Question

A student needs to determine the mass of silver that can be deposited on a silver electrode using a current of 2.5 A for 3 hours in a solution containing 0.2 M silver ions. What is the mass of silver deposited on the electrode? (Assuming 100% current efficiency and atomic weight of silver is 107.9 g/mol)

Answer 1

To determine the mass of silver deposited on the electrode, we can use Faraday's law of electrolysis. First, we need to find the total charge  Q  passed through the solution:Q = current  I   time  t Q = 2.5 A  3 hours  3600 seconds/hourQ = 27000 CoulombsNext, we need to find the number of moles of electrons  n  passed through the solution:n = Q /  Faraday's constant, F n = 27000 C /  96485 C/mol n = 0.2798 mol of electronsSince the silver ion  Ag+  requires one electron to be reduced to silver metal  Ag , the number of moles of silver deposited is equal to the number of moles of electrons:n Ag  = 0.2798 molNow we can find the mass of silver deposited:mass Ag  = n Ag   atomic weight of silvermass Ag  = 0.2798 mol  107.9 g/molmass Ag  = 30.18 gTherefore, the mass of silver deposited on the electrode is approximately 30.18 grams.