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

1 Answer
Aug 7, 2017

Answer:

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..........