In basic medium CrO oxidises S O to form SO and itself changes into Cr(OH) . The volume of 0.154 M C — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1: Determine oxidation state changes** - In CrO₄²⁻: Cr is +6 - In Cr(OH)₄⁻: Cr is +3 - Electrons gained by Cr: 6 - 3 = 3e⁻ per Cr - In SO₃²⁻: S is +4 - In SO₄²⁻: S is +6 - Electrons lost by S: 6 - 4 = 2e⁻ per S **Step 2: Balance electrons using least common multiple** - LCM(3, 2) = 6 - 2 CrO₄²⁻ gain 6e⁻ - 3 SO₃²⁻ lose 6e⁻ **Step 3: Write balanced equation** 2CrO₄²⁻ + 3SO₃²⁻ + 5H₂O → 2Cr(OH)₄⁻ + 3SO₄²⁻ Mole ratio: CrO₄²⁻ : SO₃²⁻ = 2 : 3 **Step 4: Calculate moles of SO₃²⁻** Moles of SO₃²⁻ = 0.25 M × 0.040 L = 0.010 mol **Step 5: Calculate moles of CrO₄²⁻ required** Using ratio 2:3: Moles of CrO₄²⁻ = 0.010 × (2/3) = 0.00667 mol **Step 6: Calculate volume of CrO₄²⁻** Volume = moles/molarity = 0.00667/0.154 = 0.0433 L = 43.3 mL *Note: If answer is 173 mL, verify the problem uses 3:2 ratio (3 CrO₄²⁻ : 2 SO₃²