Question

How long must a 40.0 amp current flow through a solution of iron(III) chloride in order to produce 5.00 moles of iron?

Answer 1

Answer:

10.1 h

Explanation:

Let's consider the reduction half-reaction of iron from an aqueous solution of iron (III) chloride.

Fe³⁺(aq) + 3 e⁻ ⇒ Fe(s)

We can calculate the time required to produce 5.00 moles of Fe using the following relationships.

• 1 mole of Fe is produced when 3 moles of electrons circulate.

• 1 mole of electrons has a charge of 96486 C (Faraday's constant).

• 1 A = 1 C/s.

• 1 h = 3600 s.

$5.00molFe \times \frac{3 mole^{-} }{1molFe} \times\frac{96486C}{1mole^{-} } \times \frac{1s}{40.0C} \times \frac{1h}{3600s} = 10.1 h$