Redox Reactions: Involve oxidation (loss of electrons) and reduction (gain of electrons).
Disproportionation Reactions: A single substance is both oxidized and reduced in the same reaction.
Example: Decomposition of Hydrogen Peroxide
Reaction: Hydrogen Peroxide (H₂O₂) decomposing into Water (H₂O) and Oxygen (O₂) with a catalyst like Potassium Iodide.
Chemical Equation: H₂O₂ → H₂O + O₂
Assigning Oxidation States
Methods:
Memorization of rules (found in chemistry textbooks).
Calculating through dot structures and electronegativity.
Oxidation State Assignment
Water (H₂O) Oxidation State
Oxygen in water: -2
Calculation:
Oxygen is more electronegative than hydrogen, hence attracts electrons.
Oxygen has 6 valence electrons, surrounded by 8 (6 - 8 = -2).
Oxygen (O₂) Molecule Oxidation State
Oxygen in O₂: 0
Calculation:
Bonded to another oxygen with equal electronegativity, electrons are shared equally.
Normal 6 electrons, surrounded by 6 (6 - 6 = 0).
Hydrogen Peroxide (H₂O₂) Oxidation State
Oxygen in H₂O₂: -1
Calculation:
In O-H bonds, oxygen attracts electrons due to higher electronegativity.
In O-O, electrons are shared equally.
Oxygen surrounded by 7 electrons instead of 6 (6 - 7 = -1).
Analysis of the Reaction
Oxidation:
Oxygen is oxidized from -1 in H₂O₂ to 0 in O₂.
Increase in oxidation state indicates oxidation.
Reduction:
Oxygen is reduced from -1 in H₂O₂ to -2 in H₂O.
Decrease in oxidation state indicates reduction.
Conclusion: The reaction is a disproportionation reaction, where oxygen is both oxidized and reduced.
Conclusion
Understanding oxidation states helps in analyzing redox reactions, especially disproportionation where a single substance can undergo both oxidation and reduction simultaneously.