Answer: 530 hours
Explanation:
The reduction of Nickel ions to nickel is shown as:
[tex]Ni^{2+}+2e^-\rightarrow Ni[/tex]
[tex]96500\times 2=193000Coloumb[/tex] of electricity deposits 1 mole of Nickel
1 mole of Nickel weighs = 58.7 g
Given quantity = 18.0 kg = 18000 g (1kg=1000g)
58.7 g of Nickel is deposited by 193000 C of electricity
18000 g of Nickel is deposited by = [tex]\frac{193000}{58.7}\times 18000=59182282.8C[/tex] of electricity
[tex]Q=I\times t[/tex]
where Q= quantity of electricity in coloumbs = 59182282.8C
I = current in amperes = 31.0 A
t= time in seconds = ?
[tex] 59182282.8C=31.0A\times t[/tex]
[tex]t=1909105.9sec[/tex]
(1h=3600 sec)
[tex]t=530hours[/tex]
Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A