To calculate the mass 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, we need to find the total charge passed through the solution. The charge Q can be calculated using the formula:Q = I twhere I is the current 5 A and t is the time 2 hours . Since the time is given in hours, we need to convert it to seconds:2 hours 3600 seconds/hour = 7200 secondsNow, we can calculate the charge:Q = 5 A 7200 s = 36000 CNext, we need to find the number of moles of electrons n passed through the solution using the Faraday constant F :n = Q / Fn = 36000 C / 96485 C/mol 0.373 molSince copper II ions require 2 moles of electrons to be reduced to copper metal Cu + 2e Cu , we can find the number of moles of copper deposited n_Cu as follows:n_Cu = n / 2n_Cu = 0.373 mol / 2 0.1865 molFinally, we can calculate the mass of copper deposited m_Cu using the atomic mass of copper M_Cu = 63.5 g/mol :m_Cu = n_Cu M_Cum_Cu = 0.1865 mol 63.5 g/mol 11.84 gSo, the mass of copper deposited on the cathode is approximately 11.84 grams.