Chemistry, asked by patelsneh1607, 8 months ago

explain by calculation that what masses of the product will be liberated at electrode when nickel nitrate solution is electrode using 5ampere current for 3hours

Answers

Answered by s02371joshuaprince47

0

Answer:

Explanation:

=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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