JAMB - Physics (2002 - No. 45)
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
Explication
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.