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The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is

Chemistry
JAMB 2024

The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is

  • A. 13.5
  • B. 4.5
  • C. 7.5
  • D. 1.5
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Correct Answer: Option A
Explanation

Firstly, Al\(^{3+}\) + 3e\(^-\)  →  Al

1 mole of Al =  1F X  3e\(^-\) 

   1mole   =   3F

  4.5moles =  X

X = 4.5 X 3 = 13.5 F

There is an explanation video available below.


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Explanation Video

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