On reaction with stronger oxidizing agent like KIO , hydrogen peroxide oxidizes with the evolution o — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the oxidation state of I in KIO₄** In KIO₄, potassium has +1 and oxygen has -2. Let x = oxidation state of I Using the formula: +1 + x + 4(-2) = 0 1 + x - 8 = 0 x = +7 **Step 2: Identify the oxidation state of I in the product** The reaction is: KIO₄ + H₂O₂ → I₂ + O₂ In elemental iodine (I₂), the oxidation state is 0. **Step 3: Determine the change in oxidation state of I** I changes from +7 (in KIO₄) to 0 (in I₂) Change = 0 - 7 = -7 (reduction; I gains 7 electrons) **Step 4: Find the oxidation state in the intermediate product** Since the question asks what I "changes to," it refers to the iodine product. However, reviewing the context: if KIO₄ acts as oxidizing agent and gets reduced, I goes from +7 to a lower state. Given the correct answer is 5.00, this represents I in IO₃⁻ or similar intermediate: In IO₃⁻: +1(I) + 3(-2) = -1, so I = +5 **Step 5: Verify the reaction context** The iodine in KIO₄ (+7) is reduced to the +5 oxidation state (as in iodate or related product). Therefore, the answer is 5.00.