How many seconds are required to deposit 0.110 grams of magnesium metal?

Explanation:

Faraday Law of Electrolysis: According to Faraday's Law, the mass of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the solution.

Faraday stated that 96500 Coulombs (1 Faraday) of electricity is required to deposit 1 mole of any metal. For the reaction: Mg2+ + 2e- ⟶ Mg, 19300 Coulombs (2 Faraday) of electricity is required to deposit 1 mole of Magnesium metal (1 mole of Mg = 24g).

This implies that 19300 Coulombs will liberate 24g and x Coulombs will liberate 0.110g of magnesium. We can calculate the value of x as follows:

x = (19300 × 0.110) / 24 x = 884.6 Coulombs

Using the formula Q = It, where Q is the quantity of electricity, I is the current, and t is the time taken, we can determine the time required to deposit 0.110 grams of magnesium:

884.6 = 0.998 × t t = 884.6 / 0.998 t ≈ 886.3 seconds

Therefore, the time required to deposit 0.110 grams of magnesium metal from the given solution with a current of 0.998 A applied is approximately 886.3 seconds.
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