What charge,in coulombs,is required to deposit 1.5 g Mg(s)from a solution of Mg²⁺(aq)? (F = 96480 C/mol e−)

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Quizplus · Accepted Answer

To deposit 1.5 g of Mg, calculate moles of Mg: 1.5 g / 24.305 g/mol = 0.0617 mol. Since Mg^2+ requires 2 moles of electrons per mole of Mg, the total moles of electrons needed is 0.0617 mol × 2 = 0.1234 mol e−. The charge is then 0.1234 mol e− × 96480 C/mol e− = 1.2 × 10^4 C.

What Charge,in Coulombs,is Required to Deposit 1