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 s? (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

The number of coulombs required to plate a layer of chromium metal 0.25 mm thick on an auto bumper with a total area of 0.32 m² from a solution containing \(\mathrm{CrO}_{4}^{2-}\) is 6,407,281 C. The current flow required for this electroplating, if the bumper is to be plated in 10.0 s, is 640.728 kA. If the electrolytic cell is 65% efficient and the external source has an emf of +6.0 V, the electrical power expended to electroplate the bumper is approximately 5,910 kW.

Step 1: Calculate the volume and mass of Chromium metal required for electroplating

Given the thickness and total area of the required chromium layer, we can determine its volume. Use the density of chromium to calculate the mass of the metal required. Volume = (Total area) × (Thickness) Total area = \(0.32 m^2\) Thickness = \(0.25 mm\) = \(0.025 cm\) Volume = \((0.32 × 10^4) \text{cm}^2 × 0.025 \text{cm}\) = \(80 \text{cm}^3\) Now, let's calculate the mass. Mass = Volume × Density Density of Chromium = \(7.20 \frac{\text{g}}{\text{cm}^3}\) Mass = \(80 \text{cm}^3 × 7.20 \frac{\text{g}}{\text{cm}^3}\) = \(576 \,\text{g}\)

Step 2: Calculate the moles of Chromium metal required for electroplating

To calculate the number of coulombs needed, we first need to find the moles of Chromium metal involved in the reaction. We can use the molar mass of Chromium to determine the number of moles. Moles = \(\frac{\text{Mass}}{\text{Molar Mass}}\) Molar mass of Chromium = \(51.996 \,\text{g/mol}\) Moles = \(\frac{576 \,\text{g}}{51.996 \,\text{g/mol}}\) = \(11.078 \, \text{mol}\)

Step 3: Calculate the number of coulombs required

Now we can calculate the number of coulombs needed for electroplating using Faraday's law of electrolysis. Coulombs = Moles × Faraday's Constant × Number of electrons involved in the reaction In this case, the reaction is: \(\mathrm{CrO}_{4}^{2-} + 6e^{-} \rightarrow \mathrm{Cr} + 2 \, \mathrm{O}_{2(g)}\) Faraday's constant (F) = \(96,485 \, \frac{\text{C}}{\text{mol}}\) Number of electrons (n) = 6 Coulombs = \(11.078 \, \text{mol} × 96,485 \,\frac{\text{C}}{\text{mol}} × 6\) = \(6,407,281 \,\text{C}\)

Step 4: Calculate the required current flow for electroplating

We are given the required time for plating the bumper as 10.0 s. We can calculate the current required for the process using Ohm's Law. Current (I) = \(\frac{\text{Coulombs}}{\text{Time}}\) Current = \(\frac{6,407,281 \, \text{C}}{10.0 \, \text{s}}\) = \(640,728 \, \text{A}\) or \(640.728 \, \text{kA}\)

Step 5: Calculate the electrical power expended for electroplating

We are given the cell efficiency as 65%. Using the emf, efficiency, and current flow, we can calculate the electrical power expended for electroplating. Efficiency = \(\frac{\text{Output Power}}{\text{Input Power}}\) Output Power = 6.0 V × 640.728 kA Input Power = \(\frac{\text{Output Power}}{\text{Efficiency}}\) Input Power = \(\frac{6.0 \, \text{V}\ × 640.728 \, \text{kA}}{0.65}\) = \(6.0 \times \frac{640.728}{0.65} \, \text{kW}\) ≈ \(5,910 \, \text{kW}\) The electrical power expended to electroplate the bumper is approximately 5,910 kW.

Key Concepts

Faraday's law of electrolysis

Faraday's law of electrolysis is a fundamental principle that relates the amount of material that can be deposited at an electrode to the amount of electric charge passed through the substance. Essentially, it describes how much metal can be plated onto an object.

This law highlights two key points:

• The mass of the substance produced at an electrode is proportional to the quantity of electric charge passed through the electrolyte.
• For every mole of substance produced at the electrode, a definite charge in coulombs is needed. This value is given by Faraday's constant, approximately 96,485 coulombs per mole of electrons.

When electrolyzing a solution such as chromium plating involving \(\mathrm{CrO}_4^{2-}\), electrons reduce the chromium ions to solid chromium metal. For each chromium atom deposited from \(\mathrm{CrO}_4^{2-}\), it requires involving six electrons in the reaction. Thus, the charge in coulombs required is calculated as the number of moles multiplied by Faraday's constant and the number of electrons needed per reaction.

This allows us to deduce the exact amount of electric charge necessary to plate a defined mass of metal, ensuring precision in applications such as electroplating.

current flow calculation

Calculating current flow is an essential process in electroplating, especially when determining the specifications for how much metal is to be deposited onto a surface in a given timeframe.

Current, denoted as \(I\), is the rate at which charge flows. It's measured in amperes \(\text{A}\), where one ampere equals one coulomb per second. The formula that ties this together is \(I = \frac{Q}{t}\), where \(Q\) is the charge in coulombs, and \(t\) is time in seconds.

In the context of the exercise, to plate a bumper within a 10-second window, you would need to determine the total number of coulombs required for depositing the chromium using the results from Faraday's law. Then, by dividing this charge by the given time \(10 \, \text{s}\), we arrive at the actual current required to perform the electroplating within this period.

This calculated current informs not only the practicalities of the electroplating setup but also is a cornerstone for ensuring efficiency and desired qualities in the plated layer.

electrical power expended

Power in electrical terms refers to the rate at which energy is used. During electroplating, calculating the electrical power expended gives insight into the energy efficiency and operational costs of the process.

Power \(P\) is given by the product of voltage \(V\) and current \(I\) (\(P = VI\)). However, when a system efficiency is involved, the input power tends to be higher than the output power due to inherent energy losses. This is where the efficiency ratio comes into play, calculated as \(\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}}\).

In the context of the exercise, the energy source has an efficiency of 65%, meaning not all energy supplied is effectively used for plating chromium.

• Using the formula \(\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}}\), we can compute the total energy required.
• A given external EMF of \(6.0 \, \text{V}\) and a current of \(640.728 \, \text{kA}\) provides the output power, accounting for efficiency yields the input power, which here is crucial for planning resources.

This method ensures efficient energy management during electroplating, reducing excess costs and avoiding wastage.