suppose 250 mL of a .433 M solution of CuCl2 is electrolyzed. how long will a current of 75A have to run in order to reduce the concentration of Cu2+ to .167M? What mass of Cu will be deposited on the cathode during this time?

Answer 1

You have 0.250L x 0.433M = about 0.1 mol
You want 0.250L x 0.167 = about 0.04 mol.
You want to deposit about 0.1-0.04 = about 0.06 mols Cu. g Cu = about 0.06 x 63.546 = about 4.2 g Cu

96,485 C will deposit 63.546/2 or about 31 g Cu. How many C do we need?
96,485 x 4.2/31 = about 13,000 C.
Amp x sec = C
75 x sec = 13,000
sec = about ?
You need to go through and do each step more accurately. I came up with about 171 seconds.

Answer 2

the time is supposed to be 4.6 hours. i am not having problems getting the mass of Cu, i am just struggling with how to find how long it will take.

Answer 3

Find coulombs, then amp x seconds = coulombs. How many grams Cu do you have that's deposited.

Answer 4

i get 4.207 g Cu

Answer 5

My number for g Cu = 4.226 g which I would round to 4.23 g.
(0.433-0.167)/4 x (63.546) = 4.2258 = 4.23g.

Answer 6

the book rounds the answer to 4.2 so either is fine. how do i go from here to solve for the amount of time? if you could please show a step by step process id really appreciate it!

Answer 7

I don't get 4.6 hours.

96,485 C will deposit 63.546/2 g (31.773 g) Cu. So we need how many coulombs?
96,485 C x (4.23/31.773) = ? C

amperes x seconds = coulombs
75 x seconds = ? C = about 171 seconds.

I note that 0.77 amps would take 4.6 hours. I wonder if that could be a typo in the book where they meant to type in 0.75A and not 75 A,

Answer 8

my answer is 4.7527 hr.

it is solved by this format:

time = [(0.10825-0.04175)/((0.75)(1/96485)(1/2))]/[3600]
time = 4.7527 hr

Note: for the dimensional analysis,
(mol)/(C/s)(mol/C)=s (divide by 3600 to convert to 1 hr)

hope it helps.

Answer 9

simply put,

(0.75 C/s)(time)(1 mole e-/96485 C)(1 mole Cu/2 mole e-) = (0.10825-0.04175)

time = 17,110.07 s = 4.7527 hr