A copper electrode weighs 35.42 g before the electrolysis of a CuSO4(aq) solution and weighs
36.69 g after the electrolysis has run for 20.0 min. What was the amperage (Amps) of the current
used?
this is the equation that i used is it correct
(1.27g Cu x 1 mol Cu x 2 mol e- x 96,485 c) / (63.55g Cu x 1 mol Cu x 1 mol e- x 20 min x 60 sec)
= 0.002528 A
I didn't check out our equation but I don't get that answer. INstread I plugged your answer in to see if it would plate out 1.27 grams, as follows:
Coulombs = 0.002528 x 20 x 60 = approx 3 coulombs.
96,485 coulombs will plat out 63.55/2 = 31.775 g Cu so
31.775 x about 3.0/96,485 = about 1E-3g or approx 1 mg but the problem plates out 1.27 g so 0.002528 A is not right. The correct answer is closer to 3 amperes.
To determine the amperage (Amps) of the current used during the electrolysis, we can use the equation:
Amperage (A) = (Change in mass of electrode (grams)) / (Time (min) x 60 (sec/min))
Given that the copper electrode weighed 35.42 g before electrolysis and 36.69 g after 20.0 min of electrolysis, the change in mass is:
Change in mass = 36.69 g - 35.42 g = 1.27 g
Now, we can substitute these values into the equation:
Amperage (A) = (1.27 g) / (20.0 min x 60 sec/min)
Amperage (A) = 0.002111 A
Therefore, the amperage (Amps) of the current used during the electrolysis is approximately 0.002111 A.
Your equation is mostly correct, but there is a slight error in the conversion of the mass of copper to moles of copper. Let's go through the correct calculation step by step.
First, we need to determine the change in mass of the copper electrode during the electrolysis:
Change in mass = Final mass - Initial mass
Change in mass = 36.69 g - 35.42 g
Change in mass = 1.27 g
Next, we need to convert the mass of copper to moles of copper. The molar mass of copper is 63.55 g/mol:
Moles of copper = Change in mass / molar mass
Moles of copper = 1.27 g / 63.55 g/mol
Moles of copper = 0.02 mol
Now, we need to calculate the moles of electrons transferred. In the balanced redox equation of the electrolysis of CuSO4, we know that 2 mol of electrons are transferred per 1 mol of copper:
Moles of electrons = Moles of copper x 2
Moles of electrons = 0.02 mol x 2
Moles of electrons = 0.04 mol
Next, we can calculate the total charge transferred in coulombs. Each mole of electrons corresponds to 96,485 coulombs of charge:
Total charge transferred = Moles of electrons x Faraday's constant
Total charge transferred = 0.04 mol x 96,485 C/mol
Total charge transferred = 3,859.4 C
Now, we can calculate the amperage (current) by dividing the total charge transferred by the time in seconds (convert the 20 minutes to seconds by multiplying by 60):
Amperage = Total charge transferred / (time in seconds)
Amperage = 3,859.4 C / (20 min x 60 sec/min)
Amperage = 3,859.4 C / 1200 sec
Amperage = 3.216 A
So, the correct amperage of the current used in the electrolysis is approximately 3.216 Amps.