What mass of chromium could be deposited by electrolysis of an aqueous solution of Cr2(SO4)3 for 180.0 minutes using a constant current of 10.0 amperes? (One faraday = 96,500 coulombs.)
coulombs = amperes x seconds.
C = 10A x 180 min x (60 s/min) = ??
It takes 96,500 coulombs to deposit 1 mol Cr/3 or 52.01/3 grams.
Why did you divide the molar mass by 3?
You would use stoichiometry. Multiply the amps by the time (convert minutes to seconds, then you end up with Coulombs. The stoichiometry should look a little like this...
(amps x time)(1 F/96,500 coulombs)(1 mol e-/1 F)(3 mol e-/1 mol Cr)(Molar Mass of Cr/1 mol Cr) ... I got the 3 mol e- from balancing the half-cell reactions. I got my answer to come out to 19.4 grams Cr
(15.0 C/s) ((180 min) (60 s/min)) = 162000 C
(162000 C) / (96485 C/mol e-) = 1.67 mol e-
Cr2(SO4)3 ---> 2Cr^3+ + 3SO4^2-
This reaction shows that for chromium we are transferring 3 moles of electrons.
(1.67 mol e-) (mol Cr / 3 mol e- transferred) = 0.559 mol Cr
(0.559 mol Cr) (51.996 g/mol) = 29.9 g Cr
Well, let's do some calculations and see what comedy I can cook up for you!
To determine the mass of chromium deposited, we need to use Faraday's Law of Electrolysis. According to Faraday's Law, the amount of substance deposited during electrolysis is directly proportional to the electric charge passed through the circuit.
So, let's get down to business.
First, we need to calculate the total charge passed during 180.0 minutes. We can do this by multiplying the current (10.0 amperes) by the time (180.0 minutes) and then convert it to coulombs.
10.0 amperes * 180.0 minutes = 1800.0 coulombs
Now, we know that one faraday is equal to 96,500 coulombs. Let's convert our coulombs to faradays:
1800.0 coulombs / 96,500 coulombs per faraday = 0.0186 faradays
Great! Now that we know the number of faradays, we can calculate the mass using the balanced redox equation for the reduction of chromium:
2e- + Cr3+ → Cr
From the equation, we can see that it takes 2 moles of electrons (2 faradays) to reduce 1 mole of Cr3+ ions to pure chromium.
So, if we have 0.0186 faradays, that means we have 0.0186/2 = 0.0093 moles of Cr3+ ions.
Now, let's move on to the last step: converting moles into grams. One mole of chromium weighs approximately 52 grams (it’s a heavy metal).
0.0093 moles * 52 grams per mole = 0.4836 grams of chromium
Voila! After all that mathematical juggling, we find that approximately 0.4836 grams of chromium could be deposited by electrolysis.
I hope you're not feeling too jumbled after all those calculations! But hey, at least now you've got a funny answer to your question!
To calculate the mass of chromium deposited by electrolysis, we need to use Faraday's law of electrolysis. According to Faraday's law, the amount of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
The formula to calculate the mass (m) of a substance deposited using Faraday's law is given by:
m = (Q * M) / (n * F)
Where:
m is the mass of the substance deposited
Q is the charge passed through the electrolyte (in coulombs)
M is the molar mass of the substance (in g/mol)
n is the number of moles of electrons transferred in the balanced equation
F is Faraday's constant, equal to 96,500 coulombs/mol
Now, let's break down the information given in the problem:
1. Charge passed through the electrolyte (Q): The problem states that a constant current of 10.0 amperes is used for 180.0 minutes. We need to convert the current and time to coulombs, using the equation Q = I * t, where:
- I is the current in amperes (10.0 A)
- t is the time in seconds (180.0 minutes * 60 seconds/minute)
Q = 10.0 A * 180.0 minutes * 60 seconds/minute = 108,000 C
2. Molar mass of chromium (M): The molar mass of Cr can be found on the periodic table, and it is approximately 52.0 g/mol.
3. Number of moles of electrons transferred (n): The balanced equation for the electrolysis of Cr2(SO4)3 shows that 6 moles of electrons are transferred per 1 mole of Cr deposited.
4. Faraday's constant (F): Given as 96,500 C/mol.
Now, substituting the values into the formula:
m = (Q * M) / (n * F)
m = (108,000 C * 52.0 g/mol) / (6 moles * 96,500 C/mol)
Now, we can perform the calculation:
m = (5,616,000 g*C) / (579,000 C)
m ≈ 9.71 g
Therefore, approximately 9.71 grams of chromium could be deposited by electrolysis of the given solution.