To calculate the amount of copper deposited on the cathode, we can use Faraday's law of electrolysis, which states that the mass of a substance deposited or dissolved at an electrode is directly proportional to the amount of charge passed through the electrolyte.First, let's find the total charge passed through the solution. Charge Q = Current I Time t where I = 5 A amperes and t = 30 minutes = 1800 seconds since 1 minute = 60 seconds Q = 5 A 1800 s = 9000 C coulombs Now, we can use Faraday's law to find the moles of copper deposited:Moles of copper n = Charge Q / Faraday constant F number of electrons involved in the reaction For the copper sulfate solution, the half-reaction for the deposition of copper is:Cu + 2e CuSo, the number of electrons involved in the reaction is 2.n = 9000 C / 96500 C/mol 2 = 0.0468 molNow, we can find the mass of copper deposited using the molar mass of copper:Mass of copper = Moles of copper Molar mass of copperMass of copper = 0.0468 mol 63.55 g/mol = 2.976 gTherefore, the amount of copper deposited on the cathode is approximately 2.976 grams.