This document outlines two methods for balancing redox reactions: the half-reaction method and the oxidation number method. The half-reaction method involves writing the oxidation and reduction as separate half-reactions, balancing each, and combining them. The oxidation number method assigns oxidation states and uses bracketing lines to connect oxidized and reduced atoms before balancing the overall change in oxidation. Both methods aim to balance the atoms and charges in redox reactions.

3. Equations for simple redox reactions can be
balanced by looking at them
Most redox equations require more systematic
methods
Equation-balancing process needs use of
oxidation numbers
Both charge and mass are conserved
Half-reactions balanced separately then
combined

5. STEPS OF HALF – REACTION METHOD
1. Write the formula equation if it is not given in the
problem. Then write the ionic equation.
2. Assign oxidation numbers. Delete substances
containing only elements that do not change
oxidation state.
3. Write half- reaction of oxidation.
(a). Balance the atoms.
(b). Balance the charge.
4. Write half- reaction for reduction.
(a). Balance the atoms.
(b). Balance charge.

6. 5. Conserve charge by adjusting the coefficients in front of
electrons so that the number lost in oxidation equals the
number gained in reduction.
6. Combine the half reactions and cancel out anything
common to both sides of the equation.
7. Combine ions to form the compounds shown in original
formula equation. Check to ensure that all other ions
balance.

8. Sulfur changes oxidation state from -2 to +6
Nitrogen changes from +5 to +4
Other substances deleted

9. In this example, sulfur is being oxidized

10. To balance oxygen, H2O must be added to left side
This gives 10 extra hydrogen atoms on that side
So, 10 H atoms added to right side
In basic solution, OH- ions and water can be used to balance atoms

11. Electrons added to side having greater positive net charge
Left side has no net charge
Right side has +8
Add 8 electrons to product side
(oxidation of sulfur from -2 to +6 involves loss of 8 e-)

12. Nitrogen reduced from +5 to +4

13. H2O added to product side to balance oxygen atoms
2 hydrogen ions added to reactant side to balance H atoms

14. Electrons added to side having greater positive net charge
Left side has net charge of +1
1 e- added to this side balancing the charge

16. This ratio is already in lowest terms
If not, need to reduce
Multiply oxidation half-reaction by 1
Multiple reduction half-reaction by 8
Electrons lost = electrons gained

17. Each side has 10H+, 8e-, and 4H2O
They cancel

18. The NO3- ion appeared as nitric acid in original equation
Only 6 H ions to pair with 8 nitrate ions
So, 2 H ions must be added to complete this formula
If 2 H ions added to left side, then 2 H ions must be added to the right side

19. Sulfate ion appeared as sulfuric acid in original equation
H ions added to right side used to complete formula for sulfuric acid

21. STEPS OF OXIDATION NUMBER METHOD
1. Step 1 – assign oxidation to all the atoms in the
equation.
2. Step 2 - Identifying atoms oxidized and reduced.
3. Step 3 - Use one bracketing line to connect the atoms
that undergo oxidation & another to connect
reduced.
4. Step 4 - Make the total increase in oxidation equal to the
total decrease in oxidation by using appropriate
coefficients.

22. EXAMPLE:-
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
(unbalanced)
Step 1 – assign oxidation to all the atoms
in the equation.
Step 2 -ID atoms oxidized and reduced.

23. Step 3 – Use one bracketing line to
connect the atoms that undergo
oxidation & another to connect reduced.

24. Step 4 – Make the total increase in
oxidation # equal to the total decrease in
oxidation # by using appropriate
coefficients.