Question

A 50 mL solution of 0.1 M Cu(NO3)2 was electrolyzed for 20 minutes. During this time, a current of 0.5 A was passed through the solution. What is the final concentration of Cu2+ ions in the solution after the electrolysis?

Answer 1

To find the final concentration of Cu2+ ions in the solution after electrolysis, we need to determine the amount of Cu2+ ions reduced during the process.First, let's find the total charge passed through the solution during electrolysis:Charge  Q  = Current  I   Time  t Q = 0.5 A  20 min  60 s/min = 600 CoulombsNow, we need to find the number of moles of electrons  n  transferred during the electrolysis using Faraday's constant  F = 96485 C/mol :n = Q / Fn = 600 C / 96485 C/mol  0.00622 mol of electronsSince the reduction of Cu2+ ions requires 2 moles of electrons per mole of Cu2+  Cu2+ + 2e-  Cu , we can find the moles of Cu2+ ions reduced:Moles of Cu2+ reduced = n / 2Moles of Cu2+ reduced  0.00622 mol / 2  0.00311 molNow, we can find the initial moles of Cu2+ ions in the solution:Initial moles of Cu2+ = Molarity  VolumeInitial moles of Cu2+ = 0.1 M  0.05 L = 0.005 molNext, we subtract the moles of Cu2+ reduced from the initial moles to find the final moles of Cu2+ ions:Final moles of Cu2+ = Initial moles - Moles reducedFinal moles of Cu2+ = 0.005 mol - 0.00311 mol  0.00189 molFinally, we can find the final concentration of Cu2+ ions in the solution:Final concentration of Cu2+ = Final moles / VolumeFinal concentration of Cu2+  0.00189 mol / 0.05 L  0.0378 MSo, the final concentration of Cu2+ ions in the solution after electrolysis is approximately 0.0378 M.