A salt of an aqueeous solution of copper is electrolyzed by a current of 3.0 A running for 54.0 minutes. If 3.47g Cu is produced at the cathode, what charge of the copper ions in solution?
To find the charge of the copper ions in solution, we can use Faraday's law of electrolysis, which states that the amount of substance produced or consumed at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
The formula to calculate the amount of substance produced is:
Amount of substance = (Current × Time) / (n × F)
Where:
- Current is the current passing through the electrolyte (in amperes)
- Time is the time for which the current is applied (in seconds)
- n is the number of electrons involved in the half-reaction
- F is Faraday's constant (9.65 × 10^4 C/mol)
First, let's convert the given time from minutes to seconds:
54.0 minutes × 60 seconds/minute = 3240 seconds
The number of electrons involved in the half-reaction for the reduction of copper ions to copper metal is 2.
Plugging in the values into the formula:
Amount of substance = (3.0 A × 3240 s) / (2 × 9.65 × 10^4 C/mol) ≈ 0.0527 mol
Since 3.47 g of copper is produced at the cathode, we can calculate the molar mass of copper (Cu) to convert the amount of substance to charge:
Molar mass of copper = 63.55 g/mol
Charge of copper ions in solution = Amount of substance × Molar mass = 0.0527 mol × 63.55 g/mol ≈ 3.34 g
Therefore, the charge of the copper ions in solution is approximately 3.34 g.