Problem 107 - Chemistry: The Central Science

Question

(a) How many coulombs are required to plate a layer of chromium metal \(0.25 \mathrm{~mm}\) thick on an auto bumper with a total area of \(0.32 \mathrm{~m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in \(10.0 \mathrm{~s} ?(\mathrm{c})\) If the external source has an emf of \(+6.0 \mathrm{~V}\) and the electrolytic cell is \(65 \%\) efficient, how much electrical power is expended to electroplate the bumper?

Step-by-Step Solution

Answer

To plate a layer of chromium metal with given thickness and area, (a) 320,281 C of charge are required. (b) The current flow required for this electroplating in 10.0 s is 32,028 A. (c) Considering a 65% efficient electrolytic cell, the electrical power expended to electroplate the bumper is 124.9 kW.

Step 1: (a) Calculate the volume and mass of Chromium metal

First, calculate the volume of the chromium layer to be plated by multiplying the thickness with the area of the bumper. Since the thickness is given in mm and the area is in m², we need to convert the thickness to meters before multiplication: Thickness = 0.25 mm = 0.00025 m Area = 0.32 m² Volume = Thickness × Area = 0.00025 m × 0.32 m² = 8.0 × 10⁻⁵ m³ Next, convert the volume to cm³: Volume = 8.0 × 10⁻⁵ m³ × (10⁵ cm³/m³) = 8.0 cm³ Now, we calculate the mass of chromium using its density: Density = 7.20 g/cm³ Mass = Density × Volume = 7.20 g/cm³ × 8.0 cm³ = 57.6 g

Step 2: (a) Convert the mass into moles of Chromium

Calculate the moles of Chromium using its molar mass (51.996 g/mol): Moles = Mass / Molar_mass Moles = 57.6 g / 51.996 g/mol = 1.107 mol

Step 3: (a) Calculate the required coulombs

Now, we calculate the number of coulombs needed to plate chromium. The formula for this is: Coulombs = Moles × Faraday_constant × (Molar_ratio) Faraday's constant (F) is 96,485 C/mol, and the molar ratio (charge of chromium) is 3 because the chromium ion is in the form of Cr³⁺. Coulombs = 1.107 mol × 96,485 C/mol × 3 = 320,281 C

Step 4: (b) Calculate the required current flow

To calculate the current flow required for electroplating, use the formula: Current (I) = Coulombs / Time Time is given as 10.0 s. Current = 320,281 C / 10.0 s = 32,028 A

Step 5: (c) Calculate the electrical power expended

To calculate the electrical power expended, first, we find the actual amount of power that would be used in a 65% efficient process by dividing the emf by the efficiency: Actual_emf = (emf × Efficiency) / 100 Actual_emf = (6.0 V × 65) / 100 = 3.9 V Now, we calculate the electrical power expended using the formula: Power (P) = Actual_emf × Current Power = 3.9 V × 32,028 A = 124,909.2 W or 124.9 kW

Key Concepts

Coulombs

The concept of coulombs is central to understanding electroplating. A coulomb is a unit of electrical charge, necessary to quantify the amount of electricity needed for chemical reactions like electroplating. When we aim to plate a metal, in this case, chromium, we need a certain amount of charge to deposit a specific mass.

To find out how many coulombs are required, we start by determining the moles of chromium we want to deposit. Using the molar mass, we convert the mass of chromium to moles. This helps us figure out how many moles of electrons are needed.

Finally, using the formula:

• Coulombs = Moles × Faraday's constant × (Molar_ratio)

we calculate the required charge. For chromium, in the ionic state of Cr³⁺, the molar ratio is 3. This means each mole of chromium requires three moles of electrons, translating our calculation into 320,281 coulombs.

Faraday's Constant

Faraday's constant is a key player in electrochemistry, especially in processes like electroplating. It is the charge of one mole of electrons and is approximately 96,485 coulombs per mole.

The constant is crucial because it helps relate the number of electrons, in moles, to the actual charge in coulombs needed for reactions. In our chromium plating scenario, this constant helps us understand how much electrical charge will drive the deposition of chromium on the bumper.

If one mole of electrons equates to 96,485 coulombs, and knowing the molar ratio for chromium, we can directly compute the total charge needed for the task.

Current flow

Current flow is the rate at which charge passes through a point in an electrical circuit. It's measured in amperes (A), and relates to how fast we can achieve the electroplating.

By using the formula:

• Current (I) = Coulombs / Time

we find out the amount of current needed over a given period, in this case, 10 seconds. This calculation lets us see how robust the electrical system must be to handle that amount of current in a short span.

Here, we determined a need for 32,028 amperes, showing how intense the process can be in industrial settings.

Electrical Power

Electrical power, a crucial concept in electroplating, refers to the rate of energy consumption or production in an electrical system. It's measured in watts (W).

For the job of plating the bumper, power needs are evaluated by combining voltage (emf) and current, using the formula:

• Power (P) = Actual_emf x Current

Here, power becomes a product of how much "push" (voltage) is provided and how much "flow" (current) is needed.

It's noteworthy that power efficiency impacts the actual power used, factoring in the efficiency percentage to understand the true power requirement for our plating task, calculated as 124.9 kW.

Efficiency

Efficiency in electroplating measures how well the system converts electrical energy into the desired chemical deposition. A process is rarely 100% efficient due to energy losses, often as heat.

In our problem, the electroplating process is 65% efficient, meaning only 65% of the electrical input results in chromium plating, with the rest being wasted. We calculate the effective expenditure of power by adjusting the voltage considering efficiency:

• Actual_emf = (emf × Efficiency) / 100

This allows us to accurately assess the power needed with existing inefficiencies and to understand the cost and energy implications in manufacturing settings.