Asked by Joey
A current of 156A is applied to an electrolytic cell containing molten CuCl2 for 45 minutes. What substance and how many grams of it will form on the cathode?
Answers
Answered by
DrBob222
Cu^2+ + 2e ==> Cu...........This is gain of electrons; therefore, reduction.
2Cl^- ==> Cl2 + 2e............This is a loss of electrons; therefore, oxidation.
Oxidation occurs at the anode; reduction occurs at the cathod. Therefore, we want the Cu half reaction.
How many coulombs do we generate? That's
Coulombs = amperes x seconds = 146 A x 45 min x (60 sec/min) = 394,200.
1 Faraday or 96,485 coulombs will deposit 63.54 g/2 = 31.77 g Cu metal.
So you can deposit 31.77 g Cu x (394,200/96,485) = ? g Cu
2Cl^- ==> Cl2 + 2e............This is a loss of electrons; therefore, oxidation.
Oxidation occurs at the anode; reduction occurs at the cathod. Therefore, we want the Cu half reaction.
How many coulombs do we generate? That's
Coulombs = amperes x seconds = 146 A x 45 min x (60 sec/min) = 394,200.
1 Faraday or 96,485 coulombs will deposit 63.54 g/2 = 31.77 g Cu metal.
So you can deposit 31.77 g Cu x (394,200/96,485) = ? g Cu
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