In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June \(22,1987)\) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

Short Answer

Expert verified
Attaching zinc sacrificial anodes to the USS Monitor's hull protects it from further corrosion by providing a more reactive surface for corrosive agents to attack, sacrificing the zinc instead of the iron hull. This prevents the iron atoms from losing electrons and forming rust. Periodic inspection and replacement of zinc anodes are necessary to maintain continued protection against corrosion.

Step by step solution

01

Understand Sacrificial Anodes

Sacrificial anodes are pieces of a more reactive metal that are attached to the surface of a less reactive metal structure to prevent corrosion. These anodes, usually made from zinc, magnesium, or aluminum, are more likely to corrode than the metal structure they are protecting, hence the term "sacrificial." This process works by providing a more reactive surface for the corrosive agents (like oxygen and water) to attack, sacrificing itself to save the underlying structure.
02

Electrochemical Corrosion Process

In order to understand how sacrificial anodes protect the USS Monitor hull from corrosion, we need to understand the electrochemical process that causes corrosion to occur. Metal structures corrode when they are exposed to a corrosive environment, such as seawater in this case. When the metal comes in contact with water and oxygen, an electrochemical reaction takes place that causes the metal atoms to lose electrons and form positively charged ions. These ions then react with other ions in the water to form rust (iron oxide) and other types of corrosion products.
03

The Role of Zinc Anodes

The USS Monitor's hull consists mostly of iron, which is prone to corrosion when exposed to seawater. By attaching zinc sacrificial anodes to the hull, the corrosive agents (oxygen and water) will preferentially attack the zinc anodes. This is because zinc is more reactive than iron and thus more likely to lose electrons to form zinc ions in the electrochemical reaction. As a result, the iron atoms in the hull will not lose their electrons, preventing the formation of rust.
04

Anode Replacement

Over time, the zinc anodes will corrode and lose their protective ability as they get consumed in the electrochemical process. To maintain the protection of the USS Monitor hull, the zinc sacrificial anodes must be periodically inspected and replaced when necessary. By doing so, the zinc anodes will continue to provide protection against corrosion for the iron hull. In conclusion, attaching zinc sacrificial anodes to the hull of the USS Monitor will protect it from further corrosion by offering a more reactive surface for the corrosive agents to attack, sparing the iron hull in the process. This protection method requires periodic inspection and replacement of the anodes to ensure continued protection against corrosion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{K}\) is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83} $$ a. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{g}\) blood plasma, determine the mass percent of alcohol in the blood.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free