Question
A. What is the standard potential, in volts, of a galvanic cell made of Ce4+ and Cu2+, given the standard reduction potentials below?
Ce4+ + e− → Ce3+ Eº = 1.61 V
Cu2+ + 2e− → Cu(s) Eº = 0.34 V
I think it is 1.27 V but it might be 1.95V
b. How many faradays are needed to deposit 10.5 g of copper onto the surface of an electrode from a solution of Cu2+? (Molar mass of copper = 63.55 g/mol)
I have no idea how to solve this
c. How long, in seconds, would it take to deposit 10.5 g of copper onto an electrode from a solution of Cu2+ if a current of 2.00 A was applied?
I don't know how to solve this either
Ce4+ + e− → Ce3+ Eº = 1.61 V
Cu2+ + 2e− → Cu(s) Eº = 0.34 V
I think it is 1.27 V but it might be 1.95V
b. How many faradays are needed to deposit 10.5 g of copper onto the surface of an electrode from a solution of Cu2+? (Molar mass of copper = 63.55 g/mol)
I have no idea how to solve this
c. How long, in seconds, would it take to deposit 10.5 g of copper onto an electrode from a solution of Cu2+ if a current of 2.00 A was applied?
I don't know how to solve this either
Answers
Ce4+ + e− → Ce3+ Eº = 1.61 V
Cu(s) ==> Cu^2+ + 2e Eº = -0.34 V
Ecell = 1.61 + (-0.34) = 1.27 v
You need 96,485 coulombs to deposit 1/2 gram molar mass Cu or 63. 55/2 = 31.78 g Cu. So you need 96,485 x (10.5/31.78) = ? coulombs. Convert that to Faradays. 1 F = 96,485 C.
How long?
C = amperes x seconds.
Solve for seconds.
Cu(s) ==> Cu^2+ + 2e Eº = -0.34 V
Ecell = 1.61 + (-0.34) = 1.27 v
You need 96,485 coulombs to deposit 1/2 gram molar mass Cu or 63. 55/2 = 31.78 g Cu. So you need 96,485 x (10.5/31.78) = ? coulombs. Convert that to Faradays. 1 F = 96,485 C.
How long?
C = amperes x seconds.
Solve for seconds.
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