• Know that it is a redox reaction by a change in oxidation states

Oxidation States

• Oxidation Numbers: signifies the number of charges an atom would have in a molecule or ionic compound if electrons were completely transferred
• Know which atom has been reduced/oxidized based on the changes in their oxidation states

○ Reduced = charge/oxidation state decreases (gains electrons); Oxidizing agent: electron acceptor

○ Oxidized = charge/oxidation state increases (loses electrons); Reducing agent: electron donor

○ Na is oxidized = reducing agent; chlorine is reduced = oxidizing agent

• Note: actual charges are written n+ or n-; oxidation states are written +n or –n

Oxidation State Rules

1. Any element by itself: 0
2. Monatomic ion = charge of ionO
3. Oxygen is usually -2 in its compounds

● Exception: peroxide (O²) which is -1

4. Hydrogen: +1

5. Fluorine and the rest of the halogens are -1 (most of the time)

6. Sulfur in SO4: +6

● Sum of oxidation states = 0 in compounds; sum of oxidation states = charge of the ion

● When don’t have rule for one of atoms/polyatomic ions, use the atom that does have a rule to find out oxidation state

Balancing Oxidation-Reduction Equations

Using Oxidation Numbers

1. Assign oxidation numbers to each atom
2. Determine which atoms are being reduced and oxidized
3. Write each half-reactions
4. Balance elements
5. For each half-reaction, balance charge using electrons

●  Electrons must be on opposite sides and MUST have the same coefficients

1. If necessary, multiply by integer to equalize electron count
2. Add up half-reactions and write overall equation
3. Balance remaining elements/compounds

Redox Reactions in Acidic Solutions vs Basic Solutions

Acidic Solutions

• Reaction involves H+ ions → For each half-reaction

○ Balance all elements except H and O

○ Balance oxygen using H2O

○ Balance H using H+

○ Balance the charge using electrons

Basic Solutions

• Reaction involves OH- ions → For each half-reaction

○ Use the above method to obtain the final balanced equation as if H+ were present

○ Add a number of OH- that is equal to the number of H+ ions to both sides of the equation

■ We want to eliminate H+ by forming H2O

○ Eliminate the number of H2O molecules that appear on both sides of the equation

○ Check that the elements and charges are balanced