Question

What are the equations that describe the reaction of $ Cu $ and $ HN{O_3} $ ?

Answer 1

Hint: The reactions between copper and nitric acid are examples of oxidation-reduction reactions, where gaining electrons reduces one element and losing them oxidizes the other. Nitric acid is not only a strong acid, it is an oxidizing agent.

Complete answer:
Copper can undergo one of two reactions when combined with nitric acid, depending on the concentration of the solution. If the nitric acid is dilute, the copper will be oxidized to form copper nitrate with nitric oxide as a byproduct. If the solution is concentrated, the copper will be oxidized to form copper nitrate with nitrogen dioxide as a byproduct. Nitric acid demonstrates strong oxidizing properties even at low concentrations. In the reaction between copper $ Cu $ and nitric acid $ HN{O_3} $ , copper is oxidized from $ 0 $ to $ + 2 $ while depending on the concentration of the $ N{O_3}^ - $ ion, nitrogen is reduced from $ + 5 $ to $ + 4 $ or $ + 2 $ .
The oxidation of one copper atom releases two electrons that produce either $ 1N{O_2} $ molecule or $ \dfrac{2}{3}NO $ molecule from one $ HN{O_3} $ molecule.
 $ Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O $ Which produces nitrogen dioxide and
 $ 3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O $ Which produces nitric oxide.
Both nitric oxide and nitrogen dioxide are noxious and potentially toxic at high levels; nitrogen dioxide is the ugly brown gas present in the smog haze over many cities.

Note:
Note that this reaction takes place when excess concentrated Nitric acid is reacted with copper. The equation would be balanced differently if dilute Nitric acid was used. When balancing chemical equations our goal is to have the same number of each type of atom on both sides of the equation.