Balancing Redox Equations

 

FOOL-PROOF Method Example
1. Assign oxidation numbers (o.n.) to atoms and write a net ionic equation for each half reaction. Work on each half reaction separately.

2. Balance atoms being oxidized or reduced.

3. From the change in oxidation numbers, determine the number of electrons involved and add them to the high oxidation number side of the equation.

4. Balance the charge, using either H+ (if in ACID solution) or OH- (if in BASE solution).

5. Now balance the OXYGEN by adding H2O to the appropriate side of the equation.

Check that everything balances - it will!

6. Equalize the number of electrons in each half reaction by multiplying by the appropriate factor and then add.

Zinc metal in basic solution rapidly reduces potassium nitrate to ammonia according to the unbalanced equation
Zn(s) + KNO3(aq) ----->NH3(aq) + K2Zn(OH)4
Balance this equation.

1.

Zno.n. = 0   Zno.n. = +2
Zn(s) -----> Zn(OH)42-(aq)
     
No.n. = +5   No.n. = -3
NO3-(aq) -----> NH3(aq)

2, 3.
Zn(s) -----> Zn(OH)42-(aq) + 2e-
NO3-(aq) + 8e- -----> NH3(aq)

4.
Zn(s) + 4OH-(aq) -----> Zn(OH)42-(aq) + 2e-
NO3-(aq) + 8e- -----> NH3(aq) + 9OH-(aq)

5.
Zn(s) + 4OH-(aq) -----> Zn(OH)42-(aq) + 2e-
NO3-(aq) + 6H2O(l) + 8e- -----> NH3(aq) + 9OH-(aq)

6.
[Zn(s) + 4OH-(aq) -----> Zn(OH)42-(aq) + 2e-] x 4
NO3-(aq) + 6H2O(l) + 8e- -----> NH3(aq) + 9OH-(aq)

4Zn(s) + NO3-(aq) + 7OH-(aq) + 6H2O(l) -----> 4Zn(OH)42-(aq) + NH3(aq)