The reaction of copper Cu with nitric acid HNO3 is given by the balanced equation
3Cu + 8HNO3 ⇔ 3Cu(NO3)2 + 2NO + 4H2O
where the two half-reactions are
3Cu + 6HNO3 → 3Cu(NO3)2 + 6e- + 6H+
2HNO3 + 6e- + 6H+ → 2NO + 4H2O
Considering the oxidation part of the equation to determine the number of moles of electrons
3Cu → 3Cu2+ + 6e-
a total of 6 moles of electrons is therefore lost by the copper in the reaction.
