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Chapter 21: Problem 44

With the Hall process, how many hours will it take to deposit \(664 \mathrm{~g}\) of \(\mathrm{Al}\) at a current of \(32.6 \mathrm{~A} ?\)

Short Answer

Expert verified
It will take approximately 60.52 hours to deposit 664 grams of Aluminium using the Hall process at a current of 32.6 amperes.

Step by step solution

01

Identifying constants

In the question we have been given that, the amount of Aluminium (Al) deposited is 664g and Ionisation charge of Al is 3, where each charge represents one electron absorbed. The current supplied is 32.6A and we know, 1 mole of electrons carries approximately 96485 coulombs of charge, a value known as Faraday's constant.
02

Calculate moles of Al

Firstly, calculate the moles of Al in 664g. Aluminium has a molar mass of approximately 26.98 g/mol. Therefore, moles of Al = \frac{664g}{26.98g/mol} = 24.61 mol.
03

Using Faradays First Law

We know that from Faradays first law, charge Q consumed is equal to the ionisation charge (z) of Al multiplied by the number of moles (n) and Faraday's constant (F). So, Q= n*z*F = 24.61 mol * 3 * 96485 = 7104424.5 coulombs.
04

Time Calculation

We know that charge Q = It (current * time). Therefore we can calculate time t, by rearranging the equation and finding t = \frac{Q}{I} = \frac{7104424.5 coulombs}{32.6A} = 217890.4 seconds. We can convert this to hours by dividing by 3600: \frac{217890.4s}{3600} = 60.52 hours.

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