Electrolysis is the decomposition of an electrolyte on passing current and it involves the migration of the ions of the electrolyte towards oppositely charged electrodes. Reduction occurs at cathode by the gain of electrons while oxidation at the anode by the loss of electrons. The electrical conductivity of an electrolyte increases upon dilution as well as with the increase in temperature. The nature of the products formed at the respective electrodes depends upon the nature of the electrodes as well as the nature of electrolyte whether in molten state or in aqueous solution. The mass of the substance deposited at a particular electrode is guided by the Faraday's first and second laws of electrolysis. A current of 12 amperes is passed through an electrolytic cell contianing aqueous NiSO_(4) solution. Both Ni and H_(2) are formed at cathode. The current efficiency is 60%. What is the mass of nickel deposited on the cathode per hour ? (Atomic mass of Ni=58.7).

Electrolysis is the decomposition of an electrolyte on passing current

1.	Electrolysis is the decomposition of an electrolyte on passing current and it involves the migration of the ions of the electrolyte towards oppositely charged electrodes. Reduction occurs at cathode by the gain of electrons while oxidation at the anode by the loss of electrons. The electrical conductivity of an electrolyte increases upon dilution as well as with the increase in temperature. The nature of the products formed at the respective electrodes depends upon the nature of the electrodes as well as the nature of electrolyte whether in molten state or in aqueous solution. The mass of the substance deposited at a particular electrode is guided by the Faraday's first and second laws of electrolysis. A current of 12 amperes is passed through an electrolytic cell contianing aqueous NiSO_(4) solution. Both Ni and H_(2) are formed at cathode. The current efficiency is 60%. What is the mass of nickel deposited on the cathode per hour ? (Atomic mass of Ni=58.7).
Answer» 7.883 G 3.941 g 5.91 g 2.645 g Solution :(a) Quantity of charge (Q) passed `=(12 amp)xx(60xx60 s)=43200" C"` `Ni^(2+)(AQ) +underset(2F)2e^(-) to underset(58.7 g)(Ni(s))` `2xx96500" C"` of charge deposit Ni =58.7 g 43200 C of charge deposit Ni `=((58.7 g))/((2xx96500 C))xx(43200C)=13.139 g` Mass of Ni actually DEPOSITED `=((13.139 g)xx60)/(100)=7.883" g"`.