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?

To calculate the total charge, we can use the formula: \(Q = I \times t\), where \(Q\) is the total charge, \(I\) is the current, and \(t\) is the time. Given the current \(I = 10.0 \, \mathrm{A}\), and the time \(t = 2.00 \mathrm{~h}\), we can find the total charge passed through the electrolyte. First, we need to convert the time to seconds: \(t = 2.00 \mathrm{~h} \times \dfrac{3600 \mathrm{s}}{1 \mathrm{h}} = 7200 \mathrm{s}\) Now we can calculate the total charge: \(Q = I \times t = 10.0 \, \mathrm{A} \times 7200 \mathrm{s} = 72000 \mathrm{C}\)

Now that we have the total charge, we can determine the amount of substance involved by using Faraday's law of electrolysis: \(n = \dfrac{Q}{F}\) Where \(F\) is the Faraday constant (\(96485 \mathrm{\dfrac{C}{mol}}\)) and \(n\) is the number of moles of electrons involved. Thus: \(n = \dfrac{72000 \mathrm{C}}{96485 \mathrm{\dfrac{C}{mol}}} = 0.746 \mathrm{mol}\)

For molten \(\mathrm{KF}\), the cation is \(\mathrm{K}^{+}\) and the anion is \(\mathrm{F}^{-}\). The half-reactions occurring during electrolysis are: At the anode (oxidation): \(\mathrm{2F}^{-} \rightarrow \mathrm{F}_{2} + 2e^{-}\) At the cathode (reduction): \(\mathrm{K}^{+} + e^{-} \rightarrow \mathrm{K}\)

Based on the stoichiometry, the number of moles of \(\mathrm{F}_{2}\) produced will be half the number of moles of electrons (since it takes two electrons to produce one mole of \(\mathrm{F}_{2}\)): \(n(\mathrm{F}_{2}) = \dfrac{1}{2} n = \dfrac{1}{2} \times 0.746 \mathrm{mol} = 0.373 \mathrm{mol}\) Now we can use the ideal gas law to calculate the volume of \(\mathrm{F}_{2}\) at the given temperature and pressure. The molar volume of a gas is given by: \(V_{m} = \dfrac{RT}{P}\) Where \(R\) is the gas constant (\(0.0821 \mathrm{\dfrac{L \cdot atm}{mol \cdot K}}\)), \(T\) is the temperature in Kelvin, and \(P\) is the pressure in atm. First, we need to convert the temperature to Kelvin: \(T = 25^{\circ}\mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Now we can calculate the molar volume: \(V_{m} = \dfrac{(0.0821 \mathrm{\dfrac{L \cdot atm}{mol \cdot K}})(298.15 \mathrm{K})}{1.00\mathrm{~atm}} = 24.5 \mathrm{\dfrac{L}{mol}}\) Finally, using the number of moles of \(\mathrm{F}_{2}\), we can calculate the volume of \(\mathrm{F}_{2}\) gas: \(V = 0.373 \mathrm{mol} \times 24.5 \mathrm{\dfrac{L}{mol}} = 9.14 \mathrm{L}\)

Since we know that one mole of \(\mathrm{K}^{+}\) ions is reduced to one mole of potassium metal for each electron transferred, there will be the same number of moles of potassium metal produced as moles of electrons: \(n(\mathrm{K}) = n = 0.746 \mathrm{mol}\) Now we can calculate the mass of potassium produced by using the molar mass of potassium (\(39.1 \mathrm{\dfrac{g}{mol}}\)): \(m(\mathrm{K}) = n(\mathrm{K}) \times M(\mathrm{K}) = 0.746 \mathrm{mol} \times 39.1 \mathrm{\dfrac{g}{mol}} = 29.2 \mathrm{g}\)

In summary, when molten \(\mathrm{KF}\) is electrolyzed by a current of \(10.0 \, \mathrm{A}\) for \(2.00 \mathrm{~h}\): 1. The volume of \(\mathrm{F}_{2}\) gas produced is \(9.14 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\). 2. The mass of potassium metal produced is \(29.2 \mathrm{g}\). 3. The \(\mathrm{F}_{2}\) gas is produced at the anode, while the potassium metal is formed at the cathode.