Explain by calculation that what masses of the products will be liberated at electrodes when
nickel nitrate solution is electrolysed using 5 ampere current for 3 hours.
(Atomic mass: Ni-58 g mol-1, O-16 g mol-1)
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Answer:
Explanation:
Given
=5A
Time=3×60×60=10800s [As the solution is electrolyzed for 3 hrs]
∴Charge=current×time
=5×10800
=54000C
According to the reaction.
Ni 2+
(aq.)+2e ---> Ni(s)
Nickel deposited by(2×96368)C=58.7g [Two moles of electrones are added]
∴Nickel deposited by 54000C=
(2×96368
)/(58.7×54000)
=0.04376g
∴ Hence 0.04376g of nickel will be deposited at the cathode
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