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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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Most popular questions from this chapter

A voltaic cell is based on \(\mathrm{Cu}^{2+}(a q) / \mathrm{Cu}(s)\) and \(\mathrm{Br}_{2}(l) /\) \(\mathrm{Br}^{-}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\). (Thestandard entropy of \(\mathrm{Cu}^{2+}(a q)\) is $\left.S^{\circ}=-99.6 \mathrm{~J} / \mathrm{K}\right)$

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to $0.15 \mathrm{M}\(, at what concentration of \)\mathrm{I}^{-}$ would the cell have zero potential?

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)$ (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)
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