JAMB - Chemistry (2024 - No. 96)
The amount of Faraday required to discharge 4.5 moles of Al\(^{3+}\) is
13.5
4.5
7.5
1.5
Explicaţie
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
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