1.

How many grams of nickel are deposited by a current of 100 milliampere in 20 minutes in the electrolysis of aqueous nickel sulphate solution ? (Given atomic mass of Ni =58.7 amu.)

Answer»


Solution :The reaction at the cathode is :
`underset(1MOL)Ni^(2+)(aq)+underset((2 Faradays))underset(2mol)(2e^(-))to underset((58.7 G))underset(1mol)(Ni(s))`
Quantity of charge PASSED (Q)`="Current in ampere" xx"time in seconds "`
`=(100//1000 amp.)xx(20xx60 s)=120 As =120 C`
`2xx96500 C`of charge deposit Ni=58.7 g
120 C of charge Ni`=((58.7 g))/((2xx96500C))xx(120C)=0.0365 g`.


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