Question

5. 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)​

Answer 1

ANSWER

Given

I=5A

Time=20×60=1200s

∴Charge=current×time

=5×1200

=6000C

According to the reaction.

Ni

2+

(aq.)+2e→Ni(s)

Nickeldepositeby(2×96487)C=58.7g

∴Nickeldepositeby 6000C=

2×96487

58.7×6000

=1.825g

∴ Hence 1.825g of nickel will be deposited at the cathode.

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