Question

A copper penny can be dissolved in nitric acid but not in hydrochloric acid. Using reduction potentials given in 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 placed in nitric acid? Hydrochloric acid? Support your explanations with the data from the book, and include balanced equations for all reactions.

Short answer

The product of the reaction and zinc reaction in the penny when it is placed in hydrochloric acid and nitric acid need to determined.

Step-by-step solution

Definition of oxidation reduction potential

The tendency of a molecule to process reduction is called reduction potential. This can be undergone by standard reduction potential. The measure of system to release or accept electron is called oxygen reduction potential. When there is an acceptance of electron by system called oxidizing system whereas release oxygen called reducing system.

Formation of corresponding cation

There is a formation of cation is metal dissolution. From the standard reduction potential at 25^oC,

$\begin{array}{c}C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)E^{o}red=0.34V\\ N{O_3}^{-}\left(aq\right)+4H^{+}\left(aq\right)+3e^{-}\to NO\left(g\right)+2H_{2}O\left(l\right)E^{o}red=0.96V\\ C{l}_2\left(g\right)+2e^{-}\left(aq\right)\to 2C{l}^{-}\left(aq\right)E^{o}red=1.36V\\ Z{n}^{2+}\left(aq\right)+2e^{-}\to Zn\left(s\right)E^{o}red=-0.76V\end{array}$

Zinc placed in nitric acid and hydrochloric acid

Copper cannot placed in hydrochloric acid but it can dissolved in nitric acid due to its reduction potential of Cu²⁺ which is less than reduction potential of $N{O_3}^{-}.$The reduction potential of chloride is greater oxidation potential and it can easily oxidized than copper. Copper is not dissolved in hydrochloric acid.

When a penny contains Cu and Zn and dissolved in nitric acid, it is first dissolved in solvent due to their greater oxidation potential. Zn-Cu cannot dissolved in solvent due to their higher oxidation potential of Cl^-, and it can be easily oxidized and leads to the formation of Cl₂.