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Chapter 17: Problem 4

A copper penny can be dissolved in nitric acid but not in hydrochloric acid. Using reduction potentials from the book, show why this is so. What are the products of the reaction? Newer pennies contain a mixture of zinc and copper. What happens to the zinc in the penny when the coin is placed in nitric acid? Hydrochloric acid? Support your explanations with data from the book, and include balanced equations for all reactions.

Short Answer

Expert verified
Copper pennies dissolve in nitric acid because the cell potential for this reaction is positive, making it spontaneous. In this reaction, copper nitrate, nitrogen dioxide, and water are produced. The balanced equation is \(Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)\). Zinc, found in newer pennies, also dissolves in nitric acid, forming zinc nitrate, nitrogen dioxide, and water, with the balanced equation being \(Zn(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2NO_2(g) + H_2O(l)\). Zinc also dissolves in hydrochloric acid, producing zinc chloride and hydrogen gas, according to the equation: \(Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)\).

Step by step solution

01

Write down the half-reactions for copper and zinc

We need to first write down the half-reactions for copper and zinc with nitric acid and hydrochloric acid: 1. Copper reacts with nitric acid: \(Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)\) 2. Zinc reacts with nitric acid: \(Zn(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2NO_2(g) + H_2O(l)\) 3. Copper reacts with hydrochloric acid: \(Cu(s) + 2HCl(aq) \rightarrow CuCl_2(aq) + H_2(g)\) 4. Zinc reacts with hydrochloric acid: \(Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)\)
02

Calculate the reduction potentials

Now, we need to find the reduction potentials for each of the reactions. We can look in a reference book to find the standard reduction potentials for copper and zinc: 1. Reduction potential for copper: \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\), \(Eº = +0.34 \,\text{V}\) 2. Reduction potential for zinc: \(Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)\), \(Eº = -0.76 \,\text{V}\) We also need the reduction potentials for nitric acid and hydrochloric acid: 1. Reduction potential for nitric acid: \(NO_3^-(aq) + 4H^+(aq) + 3e^- \rightarrow NO_2(g) + H_2O(l)\), \(Eº = +0.96 \,\text{V}\) NB: The reduction potential of hydrochloric acid can be found in a similar fashion.
03

Assess the spontaneity of reactions

To determine whether each reaction is spontaneous, we need to calculate the cell potential \(Eº_{cell}\) for each combination of half-reactions. If \(Eº_{cell} > 0\), the reaction is spontaneous. 1. Copper reacts with nitric acid: \(Eº_{cell} = Eº_{\text{nitric acid}} - Eº_{\text{copper}} = 0.96 - 0.34 = +0.62 \,\text{V}\) 2. Zinc reacts with nitric acid: \(Eº_{cell} = Eº_{\text{nitric acid}} - Eº_{\text{zinc}} = 0.96 - (-0.76) = +1.72 \,\text{V}\) 3. Copper reacts with hydrochloric acid: The cell potential will be negative (not spontaneous) 4. Zinc reacts with hydrochloric acid: The cell potential will be positive (spontaneous) From the above calculations, we can see that copper dissolves in nitric acid but not in hydrochloric acid because the cell potential for the reaction with nitric acid is positive (spontaneous), whereas it is negative (nonspontaneous) for the reaction with hydrochloric acid. Zinc dissolves in both nitric acid and hydrochloric acid since both have positive cell potentials.
04

Determine reaction products and balanced equations

Using the spontaneous reactions found in Step 3: 1. Reaction of copper with nitric acid (dissolution): Products: \(Cu(NO_3)_2(aq)\), \(2NO_2(g)\), and \(2H_2O(l)\) Balanced equation: \(Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l)\) 2. Reaction of zinc with nitric acid (dissolution): Products: \(Zn(NO_3)_2(aq)\), \(2NO_2(g)\), and \(H_2O(l)\) Balanced equation: \(Zn(s) + 2HNO_3(aq) \rightarrow Zn(NO_3)_2(aq) + 2NO_2(g) + H_2O(l)\) 3. Reaction of zinc with hydrochloric acid (dissolution): Products: \(ZnCl_2(aq)\) and \(H_2(g)\) Balanced equation: \(Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)\) In summary, copper pennies dissolve in nitric acid and produce copper nitrate, nitrogen dioxide, and water. Zinc present in newer pennies also dissolves in nitric acid to form zinc nitrate, nitrogen dioxide, and water, as well as in hydrochloric acid to form zinc chloride and hydrogen gas.

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Most popular questions from this chapter

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table \(17-1\) a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M},\) the cell potential is observed to be 0.31 V. Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

Copper can be plated onto a spoon by placing the spoon in an acidic solution of \(\mathrm{CuSO}_{4}(a q)\) and connecting it to a copper strip via a power source as illustrated below: a. Label the anode and cathode, and describe the direction of the electron flow. b. Write out the chemical equations for the reactions that occur at each electrode.

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$
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