# How do we represent the oxidation of copper metal to "cupric ion", with accompanying reduction of nitrate ion to NO(g)?

Jan 6, 2016

Copper metal is oxidized to cupric ion, $C {u}^{2 +}$. Nitrate ($N \left(V\right)$) is reduced to nitrous oxide ($N \left(I {I}^{+}\right)$).

#### Explanation:

$\text{Oxidation half equation (i):}$

$C u \left(s\right) \rightarrow C {u}^{2 +} + 2 {e}^{-}$

$\text{Reduction half equation (ii):}$

$N {O}_{3}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow N O + 2 {H}_{2} O$ $\left(i i\right)$

$3 \times \left(i\right) + 2 \times \left(i i\right)$:

$3 C u \left(s\right) + 2 N {O}_{3}^{-} + 8 {H}^{+} \rightarrow 3 C {u}^{2 +} + 2 N O \uparrow + 4 {H}_{2} O$

Is this balanced with respect to both mass and charge? Should it be?