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Chapter 20: Problem 83

(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 \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)\) .

Short Answer

Expert verified
For the corrosion of iron, the anode and cathode reactions are: Anode: \(Fe(s)\rightarrow Fe^{2+}(aq)\) + 2e鈦� Cathode: \(Fe^{2+}(aq)\) + 2e鈦� 鈫� Fe(s) For the air oxidation of \(Fe^{2+}(aq)\) to \(Fe_2O_3 \cdot 3H_2O(s)\), the balanced half-reactions are: Reduction: \(O_2(g) + 4e鈦� + 4H^+(aq) \rightarrow 2H_2O(l)\) Oxidation: \(4Fe^{2+}(aq) \rightarrow 4Fe^{3+}(aq) + 4e^鈦籠) Combining the half-reactions gives the overall reaction: \(4Fe^{2+}(aq) + O_2(g) + 4H^+(aq) + 6H_2O(l) \rightarrow 2H_2O(l) + 2Fe_2O_3(s) + 12H^+(aq)\)

Step by step solution

01

Determine the reactions for corrosion of iron

At the anode, iron metal loses electrons and oxidizes into aqueous ions. Conversely, at the cathode, the aqueous iron(II) ions gain electrons and form iron metal. Anode: \(Fe(s)\rightarrow Fe^{2+}(aq)\) + 2e鈦� Cathode: \(Fe^{2+}(aq)\) + 2e鈦� 鈫� Fe(s)
02

Balance the half-reactions involved in the oxidation of Fe虏鈦�(aq) to Fe鈧侽鈧兟�3H鈧侽(s)

First, we need to represent both the reduction and oxidation half-reactions. Reduction half-reaction: \(O_2(g) + 4e鈦� 鈫� 2O^{2-}(aq)\) Oxidation half-reaction: \(Fe^{2+}(aq) 鈫� Fe^{3+}(aq) + e鈦籠) We have to balance these half-reactions such that both reduce and oxidize the same amount of electrons. Therefore, multiply the oxidation half-reaction by 4 to balance the electron transfer. Balanced oxidation half-reaction (multiplied by 4): \(4Fe^{2+}(aq) 鈫� 4Fe^{3+}(aq) + 4e鈦籠) Now, we have the balanced half-reactions for air oxidation of Fe虏鈦�(aq) to Fe鈧侽鈧兟�3H鈧侽(s). Reduction half-reaction: \(O_2(g) + 4e鈦� 鈫� 2O^{2-}(aq)\) Balanced oxidation half-reaction: \(4Fe^{2+}(aq) 鈫� 4Fe^{3+}(aq) + 4e鈦籠)
03

Combine the balanced half-reactions to obtain the overall reaction

When we combine these balanced half-reactions, we must account for three moles of water in the overall reaction. To make this possible, we must include the formation of water in the reduction half-reaction, which requires the contribution of six hydrogen ions for each water molecule. \((new) Reduction half-reaction: O_2(g)+ 4e^鈦� +4H^+ (aq)鈫� 2H_2 O(l)\) Now, combining the updated (new) reduction half-reaction with the balanced oxidation half-reaction: \(4Fe^{2+}(aq) + O_2(g) + 4H^+ (aq) 鈫� 4Fe^{3+}(aq) + 2H_2 O(l)\) To produce solid Fe鈧侽鈧兟�3H鈧侽, we must turn four aqueous Fe鲁鈦� ions into two solid iron oxide (Fe鈧侽鈧�) ions, which can be done by adding 6 moles of water: \(4Fe^{3+}(aq) + 6H_2O(l) 鈫� 2Fe_2 O_3(s) + 12H^+(aq)\) Finally, adding the two reactions above, we get: \( 4Fe^{2+}(aq) + O_2(g) + 4H^+ (aq)+4Fe^{3+}(aq) + 6H_2O(l) 鈫� 4Fe^{3+}(aq) + 2H_2O(l) + 2Fe_2 O_3(s) + 12H^+(aq)\) Some terms from both sides are common and, when canceling them out, we achieve the final reaction: \(4Fe^{2+}(aq) + O_2(g) + 4H^+ (aq) + 6H_2O(l) \rightarrow 2H_2O(l) + 2Fe_2 O_3(s) + 12H^+(aq)\)

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Key Concepts

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

Anode and Cathode Reactions
In the process of corrosion, the metal undergoes a series of electrochemical reactions at different sites on its surface. At the anode, the metal is sacrificed as its atoms lose electrons and become ions, which is known as oxidation. For iron, the anode reaction is represented as:
\(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^{-}\).
On the flip side, we have the cathode, where reduction takes place - the site where electrons are gained. In the corrosion of iron, the cathode reaction could be another metal gaining these electrons or the reduction of oxygen in a moist environment. However, for this scenario, it seems to be simplified as iron completing an electrochemical circuit:
\(Fe^{2+}(aq) + 2e^{-} \rightarrow Fe(s)\).
The corrosion process of iron requires both the anode and cathode reactions to occur simultaneously, as the movement of electrons from anode to cathode is essential for the redox reaction to continue.
Oxidation and Reduction Half-Reactions
Redox reactions consist of two half-reactions: the oxidation part where a species loses electrons, and the reduction part where another species gains those electrons. Balancing these half-reactions is critical to understanding the overall chemical process. In the context of air oxidation of iron components, we have:
Oxidation half-reaction: \(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^{-}\).
To balance the charge and the number of electrons, we multiply this reaction by 4, reflecting the stoichiometry involved in the overall redox process:
\(4Fe^{2+}(aq) \rightarrow 4Fe^{3+}(aq) + 4e^{-}\).
The corresponding reduction half-reaction involves oxygen gas:\(O_2(g) + 4e^{-} \rightarrow 2O^{2-}(aq)\).
This illustrates the principle that for every electron lost in the oxidation half-reaction, one must be gained in the reduction half-reaction, ensuring charge and mass balance.
Balancing Chemical Equations
A balanced chemical equation reflects the law of conservation of mass and charge, meaning the number of atoms for each element, and the charge must be the same on both sides of the equation. When balancing redox reactions, it's often necessary to balance the electrons lost in oxidation with those gained in reduction. For example, the air oxidation of iron requires balancing both the iron and oxygen elements, ensuring that electrons lost and gained cancel each other out. The process involves identifying reactants and products and determining the coefficients that make the number of atoms of each element equal on both sides of the reaction. The final reaction for air oxidation of \(Fe^{2+}(aq)\) to \(Fe_2O_3 \cdot 3H_2O(s)\) is balanced as:
\(4Fe^{2+}(aq) + O_2(g) + 4H^+ (aq) + 6H_2O(l) \rightarrow 2H_2O(l) + 2Fe_2 O_3(s) + 12H^+(aq)\),
highlighting the importance of balancing water molecules and hydrogen ions, as well.
Air Oxidation of Iron
When iron is exposed to air, especially moist air, it undergoes a series of oxidation reactions that lead to the formation of iron oxides, commonly known as rust. The process begins with the oxidation of iron to \(Fe^{2+}\) ions, which are further oxidized to \(Fe^{3+}\) in the presence of oxygen and water.
For the case of \(Fe^{2+}(aq)\) being oxidized to \(Fe_2O_3 \cdot 3H_2O(s)\), the overall reaction incorporates both the oxidation of the iron ion and the reduction of oxygen to form water, which then reacts further to produce the hydrated iron oxide compound. This demonstrates how iron, water, and oxygen interact in a complex chemical dance to produce the familiar red-brown compound that we see on corroded iron surfaces. By understanding these processes, we can better understand the chemistry of corrosion and develop ways to protect iron and steel structures from the ravages of rust.

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Most popular questions from this chapter

The electrodes in a silver oxide battery are silver oxide \(\left(\mathrm{Ag}_{2} \mathrm{O}\right)\) and zinc. (a) Which electrode acts as the anode? (b) Which battery do you think has an energy density most similar to the silver oxide battery: a Li-ion battery, a nickel-cadmium battery, or a lead-acid battery? [ Section 20.7]

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. $$ \begin{array}{l}{\text { (a) } \mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s) \text { (acidic solution) }} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text { (acidic solution) }} \\ {\text { (c) } \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)(\text { acidic solution })} \\ {\text { (d) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (acidic solution) }} \\ {\text { (e) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (basic solution) }} \end{array} \\\ {\text { (f) } \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s) \text { (basic solution) }} \\ {\text { (g) } \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) \text { (basic solution) }} $$

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell.(b) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the ClO \(_{3}^{-}\) and \(\mathrm{Cl}^{-} .\) Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l); E^{\circ}=1.45 \mathrm{V}\).

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-28} .\) By using this value together with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$
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