What mass of copper would be deposited by a current of 1. 0 ampere passing for 965 seconds through copper (ll) tetraoxosulphate (IV) solution? [Cu = 63.5; 1F = 96500C]
-
A.
0.318g -
B.
0.635g -
C.
18g -
D.
6.35g -
E.
9.65g
Correct Answer: Option A
Explanation
Cu2+ + 2e- → Cu, From the equation 2 Faraday will liberate 63.5gm of copper The quantity of electricity passed (Q) Q = 1. 0 965 = 965 coulomb :. 2 x 96500 → 63.5, 965 → x. x = 965 / 2×96500 x 63.5 = 0.3175g (approx) 0.328