Question

Question 4: Electrochemical Cells (9 points)

Answer 1

a. What is the standard potential, in volts, of a galvanic cell made of Ce4+ and Cu2+, given the standard reduction potentials below? (3 points)
Ce4+ + e− → Ce3+ Eº = 1.61 V
Cu2+ + 2e− → Cu(s) Eº = 0.34 V
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) (3 points)
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? (3 points)

Answer 2

@Hoslos Let's start with a)

Answer 3

Are their potential both positive?

Answer 4

Yes I think so.

Answer 5

It must be cause it is metal also. From what I believe...
We have to swap the second one, so that we get opposite charged potential, to get attraction and further reaction. Also, the ratio of their electrons is 1:2 from Cs to Cu, so the potential of Cs becomes 1.61*2=3.22 and for Cu, -0.34. The resultant is then 2.88.

Answer 6

ok thanks! now b)
@Hoslos

Answer 7

That was b). You meant c).

Answer 8

My bad. It is b) now. Hehe.

Answer 9

@Hoslos just b) and c) and then we are done! :)

Answer 10

b) the formula for Faraday is F=q/n where q is charge and n is the number of moles. Then for Cu will be F= 0.34/(10.5/63.55)= 2.1C/mol.

Answer 11

thanks! last one! c)
:)

Answer 12

The conversion is that 1 F or faraday = 96500A, then 2.1=x then x= 2.1*96500/1= 202650A.
Using the formula Q=It, where I is current, t is time and Q is charge, then making t as the subjct is t= Q/I or 202650/2= 101325seconds.

Answer 13

thanks so much! you're amazing! :)

Answer 14

Heh, thanks.