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]
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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 / 2x96500 x 63.5 = 0.3175g \(\approx\) 0.328
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 / 2x96500 x 63.5 = 0.3175g \(\approx\) 0.328