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–)
(Ni = 58.7, 1F = 96 500 C mol–)
-
A.
0.20 -
B.
0.30 -
C.
0.40 -
D.
0.50
Correct Answer: Option C
Explanation
No of mole → (2.98)/(58.7) …………..0.1
2.0 ……………..x
∴ x = (0.2)/(2.98) x 58.7 x 0.40