Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 56

In an atmosphere polluted with industrial smog, \(\mathrm{Cu}\) corrodes to a basic sulfate, \(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{SO}_{4} .\) Propose a series of chemical reactions to describe this corrosion.

Short Answer

Expert verified
The series of reactions depicting the corrosion of copper in a polluted atmosphere can be summed up in three steps - reaction with oxygen to form copper oxide, reaction with water to form copper hydroxide, and reaction with sulfur dioxide to form copper sulfate.

Step by step solution

01

Reaction with Oxygen

The first step in the process of corrosion is usually the reaction of the metal with atmospheric oxygen to form an oxide. In the case of copper, this reaction forms copper (II) oxide (CuO): \[ 2 Cu(s) + O_{2}(g) \rightarrow 2 CuO(s) \] In sulfurous smog, this process can be accelerated by the presence of sulfur dioxide (SO2).
02

Reaction with Water

The second step in the process involves the reaction of the newly formed copper oxide with moisture in the air to form copper hydroxide. Here is the equation for this reaction: \[ CuO(s) + H_{2}O(l) \rightarrow Cu(OH)_2(s) \]
03

Reaction with Sulfur Dioxide

In the case of the smog-polluted atmosphere mentioned, sulfur dioxide (SO2) reacts with the copper hydroxide formed in step 2, which results in the formation of copper sulfate: \[ Cu(OH)_2(s) + SO_{2}(g) \rightarrow Cu_{2}(OH)_{2}SO_{4}(s) \]

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.

Start your free trial

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

Of the following, the two solids that will liberate \(\mathrm{Cl}_{2}(\mathrm{g})\) when heated with \(\mathrm{HCl}(\mathrm{aq})\) are \((\mathrm{a}) \mathrm{NaCl}(\mathrm{s})\) (b) \(\mathrm{ZnCl}_{2}(\mathrm{s}) ;(\mathrm{c}) \mathrm{MnO}_{2}(\mathrm{s}) ;(\mathrm{d}) \mathrm{CuO}(\mathrm{s}) ;(\mathrm{e}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s})\) (f) \(\mathrm{NaOH}(\mathrm{s})\)

Equation \((23.18),\) which represents the chromatedichromate equilibrium, is actually the sum of two equilibrium expressions. The first is an acid-base reaction, \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\). The second reaction involves elimination of a water molecule between two \(\mathrm{HCrO}_{4}^{-}\) ions (a dehydration reaction), \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} .\) If the ionization constant, \(K_{\mathrm{a}},\) for \(\mathrm{HCrO}_{4}^{-}\) is \(3.2 \times 10^{-7},\) what is the value of \(K\) for the dehydration reaction?

Why do the atomic radii vary so much more for two main-group elements that differ by one unit in atomic number than they do for two transition elements that differ by one unit?

What products are obtained when \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}^{3+}(\mathrm{aq})\) are each treated with a limited amount of NaOH(aq)? With an excess of \(\mathrm{NaOH}(\) aq)? Why are the results different in these two cases?

Briefly describe each of the following ideas, phenomena, or methods: (a) lanthanide contraction; (b) zone refining; (c) basic oxygen process; (d) slag formation.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

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.

Sign-up for free