If a current of 3.88 A is applied to a cell containing a 436 g copper anode,
determine the final mass of the copper anode after the cell runs for 40.75 h.
To determine the final mass of the copper anode, we need to calculate the amount of copper that is oxidized during the 40.75 h.
First, let's calculate the charge passed through the cell. We can use the formula:
Q = I*t
Where Q is the charge passed (in coulombs), I is the current (in amperes), and t is the time (in seconds).
Since the given time is in hours, we need to convert it to seconds:
t = 40.75 hours * 3600 seconds/hour = 146,700 seconds
Now we can calculate the charge passed:
Q = 3.88 A * 146,700 s = 570,036 C
Next, we need to convert the charge passed to the number of moles of electrons involved in the oxidation of copper. We can use Faraday's constant (F) to do this conversion:
1 mole e- = 1 F = 96,485 C
Therefore, the number of moles of electrons involved in the oxidation of copper is:
n = Q / F = 570,036 C / 96,485 C/mol = 5.911 mol
Since 1 mole of copper (Cu) corresponds to 1 mole of electrons (e-), the number of moles of copper oxidized is also 5.911 mol.
Now, we need to calculate the mass of copper oxidized. The molar mass of copper (Cu) is 63.55 g/mol. Therefore, the mass of copper oxidized is:
mass = n * molar mass = 5.911 mol * 63.55 g/mol = 375 g (rounded to three decimal places)
The final mass of the copper anode after running for 40.75 h is 375 g.