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 ≈ 0.328