Question

What are the equations that describes the reaction of Cu and HNO3?

Answer 1

`3 color(white)(l) "Cu"(s) + 8 color(white)(l) "HNO"_3 (aq) to 3 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO" (g) + 4 color(white)(l) "H"_2"O"(l)`
`1 color(white)(l) "Cu"(s) + 4 color(white)(l) "HNO"_3 (aq, "conc.") to 1 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO"_2 (g) + 2 color(white)(l) "H"_2"O"(l)`

Explanation:

Nitric (V) acid demonstrates strong oxidizing properties even at low concentrations. In the reaction between copper `"Cu"(s)` and nitric (V) acid `"HNO"_3(aq)`, copper is oxidized from `0` to `+2` while depending on the concentration of the `("NO"_3)^(-)` ion, nitrogen is reduced from `+5` to `+4` (as `"NO"_2`, high concentrations) or `+2` (as `"NO"`, low concentrations).

The oxidation of one copper atom releases `2` electrons that produce either `1 color(white)(l) stackrel(+4)("N")"O"_2` molecule or `2//3 color(white)(l) stackrel(+2)("N")"O"` molecule from one `"H"stackrel(+5)("N")"O"_3` molecule. The two equations can thus be balanced with reference to their changes in oxidation states:

• Copper `"Cu"` and `"NO"_2` are consumed/produced at a `1:2` ratio;
• Copper `"Cu"` and `"NO"` are consumed/produced at a `3:2` ratio.

Reference
"Nitric acid", https://en.wikipedia.org/wiki/Nitric_acid#Reactions_with_metals