen ratios.For example, \( \mathrm{Fe}(\mathrm{O})(\mathrm{OH}) \), known as iron oxyhydroxide, contains additional water molecules influencing its structure.
Understanding these compounds helps in developing methods to prevent and manage corrosion effectively.
Redox Chemistry
Redox chemistry is fundamental to the formation of iron corrosion products."Redox" stands for "reduction-oxidation," referring to the transfer of electrons during chemical reactions.
In these reactions, one species loses electrons (oxidation), while another gains electrons (reduction).
For iron corroding to \( \mathrm{Fe}_{2}\mathrm{O}_{3} \) and \( \mathrm{Fe}_{3}\mathrm{O}_{4} \):
  • Iron atoms lose electrons (get oxidized) to form iron ions (\( \mathrm{Fe}^{2+} \) and \( \mathrm{Fe}^{3+} \)) in the presence of oxygen.
  • In turn, oxygen molecules gain these electrons, becoming reduced and forming oxide ions (\( \mathrm{O}^{2-} \)).
The oxidation number of an element indicates the degree of oxidation or reduction that it undergoes.
For instance, in magnetite (\( \mathrm{Fe}_{3}\mathrm{O}_{4} \)), iron can exist in different oxidation states, demonstrating the complex nature of these redox processes in corrosion.
Identifying these states is crucial in solving chemical problems related to corrosion and understanding the dynamics of chemical reactions involving metals.

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Most popular questions from this chapter

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 M,\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

Disulfides are compounds that have \(S-S\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH}\), where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

Gold exists in two common positive oxidation states, +1 and +3 . The standard reduction potentials for these oxidation states are $$ \begin{array}{l} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q) &+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ \longrightarrow & 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

(a) What is the definition of the volt? (b) Do all voltaic cells produce a positive cell potential?
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