Question

A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2 M solution of `Ni(NO_(3))_(2)`. What will be the molarity of the solution at the end of electrolysis ? What will be the molarity of the solution if nickel electrodes are used ? (1 F=96500 coulomb, Ni=58.7)

Answer 1

The electrode reaction is :
`underset("1 mole")(Ni^(2+))+underset(2xx 96500 C)(2e^(-)) rarr Ni`
Quantity of electric charge passed
`=3.7xx6xx60xx60` coulomb `=79920` coulomb
Number of moles of `Ni(NO_(3))_(2)` decomposed or nickel deposited
`=1/(2xx96500)xx79920=0.4140`
Number of moles of `Ni(NO_(3))_(2)` present before electrolysis
`=0.5xx2=1.0`
Number of moles of `Ni(NO_(3))_(2)` present after electrolysis
`=(1.0-0.4140)=0.586`
Since, 0.586 moles are present in 0.5 litre,
Molarity of the solution `=2xx0.586=1.72 M`
When nickel electrodes are used anodic nickel will dissolve and get deposited at the cathode. The molarity of the solution will, thus, remain unaffected.