What current in amperes will deposit 2.7g of aluminium in 2 hours?
[Al = 27, F = 96 500 C mol-1]
Al\(^{3+}\) + 3e\(^-\) → Al
3 x 96500 → 27g
I = 3600 x 2 → 27
I = (2.7/27) x ((3 x 96500)/(3600 x 2))
= 4.025 ≈ 4
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