To calculate the efficiency of an electrochemical cell, we need to determine the actual amount of metal deposited on the electrodes and compare it to the theoretical amount that should be deposited based on the current and duration of the experiment.First, let's find the charge Q passed through the cell:Q = current timeQ = 2 mA 3 hoursQ = 2 10 A 3 3600 s 1 hour = 3600 seconds Q = 21.6 C Coulombs Now, let's find the theoretical amount of metal deposited using Faraday's law of electrolysis:M = Q Molar mass / n F For Cu copper :n = 2 Cu has a charge of +2 Molar mass = 63.5 g/molF = Faraday's constant = 96485 C/molM_Cu = 21.6 C 63.5 g/mol / 2 96485 C/mol M_Cu = 0.0071 gFor Ag silver :n = 1 Ag has a charge of +1 Molar mass = 107.9 g/molM_Ag = 21.6 C 107.9 g/mol / 1 96485 C/mol M_Ag = 0.024 gNow, let's find the actual amount of metal deposited. We'll assume that the cell is 100% efficient, meaning that all the current goes into depositing the metals. In this case, the actual amount of metal deposited is equal to the theoretical amount.Actual M_Cu = 0.0071 gActual M_Ag = 0.024 gTo find the efficiency of the cell, we'll compare the actual amount of metal deposited to the theoretical amount:Efficiency = Actual amount of metal deposited / Theoretical amount of metal deposited 100Efficiency_Cu = 0.0071 g / 0.0071 g 100 = 100%Efficiency_Ag = 0.024 g / 0.024 g 100 = 100%Since both copper and silver have 100% efficiency, the overall efficiency of the electrochemical cell is 100%.