Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?

Short Answer

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(a) An aqueous solution of \(\mathrm{MgCl}_{2}\) is not used in the electrolysis because H+ ions from the dissociation of water tend to get discharged at the cathode more readily than Mg2+ ions, resulting in the production of hydrogen gas instead of magnesium metal. (b) The mass of magnesium formed during the electrolysis process is approximately 1,029,430 g or 1,029.43 kg when passing a current of 97,000 A for a period of 24 hours with a cell efficiency of 96 %.

Step by step solution

01

Part (a): Explaining the unsuitability of aqueous MgCl2

Aqueous solutions of MgCl2 contain Mg2+ ions and Cl- ions along with the H+ ions from the dissociation of water. On passing current through the aqueous solution, H+ ions from water tend to get discharged at the cathode more readily than Mg2+ ions because of their low ionization potential. Therefore, using an aqueous solution would result in the production of hydrogen gas instead of magnesium metal.
02

Part (b): Calculating the charge passed

Given, the current is 97,000 A, and the time is 24 h. To calculate the total charge (Q) that passed through the cell, we can use the formula: Q = I × t But first, we need to convert 24 h into seconds: 1 h = 3600 s So, 24 h = 24 × 3600 s Now, we can find the charge: \(Q = 97,000 \: A \times 24 \times 3600 \: s = 8,467,200,000 \: C\)
03

Part (b): Calculating the moles of Mg formed

We will use Faraday's law of electrolysis to find the moles of magnesium formed: Moles of Mg = (Charge passed × Efficiency) / (n × Faraday's constant) Here, the efficiency of the cells is given as 0.96, and the number of electrons (n) required to reduce one magnesium ion (Mg2+) to magnesium metal is 2. The Faraday's constant (F) is 96,485 C/mol. Let's calculate the moles of Mg formed: Moles of Mg = \((8,467,200,000 \: C \times 0.96) / (2 \times 96,485 \: C/mol) = 42,372 \: mol\)
04

Part (b): Finding the mass of Mg formed

Now that we have the moles of Mg formed, we can find the mass of Mg using the molar mass of magnesium (24.305 g/mol): Mass of Mg = Moles of Mg × Molar mass of Mg Mass of Mg = \(42,372 \: mol \times 24.305 \: g/mol = 1,029,430 \: g\) So, the mass of magnesium formed during the electrolysis process is approximately 1,029,430 g or 1,029.43 kg.

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