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How much charge is required for the foll...

How much charge is required for the following reductions :
(i) 1 mol of `Al^(3+)` to Al?
(ii) 1 mol of `Cu^(2+) to Cu` ?
(iii) 1 mol of `MnO_(4)^(-)` to `Mn^(2+)` ?

Text Solution

Verified by Experts

(i) The electrode reaction is represented as:
`Al^(3+) + 3e^(-) to Al`
`therefore` Quantity fo charge required for reduction of 1 mol of `Al^(3+)`
=`3 F = 3 xx 96500 C = 289500 C`
(ii) The electrode reaction is represented by
`Cu^(2+) + 2e^(-) to Cu`
`therefore` Quantity of charge required for reduction of 1 mol of `Cu^(2+)`
`=2F= 2 xx 96500 C = 193000 C`
(iii) The electrode reaction is represented as:
`=5F = 5 xx 96500 C = 482500 C`
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