The quantity of electricity in Faraday needed to reduce 1 mol of Cr O to Cr is _______. 2 72– 3+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the redox reaction** Cr₂O₇²⁻ is reduced to Cr³⁺. Write the half-reaction: Cr₂O₇²⁻ → 2Cr³⁺ **Step 2: Determine the oxidation state changes** - In Cr₂O₇²⁻: Cr is +6 - In Cr³⁺: Cr is +3 - Each Cr gains 3 electrons: +6 → +3 **Step 3: Calculate total electrons transferred** Since there are 2 Cr atoms in Cr₂O₇²⁻: - Total electrons gained = 2 × 3 = 6 electrons per mole of Cr₂O₇²⁻ **Step 4: Apply Faraday's law** The key concept: 1 mole of electrons = 1 Faraday of charge Quantity of electricity (in Faraday) = moles of electrons transferred **Step 5: Calculate for 1 mol of Cr₂O₇²⁻** For 1 mol of Cr₂O₇²⁻ reduced: - Electrons transferred = 6 mol - Charge needed = 6 Faraday Therefore, the answer is 6.00.