Problem 76 - Chemistry: An Atoms-Focused Approach

Question

How long does it take to deposit a coating of gold \(1.00 \mu \mathrm{m}\) thick on a disk-shaped medallion \(4.0 \mathrm{cm}\) in diameter and \(2.0 \mathrm{mm}\) thick at a constant current of \(85 \mathrm{A} ?\) The density for the electroplating process is \(19.3 \mathrm{g} / \mathrm{cm}^{3} .\) The electroplating solution contains gold(111).

Step-by-Step Solution

Answer

Answer: The time required to deposit the gold coating is approximately 4.19 x 10^{-6} seconds.

Step 1: 1. Determine the volume of gold needed for the coating

First, we need to find the volume of the gold coating. The coating is on the surface of the disk-shaped medallion and has a thickness of \(1.00 \mu \mathrm{m}\). The surface area of the medallion can be calculated using the formula of circle's area: \(A = \pi r^2\), where \(r\) is the radius, and diameter \(D\) is equal to \(2r\). Radius: \(r = \frac{D}{2} = \frac{4.0\,\text{cm}}{2} = 2.0\,\text{cm}\) Surface area: \(A = \pi(2.0\,\text{cm})^2 = 12.57 \, \text{cm}^2\) Volume of gold needed: \(V = A \times t = 12.57\,\text{cm}^2 \times 1.00\,\mu \text{m} = 12.57\,\text{cm}^2\times10^{-8}\,\text{cm}=1.257\times10^{-8}\,\text{cm}^3\)

Step 2: 2. Calculate the mass of gold required

Now that we have the volume of gold needed for the coating, we can use the density of the electroplating process (\(19.3 \, \text{g/cm}^3\)) to find the mass of gold required. Mass of gold: \(m = V \times \rho = 1.257\times10^{-8}\, \text{cm}^3\times 19.3\, \text{g/cm}^3 = 2.426\times10^{-7}\,\text{g}\)

Step 3: 3. Find the number of moles of gold

Next, we need to find the number of moles of gold required. This can be done using the molar mass of gold, which is \(197.0\, \text{g/mol}\). Number of moles of gold: \(n = \frac{m}{M} = \frac{2.426\times10^{-7}\,\text{g}}{197.0\, \text{g/mol}} = 1.23\times10^{-9}\,\text{mol}\)

Step 4: 4. Determine the charges needed for the electroplating process

Now, we need to determine the total charge required for the electroplating process. Gold(III) has a charge of \(3+\). Using Faraday's Law, the charge in Coulombs needed to deposit one mole of gold(III) is: Charge: \(Q = n \times F \times Z = 1.23\times10^{-9}\, \text{mol} \times 96485\, \text{C/mol} \times 3 = 3.56\times10^{-4}\, \text{C}\)

Step 5: 5. Calculate the time required using the given current

Finally, we can find the time required for electroplating using the given current of \(85\,\text{A}\), which is the rate of flow of charge. Time: \(t = \frac{Q}{I} = \frac{3.56\times10^{-4}\,\text{C}}{85\,\text{A}} = 4.19\times10^{-6}\, \text{s}\) The time it takes to deposit a coating of gold \(1.00 \mu \mathrm{m}\) thick on the disk-shaped medallion at a constant current of \(85 \mathrm{A}\) is approximately \(4.19\times10^{-6}\) seconds.

Key Concepts

Faraday's Law

Faraday's Law is a fundamental principle in electrochemistry that helps us understand the process of electroplating. This law states that the amount of substance deposited at an electrode during electrolysis is directly proportional to the amount of electric charge passed through the circuit. Faraday's Law can be mathematically expressed as:\[ Q = n imes F \]Where:

• \( Q \) is the total charge in Coulombs
• \( n \) is the number of moles of electrons
• \( F \) is Faraday's constant, approximately 96,485 C/mol, representing the charge of one mole of electrons

This law allows us to calculate how long it will take to deposit a substance on a surface when we know the current and the required thickness of the coating.

Gold Coating Thickness

To determine the thickness of gold coating, we first need to calculate the volume of the gold that will coat the item, in this case, a medallion. The thickness given is very small, measured in micrometers (\(1 \mu \text{m} = 1 \times 10^{-6} \text{m}\)), which signifies precision in the electroplating process.The surface area of the disk-shaped medallion is calculated using the formula for the area of a circle:\[ A = \pi r^2 \]This formula helps us find how much area the gold will cover. Then, by multiplying this area by the thickness of the gold, we can determine the volume of gold needed:\[ V = A \times t \]Understanding these calculations allows for accurate predictions of how much material is necessary for desired thickness.

Mass and Density Calculations

Calculating the mass of gold required for the coating involves knowing the density of gold used in electroplating (given as \(19.3 \text{g/cm}^3\)). Density is defined as mass per unit volume and can be used to find mass when volume is known.The formula for density is:\[ \rho = \frac{m}{V} \]Rearranging the formula to solve for mass \( m \):\[ m = \rho \times V \]In our problem, multiplying the volume of gold by its density gives us the mass of gold needed for coating. This step is crucial to proceed with subsequent electrochemical calculations.

Electrochemistry Calculations

Electrochemistry combines chemistry and electricity. During electroplating, an electric current reduces cations of a desired material from a solution, coating a surface with a thin layer of that material.The last steps include calculating time for electroplating using the formula:\[ t = \frac{Q}{I} \]Where:

• \( t \) is time in seconds
• \( Q \) is total charge, calculated using Faraday’s Law
• \( I \) is current in amperes

This relation helps us determine the duration needed for the gold to be adequately deposited based on the current supplied. Understanding these calculations is key for anyone looking to explore electroplating applications.