To calculate the mass of copper deposited on the cathode, we can use Faraday's law of electrolysis. First, we need to find the total charge passed through the solution. Total charge Q = current I time t Q = 5 A 1 hour 3600 seconds/hour = 18,000 CNow, we need to find the number of moles of electrons n passed through the solution using Faraday's constant F .n = Q / Fn = 18,000 C / 96,485 C/mol = 0.1865 mol of electronsIn the case of copper sulfate, the reaction at the cathode is:Cu + 2e CuThis means that 2 moles of electrons are required to deposit 1 mole of copper. So, we need to find the number of moles of copper deposited n_Cu .n_Cu = n / 2 = 0.1865 mol of electrons / 2 = 0.09325 mol of CuFinally, we can find the mass of copper deposited m_Cu using the molar mass of copper M_Cu .m_Cu = n_Cu M_Cum_Cu = 0.09325 mol of Cu 63.55 g/mol = 5.92 gSo, the mass of copper deposited on the cathode is approximately 5.92 grams.