Dichromate ion is treated with base, the oxidation number of Cr in the product formed is — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the dichromate ion.** The dichromate ion is Cr₂O₇²⁻, where chromium's oxidation state needs to be determined. **Step 2: Determine initial oxidation state in dichromate.** Using the oxidation state rules: - Oxygen = -2 - Let Cr = x - Equation: 2x + 7(-2) = -2 - 2x - 14 = -2 - 2x = 12 - x = +6 **Step 3: Identify the reaction with base.** When dichromate ion (Cr₂O₇²⁻) is treated with base (OH⁻), it converts to chromate ion (CrO₄²⁻): Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O **Step 4: Determine oxidation state in chromate product.** In chromate ion CrO₄²⁻: - Oxygen = -2 - Let Cr = y - Equation: y + 4(-2) = -2 - y - 8 = -2 - y = +6 **Step 5: Verify no redox occurred.** The oxidation number of Cr remains +6 in both dichromate and chromate ions. This is an acid-base equilibrium, not a redox reaction. Therefore, the answer is 6.00.