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 A.

Electrolysis involves breaking down a compound through the application of an electric current, and it's at the half-cells where this process becomes particularly intriguing. Imagine slicing an electrochemical cell into two halves, each performing a unique function. One half-cell is dedicated to oxidation, where we lose electrons, while the other specializes in reduction, gaining those freely moving electrons.

In the case of hydrogen fluoride (HF) electrolysis, at the anode, HF loses electrons (oxidation) and produces hydrogen gas (H2) and fluorine gas (F2). At the cathode, water (H2O) gains electrons (reduction), creating hydroxide ions (OH-) and more hydrogen gas. These half-cell reactions are pivotal because they describe exactly what happens at each electrode, providing valuable insight into the overall chemical process.

Electrochemical cells are the wizards of chemistry, transforming chemical energy into electrical energy and vice versa. Structurally, these cells are composed of two electrodes (the cathode and the anode) immersed in an electrolyte solution that facilitates ion movement.

The purpose of these cells is to conduct redox reactions in a controlled environment. In our textbook scenario, the cell uses the liquid hydrogen fluoride with potassium fluoride to perform electrolysis. Here's a hidden gem: potassium fluoride (KF) is more than just a supporting actor. By enhancing the solution's conductivity, it ensures a smoother and more efficient flow of electricity, which effectively helps in whisking away the charges during the electrolysis of HF.

Faraday's law of electrolysis is like the accountant of electrochemistry, keeping track of all the electron transactions. It tells us that the amount of substance produced or consumed at an electrode during electrolysis is directly proportional to the amount of electrical charge passed through the electrolyte.

To put it in context with our exercise, when we pass 502 amperes of current for 15 hours through our HF solution, we are essentially adding a specific amount of charge that, thanks to Faraday, can tell us exactly how many moles of fluorine gas we will produce. It simplifies the complex world of ion exchange into a neat mathematical equation, providing a quantitative measure that bridges the gap between electricity and chemical change.

The ideal gas law is the bread and butter of understanding gases in chemistry. It relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas with a simple yet profound equation: PV = nRT. R here is the ideal gas constant.

This law becomes incredibly practical when we need to figure out how much gas is given off in a reaction like electrolysis of hydrogen fluoride. After calculating the moles of fluorine gas produced in the reaction with Faraday's law, you can forecast the volume the gas will occupy under certain temperature and pressure conditions, as exemplified in the solution. It's essentially predicting the space a batch of gas will need, just like estimating the right size of a balloon for a given amount of helium.