Question

Balance the following redox equation in basic solution.  What is the coefficient of the water?  NO ₂^-(aq) + Al(s) → NH ₃(g) + AlO ₂^-(aq)

Answer 1

The unbalanced equation

NO2- + Al = NH3 + AlO2-

Assign oxidation numbers for each atom within brackets

N(+3)O(-2)2- + Al(0) = N(-3)H(+1)3 + Al(+3)O(-2)2-

The two half cell reactions are

Oxidation reaction

O:

Al = AlO2-

Al is being oxidized

Reduction reaction

R:

NO2- = NH3 ​​​​​​

N is being reduced

Balance the oxygen atom by adding water molecules.

O:

Al + 2H2O = AlO2-

R:

NO2- = NH3 + 2H2O

Balance the hydrogen atoms by adding (H+).

O:

Al + 2H2O = AlO2- + 4H+

R:

NO2- + 7H+ = NH3 + 2H2O

In basic medium, add one OH- ion for every H+ ion

O:

Al + 2H2O + 4OH- = AlO2- + 4H2O

R:

NO2- + 7H2O = NH3 + 2H2O + 7OH-

Balance the charge.

O:

Al + 2H2O + 4OH- = AlO2- + 4H2O + 3e-

R:

NO2- + 7H2O + 6e- = NH3 + 2H2O + 7OH-

Make electron balance

O:

Al + 2H2O + 4OH- = AlO2- + 4H2O + 3e-

Multiply by 2

2Al + 4H2O + 8OH- = 2AlO2- + 8H2O + 6e-

R:

NO2- + 7H2O + 6e- = NH3 + 2H2O + 7OH-

Add the half-reactions

2Al + NO2- + 11H2O + 8OH- + 6e- = 2AlO2-+ NH3 + 10H2O + 6e- + 7OH-

The final balanced reaction

2Al + NO2- + H2O + OH- = 2AlO2- + NH3

Coefficient of water = 1