/*! 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 121 When aluminum foil is placed in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 121

When aluminum foil is placed in hydrochloric acid, nothing happens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.

Short Answer

Expert verified
In summary, the initial lack of reaction between aluminum foil and hydrochloric acid is due to the presence of a protective aluminum oxide layer on the foil's surface. After about 30 seconds, the acid dissolves this layer, allowing the reaction to occur, producing aluminum chloride and hydrogen gas. The vigorous bubbling observed is due to the formation of hydrogen gas, and the aluminum foil eventually disappears as it is consumed in the reaction.

Step by step solution

01

Understanding the chemical reaction

We need to understand the chemical reaction between aluminum (Al) and hydrochloric acid (HCl). The balanced chemical equation for this reaction is: \[ 2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2} \] Here, aluminum reacts with hydrochloric acid to form aluminum chloride (AlCl3) and hydrogen gas (H2). The hydrogen gas formed in the reaction is the cause of the bubbling observed.
02

Explaining the initial lack of reaction

Aluminum is known to form a thin layer of aluminum oxide (Al2O3) on its surface when exposed to air. This oxide layer is very stable and provides a protective barrier that prevents further corrosion of the aluminum. At the beginning of the experiment, when the aluminum foil is placed in the hydrochloric acid, the reaction does not start immediately due to the presence of the aluminum oxide layer on the foil's surface. The acid needs to dissolve this oxide layer before it reacts directly with the aluminum.
03

The start of the reaction

After about 30 seconds, the hydrochloric acid starts to dissolve or penetrate through the aluminum oxide layer on the foil's surface. This allows the acid to come into direct contact with the aluminum metal and initiates the chemical reaction.
04

Observing vigorous bubbling and the disappearance of the foil

Once the reaction starts, it proceeds vigorously. Aluminum reacts with hydrochloric acid, forming aluminum chloride and hydrogen gas, as shown in the chemical equation in Step 1. The production of hydrogen gas causes vigorous bubbling in the solution. As the reaction continues, more and more aluminum metal is consumed by the hydrochloric acid. Eventually, all of the aluminum foil reacts with the acid, causing the foil to disappear completely.
05

Conclusion

In conclusion, when aluminum foil is placed in hydrochloric acid, the initial lack of reaction is due to the presence of a protective aluminum oxide layer on the foil's surface. Once the acid dissolves the oxide layer, the reaction between aluminum and hydrochloric acid starts and proceeds vigorously, producing bubbles (hydrogen gas) and causing the aluminum foil to eventually disappear.

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

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}\)

Consider the cell described below: $$ \mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Al}^{3+}\right]\) to have changed by \(0.60 \mathrm{mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{c} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\). \(\mathrm{Ox}\) ide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C}\) ). \(\Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is -380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{CuCl}_{2}\) c. \(\mathrm{O}_{2}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) e. \(C_{6} H_{12} O_{6}\) f .\(\mathrm {Ag} \) g. \(\mathrm{PbSO}_{4}\) h. \(\mathrm{PbO}_{2}\) i. \(\quad \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) j . \(\mathrm{CO}_{2}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Chlorine gas was first prepared in 1774 by \(\mathrm{C}\). W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$ \begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) & \longrightarrow \\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) &+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned} $$ Balance this equation.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

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.