Question

Question:What volume of F₂ gas, at 25°C and 1.00 atm, is produced when molten KF is electrolyzed by a current of10.0 A for 2.00 h? What mass of potassium metal isproduced? At which electrode does each reaction occur?

Short answer

Thevolume of gas and mass produced are 9.13 L & 29.2 g.

Step-by-step solution

 Step 1: Given Information

The current is 10 A. The time value is 1 hr. The temperature is 25^oC. Pressure is 1 atm. The reduction reaction can be denoted as:

${\text{K}}^{+}{\text{+e}}^{-}\text{→K}$

The reaction at the anode can be given as:

$2{\text{F}}^{-}\left(\text{I}\right){\text{→F}}_2\left(\text{g}\right){\text{+2e}}^{-}$

The total charge and the moles of fluorine

The amount of charge in coulombs which flows in 1 second is the electric current in Amps.

The total charge passed can be given as:

$\text{Q=I×t}$

The hours can be converted to seconds as:

$$\begin{gathered} \text{Q=10.0×2×60×60} \\ =7.20\times {10}^4\text{C} \end{gathered}$$

1 mole of ${\text{F}}_2$ gives 2 moles of electrons.

The charge carried by 1 mole of electrons is:

Charge carried by one mole of electrons$=9.648\times {10}^4\text{C}$

This is known as the Faradays constant.

The number of coulombs for 1 mole of ${\text{F}}_2$ is:

$$\begin{gathered} {\text{Numbersofcoulombs=2×9.648×10}}^4 \\ {\text{=19.296×10}}^4\text{C} \end{gathered}$$

$$\begin{gathered} 19.296\times {10}^4{\text{Cdischarges1moleofF}}_2\text{.} \\ {\text{ThemolesofF}}_2{\text{dischargedby7.20×10}}^4\text{Cofcurrentcanbegivenas:} \\ {\text{MolesofF}}_2\text{=}\frac{7.20\times {10}^4}{19.296\times {10}^4} \\ \text{=0.373moles} \end{gathered}$$

Finding the volume of gas

1 mole of fluorine has a volume of 24.465 L.

The volume of fluorine for 0.373 moles is:

$$\begin{gathered} {\text{VolumeofF}}_2\text{=24.465×0.37313} \\ =9.13\text{L} \end{gathered}$$

From the cathode equation, you can see that:

1 mole of K needs 1 mole of electrons.

Hence, 1 mole of K needs $9.648\times {10}^4\text{C}$ of electrons.

The atomic weight of K is 39.098 g.

$9.648\times {10}^4\text{C}$discharges 39.098 g potassium.

$$\begin{gathered} 1\text{Cdischarge}\frac{39.098}{9.648\times {10}^4}\text{gK} \\ {\text{Theamountofchargedischargedby7.20×10}}^4\text{Ccanbegivenas:} \\ \text{Amountofcharge=}\frac{39.098}{9.648\times {10}^4}{\text{×7.2×10}}^4\text{gK} \\ =29.2\text{gK} \end{gathered}$$