Problem 92 - Chemistry The Central Science In Si Units

Question

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of 4.55 A through molten \(\mathrm{MgCl}_{2}\), for 4.50 days? (b) How many minutes are needed to plate out \(25.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\) using \(3.50 \mathrm{~A}\) of current?

Step-by-Step Solution

Answer

(a) Approximately 89.54 g of Mg is formed. (b) It takes about 12227.96 minutes to plate out 25.00 g of Mg with 3.50 A.

Step 1: Determine Charge Passed

First, calculate the total charge passed through the electrolyte. Use the formula \( Q = I \times t \) where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is time in seconds. For part (a), \( I = 4.55 \) A and \( t = 4.50 \) days. Convert days to seconds: \[ t = 4.50 \times 24 \times 60 \times 60 \text{ s} \] Now calculate the charge: \[ Q = 4.55 \times (4.50 \times 24 \times 60 \times 60) \text{ C} \]

Step 2: Calculate Moles of Electrons

Use Faraday's constant, \( F = 96500 \text{ C/mol} \), to calculate the moles of electrons. The formula is: \[ n = \frac{Q}{F} \]Substituting the charge from Step 1: \[ n = \frac{4.55 \times 4.50 \times 24 \times 60 \times 60}{96500} \text{ moles} \]

Step 3: Determine Moles of Mg Produced

In the reaction \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \), 2 moles of electrons are required for 1 mole of magnesium. Divide the moles of electrons from Step 2 by 2: \[ \text{moles of Mg} = \frac{n}{2} \]

Step 4: Calculate Mass of Mg Formed

To find the mass of magnesium formed, multiply the moles of magnesium by its molar mass (24.31 g/mol). \[ \text{mass of Mg} = \text{moles of Mg} \times 24.31 \text{ g/mol} \]

Step 5: Determine Charge for 25.00 g of Mg (Part b)

For part (b), calculate the moles of Mg required using its molar mass: \[ \text{moles of Mg} = \frac{25}{24.31} \]Then find the moles of electrons needed, which is double that of Mg moles. Calculate the total charge using: \[ Q = n \times F \]

Step 6: Calculate Time for Plating (Part b)

Use the formula \( Q = I \times t \) to find time in seconds. Given \( I = 3.50 \) A, rearrange to solve for \( t \): \[ t = \frac{Q}{3.50} \]Convert the time from seconds to minutes by dividing by 60: \[ t = \frac{Q}{3.50 \times 60} \]

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a fundamental principle in electrochemistry that relates the amount of substance produced at an electrode to the total charge passed through the electrolyte. This law is essential when calculating the amount of a chemical substance that can be produced or consumed in an electrochemical process.
Faraday's Law is given by \(m = \frac{Q}{F \cdot n}\), where:

• \(m\) is the mass of the substance (in grams),
• \(Q\) is the total electric charge (in coulombs),
• \(F\) is Faraday's constant (approximately 96500 C/mol),
• \(n\) is the number of moles of electrons required to produce one mole of the substance.

For the electrolysis of magnesium chloride to produce magnesium, Faraday's Law helps us understand the relationship between the electrical energy input and the quantity of magnesium obtained. By knowing the total charge and the stoichiometry of the reaction, we can determine the mass of magnesium produced.

Molten Salt Electrolysis

Molten Salt Electrolysis is a process used in the production of metals such as magnesium. Unlike aqueous electrolysis, where the electrolyte is dissolved in water, molten salt electrolysis involves electrolytes in a molten state. This is crucial for metals like magnesium that react with water.
In the case of magnesium chloride (\(\mathrm{MgCl}_{2}\)), the salt is melted at high temperatures. Once in the molten state, the \(\mathrm{MgCl}_{2}\) ions are free to move, allowing the current to pass through. At the cathode, magnesium ions (\(\mathrm{Mg}^{2+}\)) gain electrons (a process called reduction) to form magnesium metal \(\mathrm{Mg}\), while at the anode, chloride ions (\(\mathrm{Cl}^-\)) lose electrons (oxidation) to form chlorine gas.
This method is highly effective for extracting metallic magnesium since it avoids unnecessary chemical reactions with solvents like water. Increased temperature ensures that kinetic and electrical energies are sufficient for sustaining electrode reactions, making it a versatile industrial process.

Current and Charge Calculations

Calculating the total charge passed through an electrolyte is a straightforward process that involves the manipulation of basic formulas. Knowing the current and the duration the current flows are key components.
The formula \( Q = I \times t \) is used, where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds.

Let's consider part (a) of the given exercise, where a current of 4.55 A is applied for 4.50 days. First, we convert the time into seconds by using \( t = 4.50 \times 24 \times 60 \times 60 \). Multiplying this by the current gives us the total charge.
For part (b), involving the electroplating of magnesium, the same approach is used. However, here, the total charge is needed to be calculated based on the quantity of magnesium plated, and therefore the formula is rearranged to solve for time.

Molar Mass and Stoichiometry

Molar Mass and Stoichiometry are essential concepts when determining how much product is generated during a chemical reaction. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). For magnesium, the molar mass is approximately 24.31 g/mol. This value is vital in converting between moles and grams.
Stoichiometry allows us to relate the quantities of reactants and products in a chemical reaction using balanced equations. In the reduction of magnesium ions during electrolysis, the reaction \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \) illustrates that two moles of electrons are needed to produce one mole of magnesium.
Understanding these concepts enables an efficient transition from moles to grams. In practice, when we derive the number of moles via charge calculations, molar mass allows us to find the mass of the magnesium produced. This step is crucial in both part (a) and part (b) of the exercise to convert theoretical moles into practical, usable grams of the product.