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

Question

10061 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-07T06:17:09

Nice redox equation here.......

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

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..........