Question

How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of 0.998 A is applied?

Answer 1

Answer: The time required to deposit such amount of Mg is 886secs.

Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.

He also stated that;

96500C(1Farday) of electricity is required to deposit 1 mole of any metal.

For $Mg^2+$ +2e- ==>Mg

193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)

Which implies;

19300C will liberate 24g

xCwill liberate 0.110g

Where x is the amount of electricity to deposit 0.110g

x = (19300 × 0.110)/24

x = 884.6C

Recall that Q =It

Where Q is the quantity of electricity, I is the current and t is the time taken

884.6= 0.998 × t

t= 884.6/0.998

t= 886.3secs.

Therefore the time require is 886.3s.