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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.

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Most popular questions from this chapter

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$ \mathscr{E}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH} $$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text {meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}\) \((\pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$

It took \(150 .\) s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is $$\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \quad \mathscr{E}^{\circ}=1.10 \mathrm{V} $$ For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?
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