Question

Calculate the number of coulombs required to deposit 5.4 g of Al when the electrode reaction is Al₃ + 3e^– + Al.
(Given: Atomic mass of Al = 27 g mol⁻¹, F = 96500 C mol⁻¹)

Answer 1

\[\ce{\underset{(1 mole 27g)}{Al^3+} + \underset{3F}{3e^-} ->{Al}}\]

To deposite 27 g of Al, electricity required = 3 × 96500

= 289500 C

and to deposit 5.4 g of Al, electricity required = `289500/27 xx 5.4`

= 57900 C