How many Faraday of electricity are required to deposit 0.20 mole of nickel, if 0.10 Faraday of electricity deposited 2.98g of nickel during electrolysis of its aqueous solution?
(Ni = 58.7, 1F = 96 500 C mol–)
- 0.20
- 0.30
-
0.40
- 0.50
The correct answer is: C
Explanation
No of mole → (2.98)/(58.7) ..............0.12.0 .................x
∴ x = (0.2)/(2.98) x 58.7 x 0.40