If 2.00 liters of 0.580 M CuSO4 solution is electrolyzed by passing 1.70 amps through the solution for 2.00 hr using inert electrodes.
What is the [Cu2+] in the solution at the end of the electrolysis?
Coulombs = A x sec = 1.70 x 2 x 60 x 60 = ??
One equivalent weight will be deposited by passing 96,495 coulombs. ONe equivalent weight = 63.54 g/2 = 31.8 but you need to do that more accurately.
Then
31.8 x #coulombs/96,485 = g Cu deposited.
g Cu initially = 2L x 0.580M x 63.54 = ??
g Cu deposted from above = xx
g Cu remaining in solution = zz
M = moles/2L = zz
This assumes no electrolysis of the water (highly unlikely I think).
To find the [Cu2+] in the solution at the end of the electrolysis, we need to determine the moles of Cu2+ ions that have been consumed during the process.
First, let’s calculate the number of moles of electrons (n) that have been transferred through the solution during the electrolysis. We can use Faraday’s law, which states that the amount of charge transferred (Q) is equal to the product of the current (I), time (t), and the Faraday constant (F).
Q = I * t
In this case, the current (I) is 1.70 amps, and the time (t) is 2.00 hours. However, we need to convert the time to seconds since the Faraday constant (F) is in C/mol.
t = 2.00 hours * 60 minutes/hour * 60 seconds/minute = 7200 seconds
Now, we can calculate Q:
Q = 1.70 amps * 7200 seconds
Next, we need to determine the number of moles of electrons transferred using the equation:
n = Q / F
The Faraday constant (F) is approximately 96,485 C/mol.
n = (1.70 amps * 7200 seconds) / 96,485 C/mol
Now, let’s use the balanced redox equation to determine the moles of Cu2+ ions consumed.
Cu2+ + 2e- → Cu
From the balanced equation, we can see that for every 2 moles of electrons transferred, it consumes 1 mole of Cu2+ ions. Therefore, the moles of Cu2+ ions consumed (Cu2+ consumed) can be calculated as:
Cu2+ consumed = n / 2
Now, we need to determine the initial moles of Cu2+ ions present in the solution. This can be calculated using the volume of the solution (2.00 liters) and the molarity (0.580 M) of CuSO4.
initial moles of Cu2+ ions = volume (liters) * molarity
initial moles of Cu2+ ions = 2.00 liters * 0.580 M
Finally, we can calculate the [Cu2+] at the end of the electrolysis by subtracting the moles of Cu2+ ions consumed from the initial moles and then dividing by the final volume (2.00 liters).
[Cu2+] = (initial moles - Cu2+ consumed) / volume
Now, plug in the known values and calculate the [Cu2+].
Note: The concentration of Cu2+ will decrease as the solution is electrolyzed, given that Cu2+ is being consumed.
Please note that this answer assumes the Faraday efficiency is 100%, meaning that all the current goes exclusively to the oxidation and reduction of Cu2+ and no side reactions occur. This may not be the actual case in real-life scenarios.