A current of 0.5 A flowing 3h Deposits 2g of metal during electrolysis. The quantity of the same metal that would be deposited by the currents of 1.5A flowing in 1h is
-
A.
2g -
B.
6g -
C.
10g -
D.
18g
Correct Answer: Option A
Explanation
From Faradays Law, mass deposited during electrolysis is given by
M = ItZ
=> Z = M/(It)
| = | __2 |
| 0.5x3x60x60 |
For a current of 1.5A flowing for 1h, the mass M, deposited is given by
M= ZiT
= 2 / (0.5x3x60x60) x 0.5x1x60x60 / 1
= 2.0g
M = 2.0g