A solution of `Ni(NO_(3))_(2)` is electrolyzed between platium electrodes using a current of `5A` for `20 mi n`. What mass of `Ni` is deposited at the cathode ?

Explanation

`Ni(NO_(3))_(2)+2H^(+)+underset(2" mol")(2e^(-)) to underset(1" mol")(Ni)+2HNO_(3)`
The charge, Q on n moles of elctrons is given by Q=nF
Thus, charge required to deposit 1 mole of nickel,
`Q=2" mol"xx96500" C mol"^(-1)=193000" C"=193xx10^(5)" C"`
Quantity of electricity used `="Current in amperes"xx"Time in seconds"` ,
`=5Axx(20xx60)s=5xx20xx60" As"=600" C" (`:.`"A s"=C)`
Molar mass of nickel =58.7 g `mol^(-1)`
`1.93xx10^(5)C`of charge produce nickel `=1" mol"=1 " mol"xx58.7"g mol"^(-1)`=58.7 g
`:. 6000" C of charge produce nickel" =((58.7g))/((1.93xx10^(5)C))xx(6000" C")=1.825" g"`