Question

You pass a \(1.0\) amp current through an electrolytic cell for \(1.0 \mathrm{hr}\). There are 96,500 coul in a Faraday (F). Calculate the number of grams of each of the following that would be deposited at the cathode: (2) Cu from a \(\mathrm{Cu}^{+2}\) solution and (1) Ag from an Ag ' solution, (3) A.1 from an \(\mathrm{Al}^{3+}\) solution.

Short answer

The mass of each element deposited at the cathode when a 1.0 A current is passed through an electrolytic cell for 1.0 hr is: - Cu from a \(\mathrm{Cu}^{+2}\) solution: \(1.19 \: \text{g}\) - Ag from an Ag+ solution: \(4.03 \: \text{g}\) - Al from an \(\mathrm{Al}^{3+}\) solution: \(0.31 \: \text{g}\)

Step-by-step solution

Use Faraday’s Laws of Electrolysis to find the mol of each element

Faraday's Laws of Electrolysis states that the mass of substance deposited on an electrode is directly proportional to the amount of charge passed through the cell. First, we need to find the total charge that has been passed through the cell. The charge (Q) can be calculated using the formula: Q = current × time We are given the current (1.0 A) and time (1.0 hr = 3600 s). Q = 1.0 A × 3600 s = 3600 coulombs Now, the mol of metal deposited can be calculated as follows: mol of metal = (Q / F) / n Where F is Faraday's constant (96,500 coul/mol) and n is the number of electrons involved in the redox reaction of the specific element.

Calculate the mol of Cu, Ag, and Al

For Cu deposition from a \(\mathrm{Cu}^{+2}\) solution: n = 2 (two electrons needed for redox reaction) mol of Cu = (3600 coul / 96,500 coul/mol) / 2 For Ag deposition from an Ag+ solution: n = 1 (one electron needed for redox reaction) mol of Ag = (3600 coul / 96,500 coul/mol) / 1 For Al deposition from an \(\mathrm{Al}^{3+}\) solution: n = 3 (three electrons needed for redox reaction) mol of Al = (3600 coul / 96,500 coul/mol) / 3

Calculate the mass of Cu, Ag, and Al

Now, to find the mass of each element, we can use the formula: mass = mol × molar mass The molar masses of Cu, Ag, and Al are 63.5 g/mol, 107.9 g/mol, and 26.98 g/mol respectively. Mass of Cu = mol of Cu × molar mass of Cu Mass of Ag = mol of Ag × molar mass of Ag Mass of Al = mol of Al × molar mass of Al

Calculate the numerical values of the mass

Plug the values from Steps 2 and 3 into the equations and calculate the mass of each element: Mass of Cu = (3600 coul / 96,500 coul/mol) / 2 × 63.5 g/mol ≈ 1.19 g Mass of Ag = (3600 coul / 96,500 coul/mol) / 1 × 107.9 g/mol ≈ 4.03 g Mass of Al = (3600 coul / 96,500 coul/mol) / 3 × 26.98 g/mol ≈ 0.31 g Finally, the number of grams of each substance would be deposited at the cathode are: Cu: 1.19 g Ag: 4.03 g Al: 0.31 g

Key concepts

Electrolytic Cell

An electrolytic cell is a device that uses electrical energy to drive a non-spontaneous chemical reaction. It's a type of cell where electrolysis occurs, meaning the chemical compounds decompose into their constituents when subjected to an electric current. In the context of the problem, an electrolytic cell is used to deposit metals such as Cu, Ag, and Al at the cathode by passing a current through their solutions.
Faraday's laws of electrolysis govern this process, indicating that the amount of substance changed at the electrode is proportional to the quantity of electricity used. The importance of understanding the electrolytic cell in this experiment is to grasp how electrical energy can transform into chemical changes, a critical principle in electrochemistry.

• Electrolysis is important in industries like electroplating and electrorefining.
• It requires external power to facilitate the redox reaction.
• The cathode is the site where reduction occurs, hence, metal deposition.

Coulombs

Coulombs represent the unit of electric charge, named after Charles-Augustin de Coulomb. In the realm of electrolysis, coulombs measure the total charge passed through the electrolytic cell, determining how much metal is deposited.
The formula for calculating the total charge is given by: \[ Q = ext{current} \times ext{time} \]
For the problem at hand, with a 1.0 amp current passed over 1.0 hour (which is 3600 seconds), the total charge amounts to 3600 coulombs. This calculation is crucial as it directly informs the amount of metal that can be deposited. Knowing how to calculate the charge helps you understand how much electrical energy has been used in the process, bridging the gap between theoretical chemistry and practical applications.

• Understanding charge measurement is essential for quantitative electrolysis calculations.
• Faraday's constant (96,500 coulombs) helps relate charge to the amount of substance deposited.

Redox Reaction

In electrolysis, redox reactions are essential because they involve the transfer of electrons between species. The term "redox" stands for reduction and oxidation, the two complementary processes in these reactions.
Reduction refers to the gain of electrons, while oxidation refers to the loss of electrons. During electrolysis in an electrolytic cell, the cathode is where reduction occurs, attracting positive ions in the solution.
In the given exercise:

• For Copper (Cu): It involves 2 electrons for each Cu²⁺ ion reduced to Cu metal.
• For Silver (Ag): It involves 1 electron for each Ag⁺ ion reduced to Ag metal.
• For Aluminum (Al): It involves 3 electrons for each Al³⁺ ion reduced to Al metal.

Knowing the number of electrons involved in these redox reactions helps determine how many moles of metals will be deposited based on the charge passed through the system.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of a given substance (element or compound) divided by the amount of substance. It's usually expressed in grams per mole (g/mol).
In the context of the problem, we use molar mass to calculate the actual mass of metal deposited at the cathode after finding the moles of metal deposited using Faraday's Law. Each metal in the exercise has a known molar mass:

• Copper (Cu): 63.5 g/mol
• Silver (Ag): 107.9 g/mol
• Aluminum (Al): 26.98 g/mol

The formula used is: mass = mol × molar mass
Calculating the mass using molar mass and moles is crucial because it translates an abstract concept of chemical amounts into tangible quantities. It helps identify exactly how much of each substance is produced from a specific charge.