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What is the Faraday constant and how can it be calculated from the electrolysis of copper sulfate using a current of 2.5 amps for 2 hours?

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The Faraday constant  F  is a fundamental constant in electrochemistry, representing the amount of electric charge carried by one mole of electrons. It is approximately equal to 96,485 Coulombs/mol.To calculate the amount of copper deposited during the electrolysis of copper sulfate using a current of 2.5 amps for 2 hours, you can use Faraday's laws of electrolysis.First, convert the time to seconds: 2 hours = 2 * 60 * 60 = 7200 seconds.Next, calculate the total charge  Q  passed through the solution: Q = current  I   time  t  = 2.5 A  7200 s = 18,000 Coulombs.Now, you can use the Faraday constant to find the number of moles of electrons  n  transferred: n = Q / F = 18,000 C / 96,485 C/mol  0.1865 mol.In the electrolysis of copper sulfate, the half-reaction for copper deposition is:Cu + 2e  CuFrom the stoichiometry of the reaction, 2 moles of electrons are required to deposit 1 mole of copper. So, the number of moles of copper deposited is:0.1865 mol electrons   1 mol Cu / 2 mol electrons   0.09325 mol Cu.Finally, to find the mass of copper deposited, multiply the number of moles by the molar mass of copper  63.55 g/mol :0.09325 mol Cu  63.55 g/mol = 5.92 g  approximately .So, approximately 5.92 grams of copper will be deposited during the electrolysis of copper sulfate using a current of 2.5 amps for 2 hours.

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