Question

Question #04c20

Answer 1

`2Al + 6 H_2O+2 OH^- = + 3H_2+2 [Al(OH)_4]^-`

Explanation:

Al is an amphoter elements that can be solubilized both in acid and in basic solutions for give `Al^(3+) and [Al(OH)_4]^-`. the second form with NaOH even if diluted. However Al must be oxidized from N° of oxydation zero to three. It can be oxyded from `NO_3^-` but only in acid solutions and we are in basic solution or even from `OH^-` that give `H_2`
So you have
`2Al + 6 H_2O+2 OH^- = + 3H_2+2 [Al(OH)_4]^-`

Answer 2

We assume reduction of nitrate, `NO_3^(-)`, to nitrite, `NO_2^(-)` in basic solution.......

`2Al(s)+ 3NO_3^(-) +2HO^(-)+3H_2O rarr 2[Al(OH)_4]^(-) +3NO_2^(-)`

Explanation:

And thus we go from `N(+V)` to `N(+III)`:

`NO_3^(-) +2H^(+) + 2e^(-) rarr NO_2^(-) + H_2O` `(i)`

Meanwhile we have aluminum, a potent reductant....:

`Al(s) rarr Al^(3+) + 3e^(-)` `(ii)`

And so we add `3xx(i)+2xx(ii)`

`2Al(s)+ 3NO_3^(-) +6H^(+) + cancel(6e^(-)) rarr 2Al^(3+) + cancel(6e^(-))+3NO_2^(-) + 3H_2O`

But there is a catch; BASIC conditions were specified, and we got acid in the stoichiometric equation; we add `6xxHO^-` to each side to address this....

`2Al(s)+ 3NO_3^(-) +underbrace(6H^(+) + 6HO^(-))_(6H_2O)rarr 2Al^(3+) + 6HO^(-) +3NO_2^(-) + 3H_2O`

To give finally......(almost!)

`2Al(s)+ 3NO_3^(-) +3H_2Orarr 2Al^(3+) + 6HO^(-) +3NO_2^(-)`

Nitrate is a potent oxidant, but often its oxidizing potential is masked in aqueous solution; and of course aluminum is a potent source of electrons....In aqueous solution, the metal is probably present as aluminate ion, i.e. `[Al(OH)_4]^-`, and so to the given equation we could `2xxHO^-` (again) to each side....

`2Al(s)+ 3NO_3^(-) +2HO^(-)+3H_2O rarr 2[Al(OH)_4]^(-) +3NO_2^(-)`

And this equation represents the oxidation of aluminum in basic solution.....

Answer 3

`8Al + 5 OH^(-) + 18 H2O+ 3NO_3^(-) = 3 NH3 + 8[Al(OH)_4]^(-)`

Explanation:

Now i remember that in chemical analisis yuo can look for nitrates with Aluminium through this experience, making react nitrates with Al in basic solution for NaOH . But Al must be in dust, with dust of Copper as catalyst and Mg as reactant and catalyst. Yuo obtain directly ammonia. The same reaction can happen with nitites

`2Al + 2OH^(-) + 6 H2O= 2[Al(OH)_4]^(-) + 3H2`
`NO_3^(-) + 4 H2 = NH3 + OH^(-) + 2 H2O`

`8Al + 5 OH^(-) + 18 H2O+ 3NO_3^(-) = 3 NH3 + 8[Al(OH)_4]^(-)`