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A 100 mL solution of 1 M CuSO4 is electrolysed with 0.5A of current until its conc. becomes 0.4 M. If vol. of sol. remains constant, how long electrolysis was conducted?
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A 100 mL solution of 1 M CuSO4 is electrolysed with 0.5A of current un...
Understanding the Electrolysis of CuSO4
Electrolysis involves using an electric current to drive a chemical reaction. In this case, we are electrolyzing a 100 mL solution of 1 M CuSO4.
Initial and Final Concentration
- Initial concentration of CuSO4 = 1 M
- Final concentration after electrolysis = 0.4 M
- Volume of solution = 100 mL
Change in Concentration
- The change in concentration = Initial concentration - Final concentration
- Change = 1 M - 0.4 M = 0.6 M
Volume of CuSO4 Consumed
- Moles of CuSO4 consumed = Change in concentration × Volume (in L)
- Change in concentration = 0.6 M
- Volume = 0.1 L (100 mL)
- Moles consumed = 0.6 × 0.1 = 0.06 moles
Electrolysis and Charge Calculations
- Each mole of CuSO4 requires 2 moles of electrons (2 Faradays) for complete reduction.
- Charge required = Moles consumed × 2 (for Cu2+) × Faraday's constant (approximately 96500 C/mol)
- Charge required = 0.06 moles × 2 × 96500 C/mol = 11580 C
Time Calculation
- Current (I) = 0.5 A
- Time (t) = Charge (Q) / Current (I)
- Time = 11580 C / 0.5 A = 23160 seconds
Conclusion
- The duration of electrolysis was approximately 23160 seconds, or 6 hours and 26 minutes, to reduce the concentration of CuSO4 from 1 M to 0.4 M while keeping the volume constant.
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A 100 mL solution of 1 M CuSO4 is electrolysed with 0.5A of current un...
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A 100 mL solution of 1 M CuSO4 is electrolysed with 0.5A of current until its conc. becomes 0.4 M. If vol. of sol. remains constant, how long electrolysis was conducted?
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