A nitrate is added with sodium hydroxide, then a piece of
aluminum foil. After warming the mixture, ammonia gas is released.A source tells me that aluminum reduces nitrate ion into the ammonium ion. How is this done?Aluminum is an active metal and wants to be ionized. Al -- Al(3+) + 3e(-) ın basic media, tetrahydroxoaluminate complex will be formed: Al(3+) + 4OH(-) -- Al(OH)4(-) ======================================... Al + 4OH(-) -- Al(OH)4(-) + 3e(-) Nitrate is reduced to ammonia in basic media, ammonium in acidic media: N(5+) + 8e(-) -- N(3-) NO3(-) + 8e(-) + 10H(+) -- NH4(+) + 3H2O in basic media, we add base (OH-) to both sides: NO3(-) + 8e(-) + 11OH(-) + 10H(+) -- NH4(+) + 3H2O + 11OH(-) NO3(-) + 8e(-) + OH(-) + 10H2O --- NH3 + 4H2O + 10OH(-) ======================================... NO3(-) + 8e(-)+ 6H2O -- NH3 + 9OH(-) Now, to the balancing of these two redox half reactions: Al + 4OH(-) -- Al(OH)4(-) + 3e(-) NO3(-) + 8e(-)+ 6H2O -- NH3 + 9OH(-) ============================= 8Al + 32OH(-) -- 8Al(OH)4(-) + 24e(-) 3NO3(-) + 24e(-) + 18H2O -- 3NH3 + 27OH(-) ================================== 8Al + 3NO3(-) + 32OH(-) + 18H2O -- 8Al(OH)4(-) + 3NH3 + 27OH(-) //////////////////////////////////////... 8Al + 3NO3(-) + 5OH(-) + 18H2O -- 8Al(OH)4(-) + 3NH3 Hope this helps.Other related question