What is the quantity of charge, in Faraday units, required for the reduction of 3.5 moles of Cr ₂ O 7²⁻ in acid medium?

Step-by-step solution

The reduction half-reaction for the dichromate ion in an acidic medium is given by: Cr ₂ O ₇²⁻ + 14 H ⁺ + 6 e ⁻ 2 Cr ³⁺ + 7 H ₂ O From the balanced equation, it is observed that 1 mole of Cr ₂ O ₇²⁻ requires 6 moles of electrons for its reduction. Since 1 mole of electrons corresponds to 1 Faraday of charge, 1 mole of Cr ₂ O ₇²⁻ requires 6 Faraday of charge. For 3.5 moles of Cr ₂ O ₇²⁻ , the total charge required is: Charge = 3.5 moles 6 Faraday/mole = 21.0 Faraday Answer: 21.0