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

1 Answer
Aug 2, 2018

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)3lCu(s)+8lHNO3(aq)3lCu(NO3)2(aq)+2lNO(g)+4lH2O(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)1lCu(s)+4lHNO3(aq,conc.)1lCu(NO3)2(aq)+2lNO2(g)+2lH2O(l)

Explanation:

Nitric (V) acid demonstrates strong oxidizing properties even at low concentrations. In the reaction between copper "Cu"(s)Cu(s) and nitric (V) acid "HNO"_3(aq)HNO3(aq), copper is oxidized from 00 to +2+2 while depending on the concentration of the ("NO"_3)^(-)(NO3) ion, nitrogen is reduced from +5+5 to +4+4 (as "NO"_2NO2, high concentrations) or +2+2 (as "NO"NO, low concentrations).

The oxidation of one copper atom releases 22 electrons that produce either 1 color(white)(l) stackrel(+4)("N")"O"_2 1l+4NO2 molecule or 2//3 color(white)(l) stackrel(+2)("N")"O"2/3l+2NO molecule from one "H"stackrel(+5)("N")"O"_3H+5NO3 molecule. The two equations can thus be balanced with reference to their changes in oxidation states:

  • Copper "Cu"Cu and "NO"_2NO2 are consumed/produced at a 1:21:2 ratio;
  • Copper "Cu"Cu and "NO"NO are consumed/produced at a 3:23:2 ratio.

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