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

Aug 2, 2018

$3 \textcolor{w h i t e}{l} \text{Cu"(s) + 8 color(white)(l) "HNO"_3 (aq) to 3 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO" (g) + 4 color(white)(l) "H"_2"O} \left(l\right)$
$1 \textcolor{w h i t e}{l} \text{Cu"(s) + 4 color(white)(l) "HNO"_3 (aq, "conc.") to 1 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO"_2 (g) + 2 color(white)(l) "H"_2"O} \left(l\right)$

#### Explanation:

Nitric (V) acid demonstrates strong oxidizing properties even at low concentrations. In the reaction between copper $\text{Cu} \left(s\right)$ and nitric (V) acid ${\text{HNO}}_{3} \left(a q\right)$, copper is oxidized from $0$ to $+ 2$ while depending on the concentration of the ${\left({\text{NO}}_{3}\right)}^{-}$ ion, nitrogen is reduced from $+ 5$ to $+ 4$ (as ${\text{NO}}_{2}$, high concentrations) or $+ 2$ (as $\text{NO}$, 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 ${\text{H"stackrel(+5)("N")"O}}_{3}$ molecule. The two equations can thus be balanced with reference to their changes in oxidation states:

• Copper $\text{Cu}$ and ${\text{NO}}_{2}$ are consumed/produced at a $1 : 2$ ratio;
• Copper $\text{Cu}$ and $\text{NO}$ are consumed/produced at a $3 : 2$ ratio.

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