91Ó°ÊÓ

The reduction potentials we will need to consider are: - Copper(II) \(\rightarrow\) Copper: \(Cu^2+ + 2e^- \rightarrow Cu\); \(E^0 = 0.34 V\). - Zinc(II) \(\rightarrow\) Zinc: \(Zn^2+ + 2e^- \rightarrow Zn\); \(E^0 = -0.76 V\). - Nitrate ion \(\rightarrow\) Nitrogen(IV) oxide: \(NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O\); \(E^0 = 0.96 V\). - Hydrogen: \(2H^+ + 2e^- \rightarrow H_2\); \(E^0 = 0 V\).

To see if copper can react with nitric acid, we need to check if the following reaction is spontaneous: \[Cu(s) + NO_3^-(aq) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + NO_2(g) + H_2O(l)\] To do this, we should add the half-reactions and their standard potentials: \[\begin{cases} Cu^2+ + 2e^- \rightarrow Cu & E^0 = 0.34 V\\ NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O & E^0 = 0.96 V \end{cases}\] Adding them gives us the full reaction and the cell potential, \(E_{cell}\): \[\begin{cases} Cu(s) + NO_3^-(aq) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + NO_2(g) + H_2O(l) \\ E_{cell} = 0.34 V + 0.96 V = 1.30 V \end{cases}\] Since \(E_{cell} > 0\), the reaction between copper and nitric acid is spontaneous, and the copper penny will dissolve in nitric acid.

To test if copper can react with hydrochloric acid, we need to check if the following reaction is spontaneous: \[Cu(s)+ 2H^+(aq) \rightarrow Cu^{2+}(aq)+ H_2(g)\] Again, we add the half-reactions and their standard potentials: \[\begin{cases} Cu^2+ + 2e^- \rightarrow Cu & E^0 = 0.34 V \\ 2H^+(aq) + 2e^- \rightarrow H_2(g) & E^0 = 0 V \end{cases}\] Adding them gives us the full reaction and the cell potential, \(E_{cell}\): \[\begin{cases} Cu(s) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + H_2(g) \\ E_{cell} = 0.34 V + 0 V = 0.34 V \end{cases}\] Since \(E_{cell} > 0\), the reaction between copper and hydrochloric acid is spontaneous. However, the cell potential is significantly lower than that of the reaction with nitric acid, making the reaction with hydrochloric acid less favorable.

Now we need to determine if zinc will react with nitric acid and hydrochloric acid. We look at the reactions: \[\begin{cases} Zn(s) + NO_3^-(aq) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + NO_2(g) + H_2O(l) \\ Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g) \end{cases}\] The net cell potentials can be determined from the standard reduction potentials for zinc: \[\begin{cases} E_{cell,1}= (-0.76 V) + 0.96 V = 0.2 V \\ E_{cell,2}= (-0.76 V) + 0 V = -0.76 V \end{cases}\] Since \(E_{cell,1}\) is positive, zinc will react with nitric acid, whereas with hydrochloric acid, it will not react as \(E_{cell,2}\) is negative. In conclusion, copper pennies will dissolve in nitric acid, but their reaction with hydrochloric acid will be less favorable. Newer pennies containing zinc will react with nitric acid, but will not react with hydrochloric acid.