1.

To obtain 5.85 gm nickel how long 10 ampere current should be passed through dilute solution of NiSO_(4) during electrolysis by inert electrodes ? (atomic weight of Ni=58.5 gm)

Answer»

965 second
3860 second
1930 second
9650 second

Solution :`5.85gm` nickel `=(5.85gm)/(58.5" gm mole"^(-1))`
=0.1 mole Ni is obtained
Reaction : `Ni_((aq))^(2+)+2e^(-) to Ni_((S))`
2 mole `e^(-) to 1` mole Ni
`therefore 2F to 1` mole Ni
So, if 0.1 MOL Ni is obtained then,
`=(0.1" mole "Nixx2F)/(1" mole "Ni)`
=0.2 F electricity
=0.2`xx96500` coulomb
Time=`("coulomb")/("AMPERE")=(0.2xx96500)/(10)`
=1930 second.


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