In the titration of KMnO and oxalic acid in acidic medium, the change in oxidation number of carbon — Redox Reactions and Volumetric Analysis Chemistry Question
Question
In the titration of KMnO and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is_____ 4
💡 Solution & Explanation
**Step 1: Identify the oxidation state change of carbon in oxalic acid.** Oxalic acid (H₂C₂O₄) contains carbon with oxidation state +3. When oxidized, carbon goes to +4 (in CO₂). Change per carbon atom: +4 − (+3) = +1 **Step 2: Determine the oxidation state change of manganese.** In KMnO₄, Mn is +7. In the acidic medium product (Mn²⁺), Mn is +2. Change per Mn atom: +7 − (+2) = +5 (gains 5 electrons) **Step 3: Write the balanced redox equation.** MnO₄⁻ + 5e⁻ → Mn²⁺ (reduction) C₂O₄²⁻ → 2CO₂ + 2e⁻ (oxidation) **Step 4: Balance electrons to find the mole ratio.** Multiply first equation by 2 and second by 5: - 2MnO₄⁻ gains 10 electrons - 5C₂O₄²⁻ (containing 10 carbon atoms) loses 10 electrons **Step 5: Calculate average change per carbon atom.** Total change = 10 carbons × (+1 per carbon) = +10 Number of carbon atoms involved = 10 Average change per carbon = +10/10 = **+1** Therefore, the answer is 1.