How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten `CaCl_(2)` ?
(ii) 40.0 g of Al from molten `Al_(2) O_(3)` ?

Explanation

(i) Electrode reaction `: Ca^(2) ( aq) + underset( 2F)( 2e^(-))rarr underset( 40 g ) ( Ca(s))`
To produce 40g of Ca from molten `CaCl_(2)`, electricity required = 2F
To produce 20 g Ca from `CaCl_(2)`, electricity required = 1F
(ii) Electrode reaction `:`
`Al_(2) O_(3) ( l ) + underset( 6F) ( 6e^(-)) rarr underset(( 2 xx 27))(2Al) + 3O^(2)`
= 54 g
To produce 54 g Al from molten `Al_(2) O_(3)`, electricty required = 6F
To produce 40 g Al from molten `Al_(2) O_(3)`, electricity required `= (( 6F ) xx ( 40 g))/( ( 54g)) = 4.44F`