Question

How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI solution. (Cu = 63.5,1F=96500C)
Cu2+ + 2e- Cu

Answer 1

It take 0.54 hours to deposit 6.36g of copper

Further explanation

Faraday's Law I

"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis

W = e.i.t / 96500

$\tt \large {\boxed {\bold {W \: = \: \dfrac {e \times i \: x \: t} {96500}}}}$

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

$\tt t=\dfrac{W\times 96500}{e.i}\\\\t=\dfrac{6.36\times 96500}{31.75\times 10 }\\\\t=1933.04~s\approx 0.54~hours$