How to Balance Redox Reaction: Half-Reaction Method?

lets take example of ...
This occurs in basic solution:

Zn(s) + NO3^- (aq) ---> NH3(aq) + Zn(OH)4^2- (aq)

Answer :-

• Separate the reaction into two half-reactions, one for Zn and one for N in this case.

Zn(s) -----> Zn(OH)42- (aq)
NO31- -----> NH3
Check to make sure the main atoms, Zn and N are balanced. They already are in this case.

• Balance the O by adding water as needed.

Zn(s) 4H2O -----> Zn(OH)42-
NO31- -----> NH3 + 3H2O

• Next, balance the H by adding H+ has needed.

Zn(s) 4H2O -----> Zn(OH)42- + 4H+
NO31- + 9H+ -----> NH3 + 3H2O

• Balance the charges by adding electrons (e-) to the more positive side of each half-reaction.

Zn(s) 4H2O -----> Zn(OH)42- + 4H+ + 2e-
NO31- + 9H+ + 8e- -----> NH3 + 3H2O

• Make the electrons lost and gained equal, in this case by multiplying the first half-reaction by 4.

4Zn(s) 16H2O -----> 4Zn(OH)42- + 16H+ + 8e-
NO31- + 9H+ + 8e- -----> NH3 + 3H2O

• Add the two half-reactions together. Simplify H2O and H+ where possible.

4Zn(s) 16H2O -----> 4Zn(OH)42- + 16H+ + 8e-
NO31- + 9H+ + 8e- -----> NH3 + 3H2O
4Zn (s) + 13H2O + NO31- -----> 4Zn(OH)42- + 7H+ + NH3 (after adding)

• Notice that both sides of the equation have a -1 charge and that all atoms are balanced. The equation is balanced for an acidic (H+, H2O) solution.

• To balance for a basic solution, add the same number of OH- to each side of the equation as there are H+.

4Zn (s) + 13H2O + NO31- + 7OH- -----> 4Zn(OH)42- + 7H+ + 7OH- + NH3

• Combine the H+ and OH- to make H2O.

4Zn (s) + 13H2O + NO31- + 7OH- -----> 4Zn(OH)42- + 7H2O + NH3

• Simplify the H2O.

4Zn (s) + 6H2O + NO31- + 7OH- -----> 4Zn(OH)42- + NH3

The equation is now balanced for a basic (OH-, H2O) solution. Both sides have a net charge of -8 and all the atoms are balanced.