To calculate the mass of silver deposited on the cathode during electrolysis, we can use Faraday's laws of electrolysis. First, we need to determine the total charge Q passed through the solution. We can do this using the formula:Q = I twhere Q is the total charge, I is the current in amperes , and t is the time in seconds .Given:I = 2.5 At = 3 hours = 3 3600 seconds = 10800 secondsNow, we can calculate the total charge:Q = 2.5 A 10800 s = 27000 CoulombsNext, we need to find the number of moles of electrons n transferred during the electrolysis. We can do this using Faraday's constant F , which is approximately 96485 Coulombs per mole of electrons:n = Q / Fn = 27000 C / 96485 C/mol 0.2798 mol of electronsNow, we need to determine the number of moles of silver Ag deposited on the cathode. Silver has a valency of 1, meaning one mole of electrons will deposit one mole of silver:moles of Ag = 0.2798 molFinally, we can calculate the mass of silver deposited using the molar mass of silver Ag , which is 107.87 g/mol:mass of Ag = moles of Ag molar mass of Agmass of Ag = 0.2798 mol 107.87 g/mol 30.16 gSo, approximately 30.16 grams of silver will be deposited on the cathode during the electrolysis.