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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.

Short Answer

Expert verified
The spontaneous redox reaction that might cause corrosion between a brass pipe and a galvanized steel pipe is \(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\). This reaction occurs because zinc is more likely to be oxidized and copper is more likely to be reduced, based on their standard reduction potentials. The calculated standard emf for the reaction is 1.10 V, which indicates that the reaction is spontaneous and will cause corrosion. To prevent this corrosion, an insulating fitting should be used to connect the two pipes, as suggested in the plumber's handbook.

Step by step solution

01

Identify the half-reactions for copper and zinc

To identify the half-reactions for copper and zinc, we have to look into the standard reduction potential table: Copper (Cu): \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\), standard reduction potential, \(E^\circ_{Cu}\) = +0.34 V Zinc (Zn): \(Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)\), standard reduction potential, \(E^\circ_{Zn}\) = -0.76 V Now that we have the possible half-reactions, let's analyze the spontaneous redox reactions.
02

Determine the overall redox reaction

A redox reaction occurs when one element is oxidized while the other is reduced. From the standard reduction potentials, it can be concluded that zinc is more likely to be oxidized due to its lower standard reduction potential, while copper is more likely to be reduced because of its higher standard reduction potential. When one element is oxidized, it loses electrons and when an element is reduced, it gains electrons. So, the overall redox reaction is: Zinc is oxidized: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\) Copper is reduced: \(Cu^{2+} + 2e^- \rightarrow Cu(s)\) Combining both, we get: \(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\)
03

Calculate the standard emf for the reaction

To calculate the standard emf (\(E^\circ_{cell}\)) for the reaction, we'll use the formula: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\) In our case, the cathode is the Cu electrode, which accepts electrons, and the anode is the Zn electrode, which donates electrons. Therefore, the standard emf for the reaction is: \(E^\circ_{cell} = E^\circ_{Cu} - E^\circ_{Zn} = 0.34 V - (-0.76 V) = 1.10 V\) Since the standard emf of the cell is positive (\(E^\circ_{cell} > 0\)), the redox reaction between copper and zinc is spontaneous and will cause corrosion. #Conclusion# The spontaneous redox reaction between copper in the brass pipe and galvanized steel pipe will cause corrosion. The reaction is \(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\) and has a standard emf of 1.10 V. Therefore, we should use an insulating fitting to connect the two pipes, as recommended in the handbook.

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Key Concepts

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

Electrochemical Cells
Understanding electrochemical cells is crucial when exploring how different metals interact and can lead to corrosion. An electrochemical cell comprises two metal electrodes immersed in electrolyte solutions that conduct electricity through ion flow. These cells are integral in batteries, galvanic corrosion, and other processes that involve electron transfer between materials.

In the context of the exercise about galvanized steel and brass pipes, the direct physical connection facilitates the setup of a makeshift electrochemical cell, where zinc from the galvanized layer and copper from the brass act as anode and cathode, respectively. The moist environment surrounding the pipes serves as an electrolyte, which allows for ionic movement and electron flow from the anode to the cathode, setting the stage for corrosion through oxidation and reduction reactions.
Standard Reduction Potential
The standard reduction potential is a measure that indicates the tendency of a chemical species to gain electrons and hence be reduced. It is crucial for predicting the direction of redox reactions and is measured in volts (V). Each half-reaction has a standard reduction potential associated with it, and when comparing two potential reactions, the one with the higher potential acts as the cathode (reduction) and the other as the anode (oxidation).

For the brass and galvanized steel example, zinc has a lower standard reduction potential (\(E^\text{o}_{\text{Zn}} = -0.76\text{ V}\)) compared to copper's higher potential (\(E^\text{o}_{\text{Cu}} = 0.34\text{ V}\)), indicating that in the presence of an electrolyte, zinc will preferentially oxidize while copper will reduce, leading to the corrosion of zinc.
Corrosion Prevention
Corrosion prevention is essential to maintain the integrity of metal structures and is achieved through various methods. In pipes, where dissimilar metals come into contact, an insulating fitting can prevent the establishment of an electrochemical cell, thus eliminating the electrochemical reaction that would cause corrosion.

Other methods include coating the metals with protective layers to prevent electrode reactions, using sacrificial anodes that preferentially corrode, or applying corrosion inhibitors which can adsorb on metal surfaces and block the reactive sites. Regular maintenance and material selection, based on understanding the galvanic series, also play a significant role in preventing corrosion.
Galvanic Series
The galvanic series is a list of metals and alloys organized according to their standard reduction potentials in a given environment. Metals at the top have a greater tendency to lose electrons and corrode, acting as anodes, while those at the bottom are more likely to gain electrons and be protected, acting as cathodes.

By consulting the galvanic series, plumbers and engineers can predict which metal combinations are likely to produce galvanic corrosion. The series explains why zinc (closer to the top) would corrode when in contact with copper (closer to the bottom). It underpins the advice in the plumber's handbook, advising against direct contact between brass and galvanized steel to prevent corrosion.

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Most popular questions from this chapter

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \longrightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the fol- $$ \text { lowing: } \mathrm{pH}=0.0,\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M},\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M} $$ \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

(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 fol- lowing in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al}\).

(a) Under what circumstances is the Nernst equation applicable? (b) What is the numerical value of the reaction quotient, Q, under standard conditions? (c) What happens to the emf of a cell if the concentrations of the reactants are increased?

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