Question

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

Copper pennies can dissolve in nitric acid because the reaction between copper and nitric acid is spontaneous, with a positive cell potential ($E_{cell}=1.30V$). The products of this reaction are $C{u}^{2+}(aq),N{O}_2(g),$ and $H_{2}O(l)$. Copper doesn't dissolve in hydrochloric acid, as its reaction is less favorable ($E_{cell}=0.34V$). In newer pennies containing zinc, zinc reacts with nitric acid ($E_{cell}=0.2V$), but not with hydrochloric acid ($E_{cell}=-0.76V$).

Step-by-step solution

Identify the relevant reduction potentials

The reduction potentials we will need to consider are: - Copper(II) $\to$ Copper: $C{u}^2++2e^{-}\to Cu$; $E^0=0.34V$. - Zinc(II) $\to$ Zinc: $Z{n}^2++2e^{-}\to Zn$; $E^0=-0.76V$. - Nitrate ion $\to$ Nitrogen(IV) oxide: $N{O}_3^{-}+2H^{+}+e^{-}\to N{O}_2+H_{2}O$; $E^0=0.96V$. - Hydrogen: $2H^{+}+2e^{-}\to H_2$; $E^0=0V$.

Determine whether copper can react with nitric acid

To see if copper can react with nitric acid, we need to check if the following reaction is spontaneous: $$Cu(s)+N{O}_3^{-}(aq)+2H^{+}(aq)\to C{u}^{2+}(aq)+N{O}_2(g)+H_{2}O(l)$$ To do this, we should add the half-reactions and their standard potentials: $$\{\begin{array}{ll}C{u}^2++2e^{-}\to Cu& E^0=0.34V\\ N{O}_3^{-}+2H^{+}+e^{-}\to N{O}_2+H_{2}O& E^0=0.96V\end{array}$$ Adding them gives us the full reaction and the cell potential, $E_{cell}$: $$\{\begin{array}{l}Cu(s)+N{O}_3^{-}(aq)+2H^{+}(aq)\to C{u}^{2+}(aq)+N{O}_2(g)+H_{2}O(l)\\ E_{cell}=0.34V+0.96V=1.30V\end{array}$$ Since $E_{cell}>0$, the reaction between copper and nitric acid is spontaneous, and the copper penny will dissolve in nitric acid.

Determine whether copper can react with hydrochloric acid

To test if copper can react with hydrochloric acid, we need to check if the following reaction is spontaneous: $$Cu(s)+2H^{+}(aq)\to C{u}^{2+}(aq)+H_2(g)$$ Again, we add the half-reactions and their standard potentials: $$\{\begin{array}{ll}C{u}^2++2e^{-}\to Cu& E^0=0.34V\\ 2H^{+}(aq)+2e^{-}\to H_2(g)& E^0=0V\end{array}$$ Adding them gives us the full reaction and the cell potential, $E_{cell}$: $$\{\begin{array}{l}Cu(s)+2H^{+}(aq)\to C{u}^{2+}(aq)+H_2(g)\\ E_{cell}=0.34V+0V=0.34V\end{array}$$ 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.

Determine what happens to zinc in nitric acid and hydrochloric acid

Now we need to determine if zinc will react with nitric acid and hydrochloric acid. We look at the reactions: $$\{\begin{array}{l}Zn(s)+N{O}_3^{-}(aq)+2H^{+}(aq)\to Z{n}^{2+}(aq)+N{O}_2(g)+H_{2}O(l)\\ Zn(s)+2H^{+}(aq)\to Z{n}^{2+}(aq)+H_2(g)\end{array}$$ The net cell potentials can be determined from the standard reduction potentials for zinc: $$\{\begin{array}{l}{E}_{cell,1}=(-0.76V)+0.96V=0.2V\\ E_{cell,2}=(-0.76V)+0V=-0.76V\end{array}$$ 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.