The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and — Hydrogen Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the reactants and their oxidation states** - In KMnO₄: Mn has oxidation state +7 - In H₂O₂: O has oxidation state -1 **Step 2: Determine what happens in basic medium** In basic conditions, H₂O₂ acts as a reducing agent. Mn⁷⁺ is reduced and H₂O₂ is oxidized to O₂. **Step 3: Write the reduction half-reaction** MnO₄⁻ + 4e⁻ + 4H₂O → MnO₂ + 2OH⁻ This shows Mn⁷⁺ gains 4 electrons. **Step 4: Write the oxidation half-reaction** H₂O₂ → O₂ + 2H⁺ + 2e⁻ **Step 5: Balance electrons and combine** Multiply the oxidation half-reaction by 2: 2H₂O₂ → O₂ + 2H⁺ + 4e⁻ The electrons match, showing the reduction is consistent. **Step 6: Identify the product and manganese's oxidation state** The main reduction product is MnO₂, where manganese is bonded to 2 oxygen atoms (each -2). Oxidation state of Mn in MnO₂: x + 2(-2) = 0, so x = +4 Therefore, the answer is 4.00.