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Question

A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2M solution of Ni(NO3)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? (1F=96500 coulomb;Ni=58.7)

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Solution

The electrode reaction is:
Ni2+1 mole+2e2×96500 CNi
Quantity of electric charge passed
=3.7×6×60×60 coulomb=79920 coulomb
Number of moles of Ni(NO3)2 decomposed or nickel deposited
=12×96500×79920=0.4140
Number of moles of Ni(NO3)2 present before electrolysis
=0.5×2=1.0
Number of moles of Ni(NO3)2 present after electrolysis
=(1.00.4140)=0.586
Since, 0.586 moles are present in 0.5 litre,
Molarity of the solution =2×0.586=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.

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