First, we need to find the total charge passed through the solution during the electrolysis experiment. We can do this by multiplying the current in amperes by the time in seconds .Current = 2 ATime = 30 minutes = 30 * 60 seconds = 1800 secondsTotal charge = Current TimeTotal charge = 2 A 1800 s = 3600 CoulombsNext, we need to find the number of moles of electrons transferred during the experiment. We can do this by dividing the total charge by the Faraday constant.Moles of electrons = Total charge / Faraday constantMoles of electrons = 3600 C / 96,485 C/mol = 0.0373 molCopper sulfate CuSO4 undergoes electrolysis to produce copper Cu and sulfate ions SO4^2- . The balanced half-reaction for the reduction of copper ions at the cathode is:Cu^2+ + 2e- CuThis means that 2 moles of electrons are required to deposit 1 mole of copper. Now, we need to find the moles of copper deposited.Moles of copper = Moles of electrons / 2Moles of copper = 0.0373 mol / 2 = 0.01865 molFinally, we can find the mass of copper deposited by multiplying the moles of copper by the atomic weight of copper.Mass of copper = Moles of copper Atomic weight of copperMass of copper = 0.01865 mol 63.546 g/mol = 1.185 gSo, the mass of copper deposited on the cathode during the electrolysis experiment is approximately 1.185 grams.