To determine the mass of silver deposited on the cathode, we can use Faraday's law of electrolysis, which states that the amount of substance deposited is directly proportional to the charge passed through the solution.First, we need to calculate the total charge passed through the solution. We can do this using the formula:Charge Q = Current I Time t Given that the current I is 4 A and the time t is 30 minutes which we need to convert to seconds , we have:Q = 4 A 30 60 sQ = 4 A 1800 sQ = 7200 CNow, we can use the Faraday constant F to determine the number of moles of electrons n that have been transferred:n = Q / Fn = 7200 C / 96,485 C/moln 0.0746 molSince the reaction occurring at the cathode is the reduction of silver ions Ag+ to form silver atoms Ag , the stoichiometry of the reaction is 1:1. This means that 1 mole of electrons will deposit 1 mole of silver. Therefore, the number of moles of silver deposited is equal to the number of moles of electrons transferred:Moles of Ag = 0.0746 molNow, we can calculate the mass of silver deposited using the atomic mass of silver 107.87 g/mol :Mass of Ag = Moles of Ag Atomic mass of AgMass of Ag = 0.0746 mol 107.87 g/molMass of Ag 8.05 gSo, the mass of silver that will be deposited on the cathode is approximately 8.05 grams.