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Chapter 18: Problem 33

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

Short Answer

Expert verified
a. False. Corrosion is an electrochemical process, not an electrolytic process. b. True. c. False. Steel rusts more easily in humid Midwest states than in the dry Southwest states. d. False. Salting roads accelerates corrosion, not hindering it. e. True.

Step by step solution

01

Statement a: Electrolytic process?

False. Corrosion is not an example of an electrolytic process but an example of an electrochemical process. It involves the transfer of electrons from one species to another without an external voltage source.
02

Statement b: Reduction of iron and oxidation of oxygen

True. Corrosion of steel involves the reduction of iron (Fe) to form iron oxide (rust), coupled with the oxidation of oxygen (O). The process looks like this: \[ 4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3} \]
03

Statement c: Rusting in dry and humid areas

False. Steel rusts more easily in humid Midwest states than in the dry (arid) Southwest states. The presence of water (humidity) promotes corrosion by providing an electrolyte that enables the flow of electrons between the anode and cathode regions on the metal surface.
04

Statement d: Salting roads to hinder corrosion

False. Salting roads in the winter does not hinder the corrosion of steel but instead accelerates it. The dissolved salts in water lower the freezing point, while also creating an electrolyte, thus increasing the rate of corrosion.
05

Statement e: Cathodic protection concept

True. The key to cathodic protection is to connect a metal that is more easily oxidized (sacrificial anode) than iron to the steel surface via a wire. This will cause the sacrificial anode to corrode preferentially, leaving the steel surface protected. Common examples of sacrificial anodes are zinc and magnesium.

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Most popular questions from this chapter

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \Longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)$ b. $\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)$

A solution at \(25^{\circ} \mathrm{C}\) contains 1.0\(M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-4} M\) \(\mathrm{Ag}^{+}\). Which metal will plate out first as the voltage is gradually increased when this solution is electrolyzed? (Hint: Use the Nernst equation to calculate \(\mathscr{E}\) for each half-reaction.)

Consider the standard galvanic cell based on the following half-reactions: $$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$$ $$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$$ The electrodes in this cell are \(A g(s)\) and \(C u(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\operatorname{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) . ]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: Ag' reacts with Cl- to form AgCl(s). ] d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V}$$

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} M\) and the measured cell potential is 1.82 \(\mathrm{V}\) . Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=?$$

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)
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