Question

What is the concentration of copper ions (Cu2+) in a 100 mL sample of wastewater if 4652 coulombs of electricity are needed to completely oxidize all the copper ions present? The electrolysis is carried out using a current of 0.25 amperes for a period of 6 minutes. Assume that the only copper present is in the form of Cu2+.

Answer 1

First, let's find the number of moles of electrons transferred during the electrolysis process. We can use the formula:moles of electrons =  coulombs  /  Faraday's constant Faraday's constant  F  is approximately 96485 C/mol.moles of electrons = 4652 C / 96485 C/mol = 0.0482 molNow, let's find the number of moles of copper ions  Cu2+  present in the sample. Since each Cu2+ ion requires 2 moles of electrons to be oxidized, we can use the formula:moles of Cu2+ =  moles of electrons  / 2moles of Cu2+ = 0.0482 mol / 2 = 0.0241 molNow, we can find the concentration of Cu2+ ions in the 100 mL sample. Concentration is defined as the number of moles of solute per liter of solution:concentration of Cu2+ =  moles of Cu2+  /  volume of solution in liters First, convert the volume of the solution from mL to L:100 mL = 0.1 LNow, calculate the concentration:concentration of Cu2+ = 0.0241 mol / 0.1 L = 0.241 mol/LSo, the concentration of copper ions  Cu2+  in the 100 mL sample of wastewater is 0.241 mol/L.