# How is zinc metal oxidized, and nitrate ion reduced to give ammonium ion in aqueous by the action of nitric acid on zinc?

Jun 14, 2016

In every chemical reaction mass is conserved. Here zinc metal is oxidized, and nitrogen is reduced.

#### Explanation:

$\text{Oxidation half equation}$:

$Z n \rightarrow Z {n}^{2 +} + 2 {e}^{-}$ $\left(i\right)$

$\text{Reduction half equation}$:

$N {O}_{3}^{-} + 10 {H}^{+} + 8 {e}^{-} \rightarrow N {H}_{4}^{+} + 3 {H}_{2} O \left(l\right)$ $\left(i i\right)$

Check. ARE both equations balanced wth respect to mass and charge? Don't trust my arithmetic.

So $4 \times \left(i\right) + \left(i i\right)$ $=$

$4 Z n \left(s\right) + N {O}_{3}^{-} + 10 {H}^{+} \rightarrow 4 Z {n}^{2 +} + N {H}_{4}^{+} + 3 {H}_{2} O \left(l\right)$

We could add $9 \times N {O}_{3}^{-}$ ions to each side to give...

$4 Z n \left(s\right) + 10 H N {O}_{3} \left(a q\right) \rightarrow 4 Z n {\left(N {O}_{3}\right)}_{2} + N {H}_{4} N {O}_{3} \left(a q\right) + 3 {H}_{2} O \left(l\right)$

I had the idea that cold, dilute nitric acid oxidizes zinc metal with nitrogen oxide produced, but I follow the parameters of the question.

$3 Z n \left(s\right) + 8 H N {O}_{3} \left(a q\right) \rightarrow 3 Z n {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 N O \left(g\right) \uparrow + 4 {H}_{2} O \left(l\right)$