The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is
-
13.5
- 4.5
- 7.5
- 1.5
The correct answer is: 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
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