Question

How long will it take to plate out each of the following with a current of \(100.0 \mathrm{~A}\) ? a. \(1.0 \mathrm{~kg} \mathrm{Al}\) from aqueous \(\mathrm{Al}^{3+}\) b. \(1.0 \mathrm{~g} \mathrm{Ni}\) from aqueous \(\mathrm{Ni}^{2+}\) c. \(5.0 \mathrm{~mol} \mathrm{Ag}\) from aqueous \(\mathrm{Ag}^{+}\)

Short answer

The time it will take to plate out the given substances from their respective aqueous ions using a current of 100.0 A are as follows: a. 1.0 kg of Al will take approximately \(107,524.2~s\). b. 1.0 g of Ni will take approximately \(32.7~s\). c. 5.0 mol of Ag will take approximately \(4,824.25~s\).

Step-by-step solution

Calculate the charges per mole of the given ions

First, we need to determine the charges per mole for each ion. a. Al³⁺ → Al requires 3 moles of electrons for each mole of aluminum (3+ charge). b. Ni²⁺ → Ni requires 2 moles of electrons for each mole of nickel (2+ charge). c. Ag⁺ → Ag requires 1 mole of electron for each mole of silver (1+ charge).

Calculate the moles of the desired substances

We'll now calculate the number of moles of the given substances using their molecular weights. The atomic weights of Al, Ni, and Ag are 26.98 g/mol, 58.69 g/mol, and 107.87 g/mol, respectively. a. \(1.0~kg \mathrm{Al} = 1000~g = \frac{1000}{26.98} = 37.06~mol \mathrm{Al}\) b. \(1.0~g \mathrm{Ni} = \frac{1.0}{58.69} = 0.0170~mol \mathrm{Ni}\) c. \(5.0~mol \mathrm{Ag}\) (already given)

Calculate the total charge required for each deposition

Now, we'll multiply the moles of each substance by their respective charges per mole and Faraday's constant to get the total amount of charge required. a. \(Q_{Al} = 37.06~mol \times 3 ~mol \frac{e^-}{mol}(Al) \times 96,485~C/mol(e^-) = 10,752,421.3 ~C\) b. \(Q_{Ni} = 0.0170~mol \times 2 ~mol \frac{e^-}{mol}(Ni) \times 96,485~C/mol(e^-) = 3,273.77 ~C\) c. \(Q_{Ag} = 5.0~mol \times 1 ~mol \frac{e^-}{mol}(Ag) \times 96,485~C/mol(e^-) = 482,425 ~C\)

Calculate the time required for the deposition using the provided current

We'll now use the given current of 100.0 A to find the time it takes to deposit each substance using the calculated total charge by dividing the charge by the current. a. \(t_{Al} = \frac{Q_{Al}}{I} = \frac{10,752,421.3 ~C}{100.0 ~A} = 107,524.2~s\) b. \(t_{Ni} = \frac{Q_{Ni}}{I} = \frac{3,273.77 ~C}{100.0 ~A} = 32.7~s\) c. \(t_{Ag} = \frac{Q_{Ag}}{I} = \frac{482,425 ~C}{100.0 ~A} = 4,824.25~s\) Thus, it will take approximately 107,524.2 seconds for (a), 32.7 seconds for (b), and 4,824.25 seconds for (c) with a current of 100.0 A to plate out the given substances from their respective aqueous ions.

Key concepts

Faraday's law of electrolysis

In electroplating, Faraday's law of electrolysis is a fundamental principle that helps to quantitatively understand how much substance will be deposited during the process. This law is vitally important for calculating the amount of material that is plated out in industrial or laboratory settings.

Michael Faraday, in the 19th century, discovered that the amount of chemical change (such as plating of a metal) is directly proportional to the quantity of electricity that passes through the electrolyte solution.

The law can be summarized by:

• The mass of substance deposited or dissolved at an electrode is directly proportional to the quantity of electric charge passed through the solution.
• The mass is also dependent on the valency (or charge) of the ions being deposited.

In practical terms, to calculate how much of a metal is deposited you would typically multiply the number of moles of electrons by the Faraday's constant (approximately 96,485 coulombs per mole of electrons) and the equivalent weight of the substance. Understanding this law is crucial to solving electroplating problems like the exercise discussed.

Current and charge relationship

In electrochemical reactions, there's an integral link between the electric current used and the amount of charge that's transferred within the electrolyte solution. Let's unravel this relationship.

When an electric current flows through a solution containing dissolved ions, it carries a certain amount of charge. This is where current (I) and charge (Q) become very important.

For electrolysis, the total charge transferred is calculated using the formula \( Q = I \times t \), where:

• \( Q \) represents the total charge in coulombs.
• \( I \) represents the current in amperes.
• \( t \) represents the time in seconds for which the current flows.

This formula allows us to calculate how long an electric current must be applied to achieve a certain amount of electroplating, evident in the solving of the exercise where the required deposition time for each metal was calculated.

Moles and molar mass calculations

A good grasp of moles and molar mass is fundamental in chemistry, particularly in stoichiometric calculations like those involved in electroplating. Simply put, moles give us a way of quantifying the amount of a chemical substance.

The molar mass of an element or compound allows us to convert between grams and moles of that substance. It's effectively the mass of one mole of a given substance (usually in g/mol), reflecting the quantity of atoms as recorded from the periodic table.

In the practical electroplating exercise, calculations involved:

• Determining the moles of aluminum from 1.0 kg by dividing the given mass by the molar mass of aluminum (26.98 g/mol).
• Understanding that these moles directly relate to the number of moles of electrons needed for depositing the aluminum.

The relationship is crucial for calculating the extent of reactions and material deposition through electrochemical processes.