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Chapter 20: Problem 79

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
(a) Anode reaction: \( \mathrm{Fe(s) \rightarrow Fe^{2+}(aq) + 2e^{-}} \) Cathode reaction: \( \mathrm{2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)}\) Combined reaction: \( \mathrm{Fe(s) + 2H^{+}(aq) \rightarrow Fe^{2+}(aq) + H_{2}(g)} \) (b) Oxidation half-reaction: \( \mathrm{4Fe^{2+}(aq) \rightarrow 4Fe^{3+}(aq) + 4e^{-}} \) Reduction half-reaction: \( \mathrm{O_{2}(g) + 4e^{-} + 6H_{2}O(l) \rightarrow 12OH^{-}(aq)} \) Combined reaction: \( \mathrm{12Fe^{2+}(aq) + 6O_2(g) + 6H_2O(l) \rightarrow 4Fe_{2}O_{3}\cdot 3H_{2}O(s)}\)

Step by step solution

01

Anode reaction (Iron metal to aqueous iron(II))

For corrosion of iron to occur, iron metal (Fe) reacts with water and dissolves to produce aqueous iron(II) ions (\(\mathrm{Fe^{2+}(aq)}\)) and electrons \(\mathrm{e^{-}}\). This is an oxidation half-reaction, as the iron is losing electrons. The balanced half-reaction is: \[ \mathrm{Fe(s) \rightarrow Fe^{2+}(aq) + 2e^{-}} \]
02

Cathode reaction (Proton reduction)

In the cathode reaction, water and electrons are involved. The hydrogen ions in water are reduced, meaning they gain electrons, forming hydrogen gas (H2). The balanced half-reaction is: \[ \mathrm{2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)}\]
03

Combined anode and cathode reactions

By combining the anode and cathode reactions, we can see the overall corrosion of iron to aqueous iron(II) and hydrogen gas: \[ \mathrm{Fe(s) + 2H^{+}(aq) \rightarrow Fe^{2+}(aq) + H_{2}(g)} \]
04

Air oxidation of Fe(II) to Fe2O3.3H2O (Iron to Iron(III) oxide and water)

During the oxidation of Iron(II), it is oxidized to Iron(III) in the presence of oxygen and water, forming Iron(III) oxide with 3 water molecules attached. The balanced half-reactions are: Oxidation half-reaction (Fe(II) to Fe(III)): \[ \mathrm{4Fe^{2+}(aq) \rightarrow 4Fe^{3+}(aq) + 4e^{-}} \] Reduction half-reaction (oxygen reduction in the presence of water): \[ \mathrm{O_{2}(g) + 4e^{-} + 6H_{2}O(l) \rightarrow 12OH^{-}(aq)} \]
05

Combined oxidation and reduction of Fe(II) to Fe2O3.3H2O

By combining the oxidation and reduction half-reactions, we can see the overall oxidation of Iron(II) to form Iron(III) oxide with 3 water molecules attached: \[ \mathrm{4Fe^{2+}(aq) + O_{2}(g) + 6H_{2}O(l) \rightarrow 4Fe(OH)_{3}(s)} \] \[ \mathrm{4Fe(OH)_3\;\;(s)\rightarrow 4FeOOH\;\;(s) + 2H_2O\;\;(l)} \] \[ \mathrm{2FeOOH\;\;(s) \rightarrow Fe_{2}O_{3}\;\;(s) + H_{2}O\;\;(l)}\] The balanced equation for the air oxidation of Fe2+ to \(\mathrm{Fe_{2} O_{3} \cdot 3 H_{2} O}\): \[ \mathrm{12Fe^{2+}(aq) + 6O_2(g) + 6H_2O(l) \rightarrow 4Fe_{2}O_{3}\cdot 3H_{2}O(s)}\]

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Most popular questions from this chapter

This oxidation-reduction reaction in acidic solution is spontaneous: \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow\) \(5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(\mathrm{s})\) and \(\mathrm{Fe}^{3+}(a q) / \mathrm{Fe}^{2+}(a q)\) 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 \(C\) 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 \(25^{\circ} \mathrm{C}\).

A voltaic cell utilizes the following reaction and operates at \(298 \mathrm{~K}\) : $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s)-\rightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 M\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ce}^{4+}\right]=0.10 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=1.75 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=2.5 \mathrm{M} ?\)

An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.

(a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?
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