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Chapter 18: Problem 117

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 the reaction between aluminum foil and hydrochloric acid, the initial delay in visible reaction is due to the protective oxide layer on the aluminum surface and limited contact between reactants. Once the acid penetrates the oxide layer, the reaction produces aluminum chloride and hydrogen gas, resulting in vigorous bubbling. As aluminum is converted to aluminum chloride and dissolves in the solution, the foil eventually disappears.

Step by step solution

01

1. Understanding the reaction between aluminum and hydrochloric acid

: Aluminum reacts with hydrochloric acid (HCl) to form aluminum chloride (AlCl3) and hydrogen gas (H2). The balanced chemical equation for this reaction can be written as: \[2Al(s) + 6HCl(aq) 鈫� 2AlCl3(aq) + 3H2(g)\]
02

2. Analyzing the delay in visible reaction

: Initially, when the aluminum foil is placed in hydrochloric acid, it appears that nothing is happening. This delay in the visible reaction may be due to two factors: 1. The presence of an oxide layer on the surface of the aluminum foil; this layer forms as aluminum reacts with oxygen in the air, creating a thin layer of aluminum oxide (Al2O3) that acts as a protective barrier. 2. The initial slow reaction rate due to the relatively low concentration of aluminum and hydrochloric acid, and the fact that their contact is initially limited to the surface area of the aluminum foil.
03

3. Explaining the vigorous bubbling and disappearance of the aluminum foil

: Once the hydrochloric acid penetrates the oxide layer on the aluminum foil, it reacts with the aluminum metal, forming aluminum chloride and hydrogen gas. This reaction produces bubbles of hydrogen gas, which are visible as vigorous bubbling. As the reaction progresses, more and more aluminum is converted into aluminum chloride, which dissolves in the acidic solution. Eventually, the aluminum foil will be completely consumed in the reaction and will disappear. In conclusion, the observations made when placing aluminum foil in hydrochloric acid can be explained by the reaction between the two substances and factors that initially delay the observable effects, such as the presence of an oxide layer and the initial limited contact between the reactants. The vigorous bubbling and disappearance of the aluminum foil are the result of the production of hydrogen gas and the dissolution of aluminum chloride in the acidic solution, respectively.

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Most popular questions from this chapter

The equation \(\Delta G^{\circ}=-\mathrm{nF} \mathscr{E}^{\circ}\) also can be applied to half-reactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)\)

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ The maximum work that 1 mole of alcohol can do is \(1.32 \times 10^{3} \mathrm{kJ}\) . What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C}\) ?

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)
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