Question

Balance in basic conditions:

Al(s) + NO₂^-(aq) ---> Al(OH)₄^-(aq) + NH₄OH(aq)

Answer 1

Al -------> Al (OH)4 -

Balancing 'O' by adding H2O

Al + 4H2O -----> Al(OH)4 -

Balancing 'H'

Al + 4H2O ------> Al(OH)4 - + 4H+

Adding 4 OH- on both sides

Al + 4H2O + 4OH- -----> Al(OH)4 - + 4H2O (4H+ + 4OH- ----> 4H2O)

=> Al + 4OH- ------> Al(OH)4 -

Balancing charge by adding e-

Al + 4OH- ------> Al(OH)4 - + 3e- .................... (1)

NO2 - ---------> NH4OH

Balancing 'O'

NO2- ------> NH4OH + H2O

Balancing 'H'

NO2- + 7H+ ------> NH4OH + H2O

Adding 7 OH- on both sides

=> NO2- + 7H2O -----> NH4OH + H2O + 7OH-

=> NO2- + 6H2O ------> NH4OH + 7OH-

Balancing charge

NO2- + 6H2O + 6e- ------> NH4OH + 7OH- ................(2)

Therefore the half reactions are,

Al + 4OH- ------> Al(OH)4 - + 3e- .................... (1)

NO2- + 6H2O + 6e- ------> NH4OH + 7OH- ................(2)

2 x (1) + (2) gives,

2Al + OH- + NO2- + 6H2O ------> 2Al(OH)4 - + NH4OH ...... Balanced Equation