Calculate the mass of copper (Cu) deposited on the cathode when a 1.0 A current is passed through a copper(II) sulfate (CuSO4) solution for 2 hours.
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To calculate the mass of copper deposited on the cathode, we need to use Faraday's law of electrolysis.
According to Faraday's law, the mass of a substance deposited on an electrode is directly proportional to the amount of charge passed through the electrolyte.
The formula for calculating the mass of copper deposited (m) is given by:
m = (Q × M) / (n × F)
where:
Q is the charge passed (in coulombs),
M is the molar mass of copper (63.55 g/mol),
n is the number of electrons involved in the reaction (2 electrons for copper),
F is Faraday's constant (96,485 C/mol).
First, we need to calculate the charge passed (Q) in coulombs. We know the current (I) in amperes and the time (t) in seconds.
Q = I × t
Given:
Current, I = 1.0 A
Time, t = 2 hours = 2 × 3600 seconds
Q = 1.0 A × 2 × 3600 s = 7200 C
Now, we can substitute the values into the formula to calculate the mass of copper deposited (m):
m = (7200 C × 63.55 g/mol) / (2 mol × 96,485 C/mol)
Simplifying the equation:
m = (7200 C × 63.55 g/mol) / (2 mol × 96,485 C/mol)
= (459240 g⋅C) / 192,970 C
≈ 2.38 g
Therefore, approximately 2.38 grams of copper is deposited on the cathode after passing a 1.0 A current through a copper(II) sulfate solution for 2 hours.
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