Question

The current required to deposit 12 g of magnesium in two hours

Answer 1

Answer:

The half reaction occurring during deposition is:

Mg2+ + 2e- ----------> Mg(s)

So to produce 1 mole of Mg we need 2 moles of electrons,

1 mol of Mg = 24 g

Thus 1.2 g of Mg = (1.2/24) = 0.05 moles

Hence no. of moles of electrons required to deposit = 0.05 x 2 = 0.1 moles

Qty. of electricity passed (Q) = moles of electrons passed x 96500

= 0.1 x 96500 = 9650 C

I = Q/t

I = 9650/ 60 x 60 (converting 1 hr into seconds)

I = 9650/3600 = 2.68 A.

Explanation:

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