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?
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.
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,
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.
Find coulombs, then amp x seconds = coulombs. How many grams Cu do you have that's deposited.
i get 4.207 g Cu
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.
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!
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.
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
To find the time required for electrolysis, we can use Faraday's law of electrolysis, which states:
\(mol = \frac{Q}{F}\)
Where:
- mol is the number of moles of substance deposited or reacted
- Q is the total charge passed through the electrolytic cell
- F is Faraday's constant (the charge of one mole of electrons, approximately 96500 C/mol)
First, let's find the number of moles of Cu2+ needed to be reduced to obtain a concentration of 0.167 M.
Using the formula for molarity:
\(M = \frac{{mol}}{L}\)
Rearranging the formula to solve for moles:
\(mol = M \times L\)
\(mol = 0.167 \, M \times 0.250 \, L\)
\(mol = 0.04175 \, mol\)
Now, we need to calculate the charge (Q) required to reduce these moles of Cu2+.
The charge passed (Q) is given by:
\(Q = I \times t\)
Where:
- Q is the charge (Coulombs)
- I is the current passing through the cell (Amperes)
- t is the time (seconds)
Rearranging the formula to solve for time:
\(t = \frac{Q}{I}\)
We have I = 75 A, but we need to find Q. To find Q, we need to know the number of moles of electrons involved in the reduction of one mole of Cu2+. Through the balanced equation for the reduction of Cu2+ to Cu, we find that 2 moles of electrons are required.
Therefore, the charge (Q) is given by:
\(Q = n \times F\)
Where:
- Q is the charge (Coulombs)
- n is the number of moles of electrons
- F is Faraday's constant
Substituting the given values:
\(Q = 0.04175 \, mol \times 2 \times 96500 \, C/mol\)
\(Q = 8.033875 \, C\)
Finally, substitute the values of Q and I into the time formula:
\(t = \frac{8.033875 \, C}{75 \, A}\)
\(t = 0.107 \, s\)
Therefore, the current needs to run for approximately 0.107 seconds to reduce the concentration of Cu2+ to 0.167 M.
To calculate the mass of Cu deposited on the cathode during this time, we can use the formula:
\(mass = mol \times molar \, mass\)
The molar mass of Cu is approximately 63.55 g/mol.
\(mass = 0.04175 \, mol \times 63.55 \, g/mol\)
\(mass = 2.644 \, g\)
Therefore, approximately 2.644 grams of Cu will be deposited on the cathode during this time.