AP Chemistry Unit 4.9 Oxidation-Reduction Rates

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Oxidation-Reduction Rates

  • 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 ion
  3. Oxygen is usually -2 in its compounds
    • Exception: peroxide (O₂²⁻) which is -1
  1. Hydrogen: +1
  2. Fluorine and the rest of the halogens are -1 (most of the time)
  3. 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
    • Ex:

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
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