How much charge must pass through an electrolytic cell to reduce 0.44 mol Ca²⁺ion to calcium metal?

Question

Quizplus · Accepted Answer

Calcium (Ca) has a +2 charge in its ionic form. To reduce 0.44 mol of Ca^2+ to calcium metal, each mole requires 2 moles of electrons (2 mol e-). The charge of one mole of electrons (1 mol e-) is approximately 96,485 C (Faraday's constant). Therefore, the total charge needed is 0.44 mol * 2 mol e-/mol Ca^2+ * 96,485 C/mol e- = 8.5 × 10^4 C.

How Much Charge Must Pass Through an Electrolytic Cell to Reduce