# How does aluminum metal react with nitrate ion under basic conditions to give ammonia, and aluminum ion?

Aug 7, 2017

Nice redox equation here.......

$8 A l \left(s\right) + 3 N {O}_{3}^{-} + 5 H {O}^{-} + 18 {H}_{2} O \rightarrow 8 A l {\left(O H\right)}_{4}^{-} + 3 N {H}_{3} \left(a q\right)$

#### Explanation:

Aluminum metal is oxidized to aluminate ion.......under basic conditions.....

$A l \left(s\right) + 4 H {O}^{-} \rightarrow A l {\left(O H\right)}_{4}^{-} + 3 {e}^{-}$ $\left(i\right)$

Nitrate ion, $\stackrel{+ V}{N}$, is reduced to ammonia......$\stackrel{- I I I}{N}$.

$N {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-} \rightarrow N {H}_{3} \left(a q\right) + 3 {H}_{2} O$ $\left(i\right)$

And because we deal with a basic medium, I am simply going to add $9 \times H {O}^{-}$ to each side.........

$N {O}_{3}^{-} + {\underbrace{9 {H}^{+} + 9 H {O}^{-}}}_{9 {H}_{2} O} + 8 {e}^{-} \rightarrow N {H}_{3} \left(a q\right) + 3 {H}_{2} O + 9 H {O}^{-}$

......to give finally.....

$N {O}_{3}^{-} + 6 {H}_{2} O + 8 {e}^{-} \rightarrow N {H}_{3} \left(a q\right) + 9 H {O}^{-}$ $\left(i i\right)$

And so we take $8 \times \left(i\right) + 3 \times \left(i i\right)$

$3 N {O}_{3}^{-} + 18 {H}_{2} O + 24 {e}^{-} \rightarrow 3 N {H}_{3} \left(a q\right) + 27 H {O}^{-}$ $+$
$8 A l \left(s\right) + 32 H {O}^{-} \rightarrow 8 A l {\left(O H\right)}_{4}^{-} + 24 {e}^{-}$

And after cancelling common reagents.......

$3 N {O}_{3}^{-} + 18 {H}_{2} O + \cancel{24 {e}^{-}} \rightarrow 3 N {H}_{3} \left(a q\right) + \cancel{27 H {O}^{-}}$ $+$
$8 A l \left(s\right) + \cancel{32} 5 H {O}^{-} \rightarrow 8 A l {\left(O H\right)}_{4}^{-} + \cancel{24 {e}^{-}}$

To give finally.....

$8 A l \left(s\right) + 3 N {O}_{3}^{-} + 5 H {O}^{-} + 18 {H}_{2} O \rightarrow 8 A l {\left(O H\right)}_{4}^{-} + 3 N {H}_{3} \left(a q\right)$

Which I think is balanced with respect to mass and charge. Is it? Do not trust my 'rithmetic..........