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

It take 0.54 hours to deposit 6.36g of copper

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

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation: