Question

A chemistry student needs to calculate the concentration of copper(II) ions in a 250 mL solution, given that a current of 1.5 A was applied for 60 minutes to a copper electrode in contact with the solution. The initial concentration of copper(II) ions was 0.05 M, and the Faraday constant is 96,485 C/mol. What is the final concentration of copper(II) ions in the solution?

Answer 1

To calculate the final concentration of copper II  ions in the solution, we need to determine the amount of copper ions that were deposited on the electrode during the electrolysis process.First, we need to find the total charge  Q  that passed through the solution during the electrolysis process:Q = current  I   time  t Q = 1.5 A   60 min  60 s/min  = 1.5 A  3600 s = 5400 CNext, we need to find the number of moles of electrons  n  that were transferred during the electrolysis process:n = Q / Faraday constant  F n = 5400 C / 96,485 C/mol = 0.056 molSince the copper II  ions have a charge of +2, the number of moles of copper ions that were deposited on the electrode is half the number of moles of electrons:moles of Cu = 0.056 mol / 2 = 0.028 molNow, we need to find the change in concentration of copper II  ions in the solution:[Cu] = moles of Cu / volume of solution  in L [Cu] = 0.028 mol /  250 mL  0.001 L/mL  = 0.028 mol / 0.25 L = 0.112 MSince the copper ions were deposited on the electrode, the final concentration of copper II  ions in the solution will be the initial concentration minus the change in concentration:Final [Cu] = Initial [Cu] - [Cu]Final [Cu] = 0.05 M - 0.112 M = -0.062 MHowever, the concentration cannot be negative. This means that all the copper II  ions were deposited on the electrode, and the final concentration of copper II  ions in the solution is 0 M.