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How long will it take an electric current of 0.15A to deposit all the copper from 500mL of 0.15 CuSO4 solution?
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How long will it take an electric current of 0.15A to deposit all the ...
Calculating the Time Required for Electroplating Copper

To calculate the time required for electroplating copper, we need to consider the following factors:

Amount of Copper Deposited

The amount of copper deposited during electroplating can be calculated using Faraday's law, which states that the amount of a substance deposited during electrolysis is directly proportional to the amount of electric charge passed through the solution. The equation for Faraday's law is:

Amount of substance deposited = (Current x Time x Atomic weight of substance) / (Faraday constant x Valence)

In this case, the substance being deposited is copper (Cu), the current is 0.15A, the time is unknown, the atomic weight of copper is 63.55 g/mol, the Faraday constant is 96,485 C/mol, and the valence of copper is 2.

Volume and Concentration of Copper Sulfate Solution

To calculate the amount of copper that can be deposited from the solution, we need to know the volume and concentration of the copper sulfate solution. In this case, we are given that the volume is 500mL and the concentration is 0.15 mol/L.

Calculating the Time Required

Using the information above, we can calculate the amount of copper that can be deposited from the solution as follows:

Amount of copper deposited = (0.15A x Time x 63.55 g/mol) / (96,485 C/mol x 2)

Amount of copper deposited = (0.15 x Time x 63.55) / 192970

Amount of copper deposited = 0.000078 x Time

To find the time required for electroplating, we need to set the amount of copper deposited equal to the amount of copper in the solution:

Amount of copper deposited = Amount of copper in solution

0.000078 x Time = 0.15 mol/L x 0.5 L x 63.55 g/mol

0.000078 x Time = 4.78425 g

Time = 61,333 seconds or approximately 17 hours

Conclusion

In summary, it will take approximately 17 hours for an electric current of 0.15A to deposit all the copper from 500mL of 0.15 CuSO4 solution. This calculation is based on the volume and concentration of the solution, as well as the amount of copper that can be deposited using Faraday's law.
Community Answer
How long will it take an electric current of 0.15A to deposit all the ...
Given I=0.15A V=500ml. M=0.15M molar mass of copper sulphate=159.6g/mol. 1F=96500C
Required t=?
solution. m/E=Q/F. t=mF /EI
n=M×V=0.075mol
m=n×molar mass=11.97gm
E=molar mass/Z, where z=2
E=159.6gm/mol÷2=79.8gm/mol
t=mF÷EI
t=11.97gm×96500C÷79.8gm/mol(0.15A)
t=965sec=16min
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How long will it take an electric current of 0.15A to deposit all the copper from 500mL of 0.15 CuSO4 solution?
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