What mass of AL is deposited electronically in 30 minutes by current of 740A?
To solve this problem, we need to use the formula for electroplating:
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
First, we need to find the charge using the formula:
Charge = Current x Time
Charge = 740 A x 30 min x 60 s/min = 1,332,000 C
Next, we need to find the atomic mass of aluminum, which is 26.9815 g/mol.
Faraday's constant is 96,485 C/mol.
Now we can plug in the values into our original formula:
Mass = (Current x Time x Atomic Mass) / (Charge x Faraday's Constant)
Mass = (740 A x 30 min x 60 s/min x 26.9815 g/mol) / (1,332,000 C x 96,485 C/mol)
Mass = 1,270.87 g
Therefore, 1,270.87 grams of aluminum is deposited electronically in 30 minutes by a current of 740A.
To calculate the mass of Al deposited electronically, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrode.
The formula for calculating the mass of substance deposited can be represented as:
Mass (in grams) = (Current (in amperes) * Time (in seconds) * Molar mass (in grams)) / (Faraday's constant * Charge of the ion)
Given:
Current = 740 A
Time = 30 minutes = 1800 seconds (since 1 minute = 60 seconds)
Molar mass of Al = 26.98 g/mol (source: periodic table)
Charge of Al ion = +3 (since Al forms Al3+ ions)
Faraday's constant = 96485 C/mol
Let's calculate the mass of Al deposited electronically:
Mass = (740 A * 1800 s * 26.98 g/mol) / (96485 C/mol * 3)
Simplifying the equation:
Mass = (39876000 gC/mol) / (289455 C/mol)
Mass = 137.71 g
Therefore, the mass of Al deposited electronically in 30 minutes by a current of 740 A is 137.71 grams.