A solution of CuSO4 is electrolysed for 600 secs with a current of 1.5 amperes.what is the mass of copper deposited at cathode?

Question

Answer

W= eQ W= e*I*t

Answer

Understanding Electrolysis of CuSO4
Electrolysis involves the decomposition of a substance using an electric current. In this case, we are electrolyzing a solution of copper(II) sulfate (CuSO4).
Key Parameters
- Current (I): 1.5 A (amperes)
- Time (t): 600 seconds
- Electric Charge (Q): Calculated using the formula Q = I × t
Calculating Electric Charge
- Q = 1.5 A × 600 s = 900 Coulombs
Faraday's Law of Electrolysis
According to Faraday's law, the amount of substance deposited at the electrodes is directly proportional to the quantity of electricity passed through the electrolyte.
Molar Mass and Valency of Copper
- Molar Mass of Copper (Cu): Approximately 63.5 g/mol
- Valency of Copper in CuSO4: 2 (since Cu2+ ions are involved)
Using Faraday's Constant
- Faraday's Constant (F): Approximately 96500 C/mol (charge per mole of electrons)
Calculating Mass of Copper Deposited
1. Moles of Electrons:
- Moles of electrons = Q / F = 900 C / 96500 C/mol ≈ 0.00932 mol
2. Moles of Copper Deposited:
- Since 2 moles of electrons deposit 1 mole of Cu:
- Moles of Cu = 0.00932 mol / 2 ≈ 0.00466 mol
3. Mass of Copper:
- Mass = Moles × Molar Mass = 0.00466 mol × 63.5 g/mol ≈ 0.296 g
Conclusion
The mass of copper deposited at the cathode after 600 seconds of electrolysis with a current of 1.5 A is approximately 0.296 grams.

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