1.

How much charge is required for the following reaction? (i) 1mol of Al^(3+) to Al. (ii) 1 mol of Cu^(2+) to Cu. (iii) 10 mole of MnO_(4)^(-) to Mn^(2+)

Answer»

Solution :(i) The electrode reaction is: `Al^(3+)+3e^(-)toAl`
`THEREFORE` QUANTITY of CHARGE required for reduction of 1 MOLE of `Al^(3+)=3F=3xx96500C=289500C`.
(ii) The electrode reaction is: `Cu^(2+)+2e^(-)toCu`
`therefore`Quantity of charge required for reduction of 1 mol of `Cu^(2+)=2` faradays`=2xx96500C=193000C` ltBrgt (iii) The electrode reaction is `MnO_(4)^(-)toMn^(2+)`, i.e., `Mn^(7+)5e^(-)toMn^(2+)`
`therefore`Quantity of charge required `=5F=5xx96500C=482500C`.


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