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Chapter 18: 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 can dissolve in nitric acid but not in hydrochloric acid because the reaction between copper and nitric acid has a positive cell potential (+0.62 V), indicating a spontaneous reaction, while the reaction between copper and hydrochloric acid has a negative cell potential (-0.34 V), indicating a non-spontaneous reaction. The products of the copper and nitric acid reaction are aqueous copper(II) ions, gaseous nitrogen monoxide, and water, with the balanced equation: \(3Cu(s) + 8HNO_3(aq) \rightarrow 3Cu^{2+}(aq) + 2NO(g) + 4H_2O(l) + 6NO_3^-(aq)\). Both zinc and copper will dissolve in nitric acid, while in hydrochloric acid, only zinc will dissolve, producing zinc ions, chloride ions, and hydrogen gas.

Step by step solution

01

Gather information about reduction potentials of relevant species.

We will use the half-reaction potentials for Cu, Zn, HNO3, and HCl. We can find these values from the book or other reference materials. Copper: \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\), \(E^0 = +0.34 V\) Zinc: \(Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)\), \(E^0 = -0.76 V\) Nitric Acid: \(NO_3^-(aq) + 4H^+(aq) + 3e^- \rightarrow NO(g) + 2H_2O(l)\), \(E^0 = +0.96 V\) Hydrochloric Acid: \(2H^+(aq) + 2e^- \rightarrow H_2(g)\), \(E^0 = 0 V\)
02

Find the overall cell potential for each reaction.

Calculate the overall cell potential for both the Cu and Zn reactions with HNO3 and HCl. Cell potential (E cell): For a redox reaction, the overall cell potential can be calculated as \(E_{cell} = E_{reduction} - E_{oxidation}\). Copper and Nitric Acid: \(E_{cell} = (+0.96 V) - (+0.34 V) = +0.62 V\) Copper and Hydrochloric Acid: \(E_{cell} = (0 V) - (+0.34 V) = -0.34 V\) Zinc and Nitric Acid: \(E_{cell} = (+0.96 V) - (-0.76 V) = +1.72 V\) Zinc and Hydrochloric Acid: \(E_{cell} = (0 V) - (-0.76 V) = +0.76 V\)
03

Explain why copper dissolves in nitric acid but not in hydrochloric acid.

As observed in Step 2, the copper and nitric acid reaction has a positive cell potential (+0.62 V), indicating that the reaction is spontaneous and favors the formation of products. The copper and hydrochloric acid reaction has a negative cell potential (-0.34 V), which means the reaction is non-spontaneous and doesn't proceed in the forward direction. Thus, copper can dissolve in nitric acid but not in hydrochloric acid.
04

Identify the products of the reaction.

The products of the copper and nitric acid reaction are aqueous copper(II) ions, gaseous nitrogen monoxide, and water. The balanced equation for this reaction is: \(3Cu(s) + 8HNO_3(aq) \rightarrow 3Cu^{2+}(aq) + 2NO(g) + 4H_2O(l) + 6NO_3^-(aq)\)
05

Describe the behavior of zinc in both acids.

As calculated in Step 2, both the zinc and nitric acid, and zinc and hydrochloric acid reactions have positive cell potentials (+1.72 V and +0.76 V, respectively), which indicates that both reactions are spontaneous and will proceed in the forward direction. The balanced equation for the zinc and nitric acid reaction is: \(Zn(s) + 2HNO_3(aq) \rightarrow Zn^{2+}(aq) + 2NO_3^-(aq) + NO(g) + H_2O(l)\) The balanced equation for the zinc and hydrochloric acid reaction is: \(Zn(s) + 2HCl(aq) \rightarrow Zn^{2+}(aq) + 2Cl^-(aq) + H_2(g)\) Therefore, both zinc and copper will dissolve in nitric acid, with the products being the corresponding metal ions, gaseous nitrogen monoxide, water, and nitrate ions. In hydrochloric acid, only zinc will dissolve, producing zinc ions, chloride ions, and hydrogen gas.

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Most popular questions from this chapter

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{~kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) b. \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\)

Consider only the species (at standard conditions) $$\mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd}$$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 \mathrm{M} \mathrm{Cd}^{2+}, 1.0 \mathrm{MAg}^{+}, 1.0 \mathrm{M} \mathrm{Au}^{3+}\), and \(1.0 \mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\). b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI. c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\). d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) at \(25^{\circ} \mathrm{C}\).
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