Question

A Electrolysis of molten NaCl is done in cells operating at \(7.0 \mathrm{V}\) and \(4.0 \times 10^{4} \mathrm{A}\). What mass of \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) can be produced in 1 day in such a cell? What is the energy consumption in kilowatt-hours? ( \(1 \mathrm{kWh}=3.6 \times 10^{6} \mathrm{J}\) and \(1 \mathrm{J}=\) \(1 \mathrm{C} \cdot \mathrm{V}\)

Short answer

The cell produces 824.964 kg of Na(s) and 1269.624 kg of Cl2(g) with an energy consumption of 6710.0 kWh in one day.

Step-by-step solution

Calculate Total Charge

First, we need to find the total charge passed through the electrolysis cell in one day. The formula for charge \(Q\) is given by \(Q = I \times t\), where \(I\) is the current in amperes and \(t\) is the time in seconds. Given \(I = 4.0 \times 10^4 \) A and \(t = 1\) day = \(24 \times 60 \times 60\) s: \[ Q = 4.0 \times 10^4 \, \text{A} \times 24 \times 60 \times 60 \, \text{s} = 3.456 \times 10^9 \, \text{C}\]

Calculate Moles of Electrons

Electrolysis reactions require electrons, and the number of moles of electrons can be calculated using the charge from Step 1. By Faraday's constant \(F = 96485\) C/mol, the number of moles of electrons \(n_e\) is:\[ n_e = \frac{Q}{F} = \frac{3.456 \times 10^9 \, \text{C}}{96485 \, \text{C/mol}} \approx 35824.5 \, \text{mol}\]

Calculate Mass of \(\mathrm{Na}\) Produced

From the reaction \(2 \mathrm{Na}^+ + 2 e^- \rightarrow 2 \mathrm{Na(s)}\), 1 mole of \( \mathrm{Na} \) requires 1 mole of electrons. Thus, using the number of moles of electrons:\[ \text{Moles of Na} = 35824.5 \, \text{mol}\]Since the molar mass of \( \mathrm{Na} \) is 23.0 g/mol, the mass produced is:\[ \text{Mass of Na} = 35824.5 \, \text{mol} \times 23.0 \, \text{g/mol} \approx 824964 \, \text{g} = 824.964 \, \text{kg}\]

Calculate Mass of \(\mathrm{Cl}_2\) Produced

The reaction for chlorine gas is \(2 \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2 e^-\), so 1 mole of \(\mathrm{Cl}_2\) requires 2 moles of electrons. The moles of \(\mathrm{Cl}_2\) produced are:\[ \text{Moles of Cl}_2 = \frac{35824.5}{2} \approx 17912.25 \, \text{mol}\]With a molar mass of \( \mathrm{Cl}_2 = 70.9 \) g/mol:\[ \text{Mass of Cl}_2 = 17912.25 \, \text{mol} \times 70.9 \, \text{g/mol} \approx 1269624 \, \text{g} = 1269.624 \, \text{kg}\]

Calculate Energy Consumption

The energy consumed (in Joules) is calculated by \(W = V \times Q\), where \(V = 7.0\) V is the voltage. Thus:\[ W = 7.0 \, \text{V} \times 3.456 \times 10^9 \, \text{C} = 2.4192 \times 10^{10} \, \text{J}\]To convert Joules to kWh (where \(1 \text{kWh} = 3.6 \times 10^6 \, \text{J}\)):\[ \text{Energy in kWh} = \frac{2.4192 \times 10^{10} \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} \approx 6710.0 \, \text{kWh}\]

Key concepts

Faraday's Constant

Faraday's Constant is a critical number in electrochemistry, representing the charge of one mole of electrons. It is typically expressed as \( 96485 \) C/mol. This means that every mole of electrons carries \( 96485 \) coulombs of electric charge.

When calculating the amount of substance involved in an electrochemical reaction, Faraday's Constant is used to convert electrical charge into moles of electrons. For instance, if you know the total charge passed through an electrolysis cell, you can determine how many moles of electrons were used by dividing the charge by Faraday’s Constant.

Understanding Faraday's Constant is essential for anyone studying chemistry at an advanced level. It allows us to directly relate the macroscopic current flowing in a circuit to the microscopic process of electron transfer.

Moles of Electrons

The term "moles of electrons" refers to the amount of electron transport during an electrochemical reaction. In electrolysis, this concept is fundamental for determining how much of a substance can be produced from electrical energy.

Using the previously calculated total charge from an electrolysis process, we can find out the moles of electrons by applying Faraday's Constant. The formula is simple: divide the total charge \( Q \) by \( F \), Faraday’s Constant. Here, \( n_e = \frac{Q}{F} \).

This value is crucial in stoichiometry of electrolysis reactions. It lets us calculate how much of a chemical element can be produced or consumed in a given time under specified electrical conditions. In our exercise, understanding this helped us find the moles of sodium and chlorine gas produced.

Energy Consumption

Energy consumption in electrolysis processes refers to the amount of energy required to drive a chemical reaction. It is particularly relevant in industrial applications where minimizing energy use is essential for cost-effectiveness.

In the exercise, we calculated the energy consumed during the electrolysis of molten sodium chloride by using the formula \( W = V \times Q \), where \( W \) is the work done (or energy), \( V \) stands for the voltage, and \( Q \) is the charge. The resulting energy in joules can be converted into kilowatt-hours (kWh), a more common unit of energy in everyday life. For this, divide the total energy in joules by \( 3.6 \times 10^6 \) (since 1 kWh = 3.6 million joules).

Understanding this concept allows industries to optimize their processes and reduce environmental impact by minimizing energy usage.

Sodium Production

Sodium production through electrolysis involves converting sodium ions into solid sodium by exchanging electrons. In the reaction given by \( 2 \mathrm{Na}^+ + 2 e^- \rightarrow 2 \mathrm{Na(s)} \), each sodium ion gains an electron to become neutral sodium.

The measure of moles of electrons tells us directly how many moles of sodium ions can be converted. Since each mole of sodium requires one mole of electrons, this is a straightforward calculation.

After determining the moles of sodium produced, we can further calculate the mass by multiplying the moles by the molar mass of sodium (23.0 g/mol). Knowing the mass allows industries to determine not only the feasibility of a process but also calculate costs and potential yields. In our discussion, understanding this conversion helps us identify how much elemental sodium can be expected after a day of continuous electrolysis.