Question

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

Answer 1

Nice redox equation here.......

`8Al(s)+3NO_3^(-) + 5HO^(-) + 18H_2O rarr 8Al(OH)_4^(-)+3NH_3(aq)`

Explanation:

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

`Al(s) +4HO^(-) rarr Al(OH)_4^(-)+3e^(-)` `(i)`

Nitrate ion, `stackrel(+V)N`, is reduced to ammonia......`stackrel(-III)N`.

`NO_3^(-) +9H^(+) + 8e^(-) rarr NH_3(aq) + 3H_2O` `(i)`

And because we deal with a basic medium, I am simply going to add `9xxHO^-` to each side.........

`NO_3^(-) +underbrace(9H^(+) + 9HO^(-))_(9H_2O) + 8e^(-) rarr NH_3(aq) + 3H_2O+9HO^-`

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

`NO_3^(-) +6H_2O + 8e^(-) rarr NH_3(aq) +9HO^-` `(ii)`

And so we take `8xx(i) + 3xx(ii)`

`3NO_3^(-) +18H_2O + 24e^(-) rarr 3NH_3(aq) +27HO^-` `+`
`8Al(s) +32HO^(-) rarr 8Al(OH)_4^(-)+24e^(-)`

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

`3NO_3^(-) +18H_2O + cancel(24e^(-)) rarr 3NH_3(aq) +cancel(27HO^-)` `+`
`8Al(s) +cancel(32)5HO^(-) rarr 8Al(OH)_4^(-)+cancel(24e^(-))`

To give finally.....

`8Al(s)+3NO_3^(-) + 5HO^(-) + 18H_2O rarr 8Al(OH)_4^(-)+3NH_3(aq)`

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