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This lesson covers oxidation states, identifying redox reactions, writing and balancing half-equations, combining half-equations, and redox titrations.
Oxidation states (oxidation numbers) are assigned using these rules:
Example 1: Find the oxidation state of Mn in KMnO₄.
K = +1, each O = −2, total = 0 +1 + Mn + 4(−2) = 0 → Mn = +7
Example 2: Find the oxidation state of S in SO₄²⁻.
S + 4(−2) = −2 → S = +6
Oxidation is the loss of electrons (increase in oxidation state). Reduction is the gain of electrons (decrease in oxidation state). Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
An oxidising agent (oxidant) causes oxidation by accepting electrons — it is itself reduced. A reducing agent (reductant) causes reduction by donating electrons — it is itself oxidised.
Steps for writing half-equations in acidic solution:
Example 3: Write the half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution.
Step 1: MnO₄⁻ → Mn²⁺ Step 2: Mn is already balanced. Step 3: Add 4H₂O to balance O: MnO₄⁻ → Mn²⁺ + 4H₂O Step 4: Add 8H⁺ to balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Step 5: Left charge = −1 + 8 = +7. Right charge = +2. Add 5e⁻ to the left:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Mn goes from +7 to +2 (gain of 5 electrons) — this is reduction.
To combine two half-equations into a full redox equation:
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