Problem 80 - Chemistry The Science in Context

Question

How long does it take to deposit a coating of gold \(1.00 \mu \mathrm{m}\) thick on a disk-shaped medallion \(2.0 \mathrm{cm}\) in diameter and \(3.0 \mathrm{mm}\) thick at a constant current of \(45 \mathrm{A} ?\) The density of gold is \(18.3 \mathrm{g} / \mathrm{cm}^{3} .\) The gold solution contains gold( 111 ).

Step-by-Step Solution

Answer

Answer: It takes approximately 515 seconds to deposit the gold layer.

Step 1: Find the volume of gold deposited

Using the dimensions and thickness of the medallion and the required thickness of gold coating, we can calculate the volume of gold deposited: Volume = (Surface area of the medallion) × (Thickness of gold coating) The surface area of a disk is given by the formula \(A = \pi r^2\), where A is the surface area and r is the radius of the disk. Radius = Diameter/2 = 2.0 cm / 2 = 1.0 cm Surface area of the medallion, \(A = \pi (1.0)^2 = \pi \mathrm{cm}^2\) Thickness of gold coating = 1.00 µm = 1.00×10^{-6} cm Volume of gold = \(\pi \mathrm{cm}^2 × 1.00×10^{-6} \mathrm{cm} = \pi × 10^{-6} \mathrm{cm}^3\)

Step 2: Find the mass of gold deposited

Given the density of gold, we can now find the mass of gold deposited: Mass = (Volume of gold) × (Density of gold) Density of gold = 18.3 g/cm³ Mass = \((\pi × 10^{-6} \mathrm{cm}^3) × (18.3 \mathrm{g}/\mathrm{cm}^3) = 18.3\pi × 10^{-6} \mathrm{g}\)

Step 3: Calculate the amount of gold in moles

To find the time taken to deposit the gold, we need to determine the amount of gold in moles. We will use the molar mass of gold: Molar mass of gold = 197.0 g/mol Amount of gold in moles = (Mass of gold) / (Molar mass of gold) Moles of gold = \((18.3\pi × 10^{-6} \mathrm{g}) / (197.0 \mathrm{g/mol}) = \frac{18.3\pi}{197.0} × 10^{-6} \mathrm{mol}\)

Step 4: Apply Faraday's Law

Using Faraday's law, we can find the time required to deposit the gold: Charge = (Amount of gold in moles) × (Faraday's constant) × (Number of electrons involved in the redox reaction) Faraday's constant = 96,485 C/mol Number of electrons involved in the redox reaction for gold(III) = 3 Charge = \((\frac{18.3\pi}{197.0} × 10^{-6} \mathrm{mol}) × (96,485 \mathrm{C/mol}) × 3 = \frac{54.9\pi}{197.0} × 10^{-6} × 96,485 \mathrm{C}\)

Step 5: Find the time required

Now we can use the current and the amount of charge required to deposit the gold layer to find the time taken: Time = (Charge) / (Current) Current = 45 A Time = \((\frac{54.9\pi}{197.0} × 10^{-6} × 96,485 \mathrm{C}) / (45 \mathrm{A}) = \frac{54.9\pi}{197.0} × 10^{-6} × 96,485/45 \mathrm{s} \approx 515 \mathrm{s}\) It takes approximately 515 seconds to deposit a coating of gold 1.00 µm thick on the disk-shaped medallion.

Key Concepts

Understanding Faraday's Law

Faraday's Law of electrolysis is a fundamental principle used to determine the quantity of a substance produced at an electrode during the process of electrolysis. It states that the amount of substance discharged at an electrode is directly proportional to the amount of electric charge passed through the electrolyte. Considering the exercise in question, Faraday's Law allows us to calculate how long it will take to electroplate a specific thickness of gold onto a medallion.

Faraday's Law can be mathematically expressed as: \[ m = \frac{Q}{F} \times \frac{M}{z} \] where:

• \( m \) is the mass of the substance deposited,
• \( Q \) is the total electric charge passed through the solution,
• \( F \) is Faraday's constant (approximately 96,485 C/mol),
• \( M \) is the molar mass of the substance, and
• \( z \) is the valency number of ions of the substance (number of electrons transferred per ion).

In the gold plating problem, Faraday’s Law connects the amount of gold deposited with the charge transferred by the electrical current, ultimately giving us the time required for the deposition.

Calculating Molar Mass

The molar mass of a substance is the mass of one mole of that substance. It's an essential parameter in chemistry, particularly in stoichiometry, which relates chemical reactions to their respective quantities. When we talk about molar mass calculation, it's the process of determining the mass (in grams) per mole of a given element or compound. In this case, the molar mass of gold (Au) is 197.0 g/mol.

This value allows us to convert the mass of the gold deposited during electroplating to the amount in moles, which is critical for using Faraday’s Law. The conversion is straightforward and can be done using the equation: \[ \text{moles of substance} = \frac{\text{mass of substance}}{\text{molar mass of substance}} \] This step is crucial in the solution because it acts as a bridge between the mass of gold and the amount of electric charge required to deposit this mass, according to Faraday's Law.

Volume and Mass Relationship

The relationship between volume and mass is of paramount importance in chemistry and is used for various applications, including electroplating calculations. Volume dictates how much space an object or substance occupies, while mass measures the amount of matter it contains.

By knowing the relationship between the volume of a substance and its mass, we can quantify the mass of a substance deposited on an item during electroplating. Particularly, the density (\( \rho \)) of a material, which is mass per unit volume, comes into play to bridge these two properties. The formula to connect these two measurements is:\[ \text{mass} = \text{density} \times \text{volume} \] Within our exercise, this is used to determine the mass of gold that will be deposited based on the volume of gold needed to achieve the desired thickness and the known density of gold.

Redox Reactions in Electroplating

Redox reactions, or oxidation-reduction reactions, are the foundation of electroplating processes where one metal is deposited onto another. A redox reaction involves the movement of electrons between two substances; one species undergoes oxidation (loses electrons) while the other undergoes reduction (gains electrons).

Understanding redox reactions is necessary when applying Faraday's Law in electroplating, as the number of electrons transferred during the reaction (\( z \)) impacts the overall amount of charge needed for the deposition of a metal. For instance, in the gold electroplating scenario stated in the exercise, gold ions (\( Au^{3+} \)) are reduced (gain electrons) to form solid gold (\( Au \)) at the cathode, with three electrons involved in the process for each ion of gold. Knowing this allows us to calculate the total charge passed during the electroplating which is essential to determine the time required for the process.