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

Given that the current flows for 2.00 hours, we can convert this time to seconds by using the following conversion factor: \(1 \thinspace hour = 3600 \thinspace seconds\). Now, we can multiply 2.00 hours with the conversion factor: \[2.00 \thinspace hours * 3600 \thinspace \frac{seconds}{hour} = 7200 \thinspace seconds\]

Now we need to find the total charge passed through the electrolyte during electrolysis. The given current is 10.0 A, and current is the rate of flow of electric charge. We can use the formula: \[Charge = Current * Time\] Now we can plug in our values for current (10.0 A) and time (7200 s): \[Charge = 10.0 \thinspace A \times 7200 \thinspace s = 72000 \thinspace C\]

We know that the formation of F2 gas involves two electrons because: \[2 \thinspace F^{-1} \rightarrow 1 \thinspace F_{2} + 2 \thinspace e^{-}\] Using Faraday's Law, which states that the amount of substance produced is directly proportional to the charge (in Columbs) passed through the electrolyte, we can determine the number of moles of produced F2. \[moles \thinspace of \thinspace F2 = \frac{Charge}{2 \times Faraday's \thinspace constant}\] Faraday's constant is approximately 96485 C/mol. Now we can plug in our values for charge (72000 C) and Faraday's constant: \[moles \thinspace of \thinspace F2 = \frac{72000 \thinspace C}{2 \times 96485 \thinspace C/mol (e^{-})} = 0.373 \thinspace moles\]

Now, we can use the Ideal Gas Law to find the volume of F2 gas produced: \[PV = nRT\] In this formula, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. We are given the pressure as 1.00 atm, the temperature as 25°C, and we've found the number of moles to be 0.373 moles. We need to convert the temperature to Kelvin by adding 273.15: \[T_{K} = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \thinspace K\] The gas constant is 0.0821 L atm/mol K. Now we can plug in our values for pressure, moles, gas constant, and temperature into the Ideal Gas Law: \[1.00 \thinspace atm \times V = 0.373 \thinspace moles \times 0.0821 \thinspace \frac{L \thinspace atm}{mol \thinspace K} \times 298.15 \thinspace K\] Solve for volume, V: \[V = \frac{0.373 \thinspace moles \times 0.0821 \thinspace L \thinspace atm/mol \thinspace K \times 298.15 \thinspace K}{1.00 \thinspace atm} = 9.13 \thinspace L\]

During the electrolysis of KF, potassium ions are reduced to potassium metal: \[K^{+} + e^{-} \rightarrow K\] Moles of potassium metal produced will be the same as the moles of F2 produced, which is 0.373 moles. Since the molar mass of potassium is approximately 39.10 g/mol, we can calculate the mass of potassium metal produced: \[Mass \thinspace of \thinspace potassium \thinspace metal = moles \thinspace of \thinspace potassium \times molar \thinspace mass\] Now plug in our values for moles of potassium metal (0.373 moles) and the molar mass of potassium: \[Mass \thinspace of \thinspace potassium \thinspace metal = 0.373 \thinspace moles \times 39.10 \thinspace \frac{g}{mol} = 14.57 \thinspace g\]

In the electrolysis of molten KF, the following reduction half-reactions occur: At the cathode (negative electrode), potassium ions are reduced to potassium metal, while at the anode (positive electrode), fluoride ions are oxidized to form F2 gas. In conclusion: - The volume of \(\mathrm{F}_{2}\) gas produced is 9.13 L - The mass of potassium metal produced is 14.57 g - Potassium ions are reduced to potassium metal at the cathode, while fluoride ions are oxidized to form F2 gas at the anode.