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 ?

A solution of `Ni(NO_(3))_(2)` is electrolyzed between platium electro

1.	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 ?
Answer» `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"`

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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 ?

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