Question

What volume of \(\mathrm{F}_{2}\) gas, at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\), is produced when molten \(\mathrm{KF}\) is electrolyzed by a current of \(10.0 \mathrm{~A}\) for \(2.00 \mathrm{~h} ?\) What mass of potassium metal is produced? At which electrode does each reaction occur?

Short answer

In the electrolysis of molten KF, approximately 9.13 L of F2 gas is produced, and approximately 29.2 g of potassium metal is formed when a current of 10.0 A is applied for 2.00 hours. The potassium ions (K+) are reduced to potassium metal (K) at the cathode, while the fluoride ions (F-) are oxidized to form fluorine gas (F2) at the anode.

Step-by-step solution

1. Convert time to seconds

We are given the time in hours, which needs to be converted to seconds: \(2.00~h × \frac{3600~s}{1~h} = 7200~s\)

2. Calculate the total charge

Next, we will calculate the total charge that passed through the electrolytic cell: Charge = Current × Time Charge = \(10.0~A × 7200~s = 72000~C\)

3. Determine moles of electrons transferred

Using Faraday's Constant, we can determine the moles of electrons transferred: Moles of electrons = \(\frac{72000~C}{96485~C/mol} ≈ 0.746~mol\)

4. Find moles of F2 and K produced

The half-reactions for the electrolysis of molten KF are as follows: Cathode: \(K^+ + e^- \rightarrow K\) Anode: \(2F^- \rightarrow F_{2} + 2e^-\) For every 2 moles of electrons, 1 mole of F2 is produced at the anode, while 1 mole of K is produced at the cathode for every mole of electrons. Therefore: Moles of F2 = \(\frac{0.746}{2} ≈ 0.373~mol\) Moles of K = \(0.746~mol\)

5. Determine the volume of F2 gas

Using the Ideal Gas Law (PV = nRT) and the conditions given (T = 25°C = 298.15 K and P = 1.00 atm), we can calculate the volume of F2 gas: Volume = \(\frac{nRT}{P}\) Volume = \(\frac{0.373~mol × 0.0821~\frac{L·atm}{mol·K} × 298.15~K}{1.00~atm} ≈ 9.13~L\)

6. Calculate the mass of potassium metal

We can now determine the mass of potassium metal produced using the molar mass of K (39.1 g/mol): Mass of K = moles of K × molar mass of K Mass of K = \(0.746~mol × 39.1~\frac{g}{mol} ≈ 29.2~g\)

7. Identify the electrodes

At the cathode, potassium ions (K+) are reduced to potassium metal (K). At the anode, fluoride ions (F-) are oxidized to form fluorine gas (F2). In summary, the volume of F2 gas produced is approximately 9.13 L, the mass of potassium metal produced is approximately 29.2 g, and the reactions occur at the cathode and anode as described above.

Key concepts

Faraday's Constant

The intricacies of electrolysis depend largely on an essential constant known as Faraday's Constant. In essence, this constant represents the charge on one mole of electrons, which is approximately 96,485 coulombs. This may sound complex, but imagine it as a bridge that connects the world of electricity (coulombs) with chemical substances (moles).

During electrolysis, like the one involving molten potassium fluoride (KF), we track the electric charge that has been utilized to cause a chemical change. This transition from electrical energy to chemical transformation can be quantified using Faraday's Constant. To calculate the amount of substance altered in a reaction at an electrode, you'd use the formula:
Mo0;@&\(\text{Moles of electrons} = \frac{\text{Total Charge (C)}}{\text{Faraday's Constant (C/mol)}}\).

For instance, with the total charge of 72,000 coulombs passed during the electrolysis of KF, you can determine how many moles of electrons contributed to the reaction. Soon, that calculation leads you to the amount of F2 gas and potassium metal produced. Remember, each mole of electrons reduces to a certain quantity of the chemical substance at the electrodes, providing a predictable and measurable outcome.

Ideal Gas Law

In the context of electrolysis, gases can form, and to grasp their behavior under certain conditions, we turn to the Ideal Gas Law. This formula, PV = nRT, is the cornerstone for understanding gases in chemistry. It's a phenomenally flexible and simple tool allowing us to connect the dots between pressure (P), volume (V), the amount of substance (n in moles), the ideal gas constant (R), and temperature (T in Kelvin).

For your electrolysis problem, determining the volume of F2 gas produced at a specific temperature and pressure requires using this law. By plugging in the moles of F2 gas, the temperature in Kelvin (remember to convert from Celsius!), the pressure in atm, and the ideal gas constant (0.0821 L·atm/mol·K), the volume the gas would occupy can be calculated. It's like solving a puzzle - if you have four pieces and are looking for the fifth, the Ideal Gas Law gives you the complete picture of how a gas behaves under set conditions.

Redox Reactions

Electrolysis involves fascinating chemical reactions called redox reactions, where reduction and oxidation occur simultaneously. To put it simply, one substance gives away electrons (oxidation), while another gains electrons (reduction). In your electrolysis example, the fluoride ions lose electrons to become fluorine gas (oxidation), while potassium ions gain electrons to form potassium metal (reduction).

The reactions at both the cathode and the anode are classic examples of half-reactions in a redox process - one cannot occur without the other. By observing the transfer of electrons, we notice that for every two electrons that move, one F2 molecule forms, and for each electron, a K atom appears. It's like a dance of atoms and charges, elegant and perfectly balanced, ensuring the conservation of charge and mass in a chemical process. Redox reactions are the heart of many such practical applications beyond the textbook, driving processes from battery operation to metabolic reactions within our cells.