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Question
Calculate the number of coulombs required to deposit 5.4 g of Al when the electrode reaction
is: Al3 + 3e– + Al [Atomic Weight of Al 27 g/mol].
Solution
\[\ce{\underset{\text{(1 mole 27g)}}{Al^3^+} + \underset{\text{3F}}{3e^-} ->{Al}}\]
To deposite 27 g of Al, electricity required = `3 xx 96500 = 289500` C
and to deposit 5.4 g of Al, electricity required = `289500/27 xx 5.4`
= 57900 C
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