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
-
2g
- 6g
- 10g
- 18g
The correct answer is: A
Explanation
From Faraday's Law, mass deposited during electrolysis is given by
\(M_1 = I_1t_1Z\)
=> Z = \(\frac{M_1}{(I_1t_1)}\)
⇒ Z =\(\frac{2}{0.5 \times3}\) = 1.333Kg/C
For a current of 1.5A flowing for 1h, the mass M, deposited is given by
\(M_2 = ZI_2T_2\) = 1.333 x 1 x 1.5 = 2.0g
\(M_2\) = 2.0g
note: converting time to seconds here is not necessary.