Question

Question: A student wants to quantify the amount of iron (Fe) in a water sample using coulometry. The sample has a mass of 50.0 g and contains iron in the form of Fe(II) ions. The student sets up the electrochemical cell and begins the oxidation reaction, which requires a current of 16.0 A for 1200 s. What is the concentration of Fe(II) ions in the water sample in mol/L?

Answer 1

To solve this problem, we will first calculate the moles of electrons transferred during the oxidation reaction, then determine the moles of Fe II  ions, and finally calculate the concentration of Fe II  ions in the water sample.1. Calculate the charge  in coulombs  passed through the electrochemical cell:Charge  Q  = Current  I   Time  t Q = 16.0 A  1200 s = 19200 C2. Calculate the moles of electrons transferred:Moles of electrons  n  = Charge  Q  / Faraday constant  F Faraday constant  F  = 96485 C/moln = 19200 C / 96485 C/mol  0.1988 mol of electrons3. Determine the moles of Fe II  ions:The balanced oxidation half-reaction for Fe II  to Fe III  is:Fe II   Fe III  + e^-So, 1 mole of Fe II  ions requires 1 mole of electrons for oxidation.Moles of Fe II  ions = Moles of electrons = 0.1988 mol4. Calculate the concentration of Fe II  ions in the water sample:First, convert the mass of the water sample to volume, assuming the density of water is 1 g/mL or 1 g/cm:Volume = Mass / Density = 50.0 g / 1 g/mL = 50.0 mLNow, convert the volume from mL to L:Volume = 50.0 mL   1 L / 1000 mL  = 0.050 LConcentration of Fe II  ions = Moles of Fe II  ions / VolumeConcentration = 0.1988 mol / 0.050 L  3.976 mol/LThe concentration of Fe II  ions in the water sample is approximately 3.976 mol/L.