Problem 31 - Chemistry The Molecular Science

Question

To produce magnesium metal, \(1000 . \mathrm{kg}\) molten \(\mathrm{MgCl}_{2}\) are electrolyzed. (a) At which electrode is magnesium produced? (b) What is produced at the other electrode? (c) Calculate the amount (mol) of electrons used in the process. (d) An industrial process uses \(8.4 \mathrm{kWh}\) per pound of \(\mathrm{Mg}\). Calculate the energy (J) required per mole of magnesium.

Step-by-Step Solution

Answer

(a) Magnesium is produced at the cathode. (b) Chlorine gas is produced at the anode. (c) 21000 mol of electrons are used. (d) 1620000 J/mol are required.

Step 1: Identify Magnesium Production Electrode

During electrolysis, magnesium metal is produced at the cathode. This is because the cathode is the site where reduction occurs, and magnesium ions \((\text{Mg}^{2+})\) gain electrons to form magnesium metal \((\text{Mg})\).

Step 2: Determine Product at Other Electrode

At the anode, oxidation takes place. With \(\text{MgCl}_2\), chloride ions \((\text{Cl}^- )\) lose electrons to form chlorine gas \((\text{Cl}_2)\).

Step 3: Calculate the Electron Moles Used

For every mole of \(\text{MgCl}_2\), 2 moles of electrons are necessary because \(\text{Mg}^{2+}\) requires 2 electrons for reduction. If we have 1000 kg of \(\text{MgCl}_2\), first convert it to moles: \[\text{Molar mass of } \text{MgCl}_2 = 24.31 + 2 \times 35.45 = 95.21 \text{ g/mol}\]\[\text{Moles of } \text{MgCl}_2 = \frac{1000 \times 10^3}{95.21} \approx 10500 \text{ mol}\]Thus, \[\text{Moles of electrons} = 2 \times 10500 = 21000 \text{ mol}\]

Step 4: Calculate Energy per Mole of Magnesium

The energy usage is given as \(8.4\text{ kWh per pound}.\) First, convert this to joules per pound:\[8.4 \text{ kWh} = 8.4 \times 3600 \times 1000 \text{ J} = 30240 \text{ kJ/pound}\]Convert pounds to grams:\[1 \text{ lb} = 453.592 \text{ g}\]Energy per gram: \[\frac{30240}{453.592} \approx 66.67 \text{ kJ/g} \]Finally, convert to moles using the molar mass of magnesium \((24.31 \text{ g/mol})\):\[66.67 \text{ kJ/g} \times 24.31 \text{ g/mol} \approx 1620 \text{ kJ/mol} = 1620000 \text{ J/mol}\]

Key Concepts

Electrode Reactions

Electrode reactions are fundamental in understanding electrolysis processes, especially when producing metals like magnesium. In the electrolysis of molten magnesium chloride (\(\text{MgCl}_2\)), the reactions at the electrodes involve the movement of electrons to create products. At the cathode, which carries a negative charge, reduction occurs. Magnesium ions \((\text{Mg}^{2+})\) gain electrons to form magnesium metal \((\text{Mg})\). This reaction can be simplified as \[\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}\].
On the other hand, at the anode, which carries a positive charge, oxidation takes place. Chloride ions \((\text{Cl}^- )\) lose electrons to form chlorine gas \((\text{Cl}_2)\). This is represented by the equation \[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\].
Through these electrode reactions, the transfer of electrons not only facilitates chemical changes but also results in the separation of magnesium metal and chlorine gas, showcasing the role of electrochemical processes in industrial production.

Chemical Calculations

Understanding the chemical calculations involved in the electrolysis of magnesium chloride is crucial for determining quantities like the amount of electrons used. In this process, every mole of magnesium chloride \((\text{MgCl}_2)\) requires two moles of electrons for the reduction of magnesium ions. Calculations begin with determining how many moles of \(\text{MgCl}_2\) are present in 1000 kg of the compound.
The molar mass of \(\text{MgCl}_2\) is \(95.21\text{ g/mol}\), calculated by adding the atomic mass of magnesium (24.31 g/mol) and twice the atomic mass of chlorine (2 \(\times\) 35.45 g/mol). With 1000 kg of \(\text{MgCl}_2\) corresponding to 1,000,000 grams, we find the number of moles of \(\text{MgCl}_2\) by dividing the mass by its molar mass: \[\frac{1000 \times 10^3}{95.21} \approx 10500 \text{ mol}\].
This results in 21000 moles of electrons, as each mole of \(\text{MgCl}_2\) provides 2 moles of electrons. Such calculations are integral in understanding the material and energy balance of the industrial electrolysis process.

Industrial Electrolysis

Industrial electrolysis is a pivotal technique in extracting and producing valuable metals. For magnesium production, electrolysis of magnesium chloride provides an efficient method. This process is energy-intensive, and understanding the energy consumption is key to optimizing and scaling industrial applications. According to the exercise, the process consumes \(8.4\text{ kWh}\) per pound of magnesium produced.
To compare this with energy consumed per mole, we convert \(8.4\text{ kWh}\) into joules: \(8.4 \times 3600 \times 1000 \text{ J} = 30240 \text{ kJ/pound}\). A pound is equivalent to approximately \(453.592\text{ g}\), and calculating energy per gram gives us \(\frac{30240}{453.592} \approx 66.67 \text{ kJ/g}\).
Considering the molar mass of magnesium as \(24.31 \text{ g/mol}\), the energy per mole is \(1620 \text{ kJ/mol} \equiv 1620000 \text{ J/mol}\). This contextualizes the high energy requirements of industrial electrolysis and emphasizes the need for efficiencies and innovations in energy usage within the industry.