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In the construction of the Statue of Liberty, a framework of iron ribs was covered with thin sheets of copper less than \(2.5 \mathrm{mm}\) thick. A layer of asbestos separated the copper skin and iron framework. Over time, the asbestos wore away and the iron ribs corroded. Some of the ribs lost more than half their mass in the 100 years before the statue was restored. At the same time, the copper skin lost only about \(4 \%\) of its thickness. Use electrochemical principles to explain these observations.

Short Answer

Expert verified
The iron framework and copper skin's reaction is an instance of galvanic corrosion, an electrochemical process. Iron, which is more reactive and situated higher in the electrochemical series, tends to oxidize and corrode. Copper, which is less reactive, largely remains unaffected. The corrosion accelerated when asbestos, acting as insulation between the two metals, wore away and directly exposed iron to the environment.

Step by step solution

01

Understanding the Behavior of Copper and Iron in Contact

When two different metals, in this case, copper and iron, come into contact, galvanic corrosion can happen. In such a pairing, iron is more reactive, meaning it tends to corrode, while copper is less reactive, so it remains largely unaffected.
02

Role of Electrochemical Series

In the electrochemical series, iron lies above copper. This means that iron has a greater tendency to lose electrons and form positive ions. When the asbestos wore away, the iron frameworks were exposed to the environment, facilitating the oxidation of iron, while copper behaves as a cathode in this reaction.
03

Effect of Asbestos

Asbestos initially acted as an insulation material, preventing direct contact between iron and copper, therefore preventing galvanic corrosion. But as it wore away, it exposed iron to the air and water, which given its nature, started oxidizing.

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Key Concepts

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

Electrochemical Principles
Electrochemical principles are at the core of understanding why materials like the iron ribs in the Statue of Liberty corrode over time. Put simply, it involves a process in which chemicals react through the exchange of electrons. This interaction is driven by the tendency of some materials to donate electrons, known as oxidation, and others to receive them, known as reduction.

In the case of the Statue of Liberty, the electrochemical reactions occurred as the iron ribs came into direct contact with the copper skin after the layer of asbestos wore away. With water and oxygen from the environment acting as electrolytes, a spontaneous flow of electrons began from the more reactive iron to the less reactive copper. This flow of electrons is what underpins the galvanic corrosion seen in the iron ribs.

To mitigate such problems in the future, a deep understanding of how different materials react with one another in the presence of electrolytes is crucial. Protective coatings, physical barriers, or using similarly reactive materials in construction are common strategies to prevent unwanted electrochemical reactions.
Electrochemical Series
The electrochemical series is akin to a leaderboard that shows the reactivity of various elements when they come into contact with each other. Elements at the top of the series, like iron in the context of the Statue of Liberty, are more likely to lose electrons, experience oxidation, and subsequently undergo corrosion. Conversely, elements like copper sit lower on this hierarchy and are less likely to corrode as they tend to gain electrons, serving as a reduction agent in electrochemical reactions.

It's this ranking in the electrochemical series that explains why, over time, the iron ribs suffered significant corrosion while the copper skin remained relatively intact. As educators, we emphasize the importance of consulting the electrochemical series when selecting materials for construction, especially when those materials will be in contact with one another in a potentially corrosive environment. It's a preventative measure for engineers to predict and avoid galvanic corrosion.
Oxidation and Reduction
Understanding oxidation and reduction is fundamental to grasp why metals corrode. Oxidation is the process by which an atom loses electrons, while reduction is the gain of electrons. These two halves of a whole—where one substance gives up electrons and another accepts them—are often remembered by the mnemonic 'OIL RIG': Oxidation Is Loss, Reduction Is Gain.

In our Statue of Liberty example, iron's oxidation corresponds to its corrosion—iron atoms lose electrons and form rust when exposed to oxygen and moisture. Meanwhile, copper undergoes the complementary process of reduction, acquiring the electrons lost by iron. It's a testament to the strength of these electrochemical principles that after a hundred years, the copper lost only a tiny fraction of its thickness, whereas iron was profoundly affected.

When analyzing structures subjected to such electrochemical challenges, it's not only the knowledge of oxidation and reduction that matters. Understanding the environmental conditions and the protective measures, like the asbestos initially used, is equally pivotal in preserving the integrity of metallic constructions.

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Most popular questions from this chapter

From the observations listed, estimate the value of \(E^{\circ}\) for the half- reaction \(\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}(\mathrm{s})\) (a) The metal M reacts with HNO \(_{3}(\text { aq })\), but not with \(\mathrm{HCl}(\mathrm{aq}) ; \mathrm{M}\) displaces \(\mathrm{Ag}^{+}(\mathrm{aq}),\) but not \(\mathrm{Cu}^{2+}(\mathrm{aq})\) (b) The metal \(M\) reacts with \(\mathrm{HCl}(\mathrm{aq}),\) producing \(\mathrm{H}_{2}(\mathrm{g}),\) but displaces neither \(\mathrm{Zn}^{2+}(\text { aq })\) nor \(\mathrm{Fe}^{2+}(\mathrm{aq})\).

The quantity of electric charge that will deposit \(4.5 \mathrm{g}\) Al at a cathode will also produce the following volume at STP of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}^{+}(\) aq) at a cathode: (a) \(44.8 \mathrm{L} ;\) (b) \(22.4 \mathrm{L} ;\) (c) \(11.2 \mathrm{L} ;\) (d) \(5.6 \mathrm{L}\).

A solution containing both \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) ions is subjected to electrolysis. (a) Which metal should plate out first? (b) Plating out is finished after a current of \(0.75 \mathrm{A}\) is passed through the solution for 2.50 hours. If the total mass of metal is \(3.50 \mathrm{g},\) what is the mass percent of silver in the product?

The gas evolved at the anode when \(\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is electrolyzed between Pt electrodes is most likely to be (a) \(\mathrm{O}_{2} ;\) (b) \(\mathrm{H}_{2} ;\) (c) \(\mathrm{SO}_{2} ;\) (d) \(\mathrm{SO}_{3} ;\) (e) a mixture of sulfur oxides.

Show that for a combination of half-cell reactions that produce a standard reduction potential for a half-cell that is not directly observable, the standard reduction potential is $$E^{\circ}=\frac{\sum n_{i} E_{i}^{\circ}}{\sum n_{i}}$$ where \(n_{i}\) is the number of electrons in each half-reaction of potential \(E_{i}^{\circ} .\) Use the following half-reactions: $$ \begin{array}{c} \mathrm{H}_{5} \mathrm{IO}_{6}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+ \\ 3 \mathrm{H}_{2} \mathrm{O}(1) \quad E^{\circ}=1.60 \mathrm{V} \\ \mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{I}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.19 \mathrm{V} \\ 2 \mathrm{HIO}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.45 \mathrm{V} \\ \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \quad E^{\circ}=0.535 \mathrm{V} \end{array} $$ Calculate the standard reduction potential for $$ \mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow $$ $$ \frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO} $$

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