Question

What happens when $Cu$ reacts with cold dilute $HN{O_3}$?

Answer 1

Hint: Copper is an inaccessible metal and does not react with dilute acid under normal conditions. However, it reacts with nitric acid. ... Nitric acid is an oxidizing agent and the reaction is not a normal acid + metal reaction. The products are oxides of nitrogen instead of hydrogen.

Complete step by step answer:
-Copper is a reddish-brown metal, widely used in plumbing and electrical wiring; It is perhaps most familiar to people in the United States as Penny. (Although from $1983$ onwards, penises are actually made of zinc, surrounded by a thin copper foil of paper to give them a traditional form of penny.)
-Copper is oxidized by concentrated nitric acid, HNO3 to produce $Cu^{2+}$ ions; nitric acid is reduced to nitrogen dioxide, a poisonous brown gas with a irritating odor:
                $Cu(s) + 4N{O_3}(aq) \to Cu{(N{O_3})_2}(aq) + 2N{O_2}(g) + 2{H_2}O$
-When copper is first oxidized, the solution is very concentrated, and the $Cu^{2+}$ product is initially coordinated from nitric acid to nitrate ions, giving the solution a green, and then a greenish-brown color.
-When the solution is diluted with water, the water molecules displace the nitrate ions in the coordination sites around the copper ions, causing the solution to turn blue.
-In dilute nitric acid, the reaction produces nitric oxide,
                $3Cu(s) + 8HN{O_3}(aq) \to 3Cu{(N{O_3})_2}(aq) + 2N{O_{}}(g) + 4{H_2}O$

Note: Copper reacts with cold and dilute nitric acid to remove copper nitrate, water, and nitric oxide.
\[3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 4{H_2}O + 2NO\]
The formed nitric oxide combines with oxygen of air to give brown fumes of $N{O_2}$.
$2NO + {O_2} \to 2N{O_2}$