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If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.

Short Answer

Expert verified
To inhibit corrosion on a steel ship resting in water, a negative charge should be applied. This makes the steel ship act as a cathode, suppressing the oxidation (anodic) reaction that leads to corrosion. By applying a negative charge, we effectively utilize the principle of cathodic protection to prevent the metal surface from corroding.

Step by step solution

01

Identify the electrode reactions involved in corrosion

The primary reactions in the corrosion process are anodic and cathodic half-cell reactions. At the anode, metal atoms are oxidized to form metal ions that dissolve in the environment. At the cathode, reduction occurs, which can involve different reactants such as oxygen. The overall corrosion process involves the transfer of electrons from the anode to the cathode through the metal.
02

Recognize the principle of cathodic protection

Cathodic protection is a technique used to prevent corrosion of a metal surface (like a steel ship in water) by making the metal surface act as a cathode and avoiding the anodic reaction (which leads to corrosion). This can be achieved by applying an external current or by connecting the protected metal to a more reactive (easily-corroded) metal called the sacrificial anode.
03

Determine the type of charge that should be applied

The goal is to ensure that the steel ship acts as a cathode, which means that we want to suppress the oxidation (anodic) reaction that leads to corrosion. Hence, we should apply a negative charge to the steel ship, since this charge will force the metal to act as a cathode and inhibit any further corrosion.
04

Conclusion

To inhibit corrosion on a steel ship resting in water, a negative charge should be applied. This will ensure that the steel ship acts as a cathode, suppressing the oxidation reaction that leads to corrosion.

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Most popular questions from this chapter

A plumber's handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{Cl}_{2}, \mathrm{O}_{2} .\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al}\).

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q)-\cdots\) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l)-\cdots \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (b) \(\operatorname{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots+\mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)-\cdots\) \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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