/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A plumber's handbook states that... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 83

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 reactions that might cause corrosion when a brass pipe (mixture of copper and zinc) is connected to a galvanized steel pipe are: 1. Cu虏鈦簗Cu (cathode) and Zn虏鈦簗Zn (anode), with E掳(cell) = 1.10 V 2. Cu虏鈦簗Cu (cathode) and Fe虏鈦簗Fe (anode), with E掳(cell) = 0.78 V These reactions are spontaneous based on the positive standard emf values, which justifies the recommendation in the plumber's handbook to use an insulating fitting to connect brass to galvanized steel pipes.

Step by step solution

01

Write possible half-reactions for each metal

First, we should write down the possible reduction half-reactions for each metal involved: copper (Cu), zinc (Zn), and iron (Fe, as galvanized steel is mostly composed of iron and coated with zinc). Reduction half-reactions: Cu虏鈦� + 2e鈦� 鈫� Cu Zn虏鈦� + 2e鈦� 鈫� Zn Fe虏鈦� + 2e鈦� 鈫� Fe
02

Find standard reduction potentials

Next, we need to find the standard reduction potentials for each half-reaction. Standard reduction potentials can be found in standard reference tables. Standard reduction potentials (E掳): Cu虏鈦簗Cu: +0.34 V Zn虏鈦簗Zn: -0.76 V Fe虏鈦簗Fe: -0.44 V
03

Calculate emf for each possible redox reaction

Now, for each possible redox reaction, we need to calculate the standard emf. Subtract each anode's standard reduction potential from the reduction potential of each cathode. 1. Cu虏鈦簗Cu (cathode) and Zn虏鈦簗Zn (anode) E掳(cell) = E掳(cathode) - E掳(anode) = 0.34 - (-0.76) = 1.10 V 2. Cu虏鈦簗Cu (cathode) and Fe虏鈦簗Fe (anode) E掳(cell) = E掳(cathode) - E掳(anode) = 0.34 - (-0.44) = 0.78 V 3. Zn虏鈦簗Zn (cathode) and Fe虏鈦簗Fe (anode) E掳(cell) = E掳(cathode) - E掳(anode) = -0.76 - (-0.44) = -0.32 V
04

Identify spontaneous redox reactions

A positive emf indicates that the reaction is spontaneous. Based on our calculations, the spontaneous redox reactions are: 1. Cu虏鈦簗Cu (cathode) and Zn虏鈦簗Zn (anode), with E掳(cell) = 1.10 V 2. Cu虏鈦簗Cu (cathode) and Fe虏鈦簗Fe (anode), with E掳(cell) = 0.78 V The reaction between Zn虏鈦簗Zn (cathode) and Fe虏鈦簗Fe (anode) is not spontaneous, with E掳(cell) = -0.32 V.
05

Conclusion

Spontaneous redox reactions that might cause corrosion when a brass pipe (mixture of copper and zinc) is connected to a galvanized steel pipe are: 1. Cu虏鈦簗Cu (cathode) and Zn虏鈦簗Zn (anode), with E掳(cell) = 1.10 V 2. Cu虏鈦簗Cu (cathode) and Fe虏鈦簗Fe (anode), with E掳(cell) = 0.78 V These reactions are spontaneous based on the positive standard emf values. This provides the justification for not connecting brass pipes directly to galvanized steel pipes.

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 91影视!

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

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s)--\rightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode compartment is \(1.00 \mathrm{M}\) and the cell generates an emf of \(+0.22 \mathrm{~V}\), what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode compartment? (b) If the anode compartment contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s)\), what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(\mathrm{s})\) and \(\mathrm{Fe}^{3+}(a q) / \mathrm{Fe}^{2+}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode, and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(C\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\).

The following quotation is taken from an article dealing with corrosion of electronic materials: "Sulfur dioxide, its acidic oxidation products, and moisture are well established as the principal causes of outdoor corrosion of many metals." Using \(\mathrm{Ni}\) as an example, explain why the factors cited affect the rate of corrosion. Write chemical equations to illustrate your points. (Note: \(\mathrm{NiO}(s)\) is soluble in acidic solution.)

For each of the following reactions, write a baFor each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)lanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation
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.