Fluorine \(\left(\mathrm{F}_{2}\right)\) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process. (b) What is the purpose of KF? (c) Calculate the volume of \(\mathrm{F}_{2}\) (in liters) collected at \(24.0^{\circ} \mathrm{C}\) and 1.2 atm after electrolyzing the solution for \(15 \mathrm{~h}\) at a current of \(502 \mathrm{~A}\).

The dissociation of Hydrogen Fluoride (HF) to Hydrogen (\(H^+\)) and Fluorine (\(F^-\)) ion, is represented as: \(2HF \rightarrow 2H^+ + 2F^-\). During electrolysis, at the anode (oxidation), Fluorine ions gain electrons to form Fluorine gas: \(2F^- \rightarrow F_2 + 2e^-\). At the cathode (reduction), Hydrogen ions lose electrons to form Hydrogen gas: \(2H^+ + 2e^- \rightarrow H_2\).

The overall reaction involves combining the half-cell reactions (ignoring the electrons as they are the same in both half-cell reactions), which gives: \(2HF \rightarrow H_2 + F_2\).

Potassium fluoride (KF) is used to increase the concentration of \(F^-\) ions in the solution, thus facilitating the electrolysis process by allowing more fluorine gas to be produced.

The volume of \(F_2\) can be calculated using Faraday's law of electrolysis and the ideal gas law. Faraday's law states that the amount of substance produced at an electrode during electrolysis is directly proportional to the number of moles of electrons (amount of electricity) transferred at that electrode. The total charge (Q) passed during electrolysis can be found by multiplying the current (I) in amperes by the time (t) in seconds: \(Q=It\). Here, \(I = 502 A\) and \(t = 15 \times 3600 = 54000 s\). So \(Q = 502 A \times 54000 s = 27.108 \times 10^6 coulombs = 27.108 \times 10^6 / 96485 = 280.84\) mol of electrons. Since the production of one mole of \(F_2\) requires two moles of electrons according to the half-reaction, we get \(140.42\) mol of \(F_2\) produced. Finally, we use the ideal gas law (\(PV = nRT\)) where \(P = 1.2 atm, n = 140.42 mol, R = 0.0821 L.atm/mol.K\) and \(T = 24.0+273.15 = 297.15 K\). Solving for V, we get \(V = nRT/P = 140.42 \times 0.0821 \times 297.15 / 1.2 = 2850 L\).