What quantity of charge is required to cause reduction of 0.20 mole of Cr³⁺ to Cr?

Question

Quizplus · Accepted Answer

The reduction of Cr^3+ to Cr requires 3 moles of electrons per mole of Cr^3+. The charge of one mole of electrons (Faraday's constant) is approximately 96,485 C (Coulombs). Therefore, for 0.20 mole of Cr^3+, the required charge is 0.20 mole * 3 moles of electrons/mole of Cr^3+ * 96,485 C/mole of electrons = 57,891 C, which is approximately 5.8 × 10^4 C.

What Quantity of Charge Is Required to Cause Reduction of 0.20