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How much electricity in terms of Faraday is required to produce:

(i) 20.0 g of Ca from molten CaCl₂?
(ii) 40.0 g of Al from molten Al₂O₃?

1 Answer

1 vote

Final answer:

To produce 20.0 g of calcium from CaCl₂, 0.998 Faraday of electricity is needed. To produce 40.0 g of aluminum from Al₂O₃, 4.446 Faraday of electricity is required. These calculations are based on Faraday's laws of electrolysis and the valency of each metal.

Step-by-step explanation:

The question involves calculating the amount of electricity, in terms of Faraday, needed to produce a certain mass of an element through the process of electrolysis. Faraday's laws of electrolysis states that the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

Calcium (Ca) Production:

First, we will calculate the amount of electricity needed to produce 20.0 g of calcium (Ca) from molten CaCl2. The molar mass of Ca is approximately 40.08 g/mol.

20.0 g Ca × (1 mol Ca/40.08 g Ca) = 0.499 mol Ca

Since the valency of Ca is +2, it will require 2 Faraday of charge to discharge one mole of Ca according to Faraday’s laws of electrolysis. Therefore:

0.499 mol Ca × 2 Faraday/mol Ca = 0.998 Faraday

Aluminum (Al) Production:

Next, we will calculate the amount of electricity needed to produce 40.0 g of aluminum (Al) from molten Al2O3. The molar mass of Al is approximately 26.98 g/mol.

40.0 g Al × (1 mol Al/26.98 g Al) = 1.482 mol Al

Aluminum has a valency of +3, so 3 Faraday of charge is needed to discharge one mole of Al atoms. Hence:

1.482 mol Al × 3 Faraday/mol Al = 4.446 Faraday

answered
User Teoman Kirac
by
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