Question

It took 150 . s for a current of 1.25 \(\mathrm{A}\) to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)

Short answer

Using Faraday's law of electrolysis and the given data, we calculate the theoretical mass of the metal deposited for a \(1+\) cation by first finding the total charge passed through the cell (187.5 C) and then the number of moles of metal deposited (0.00194 mol). Without knowing the molar mass of the metal, we can't reach a definite conclusion. However, if we assume that the metal is from Group 1 of the periodic table (e.g., Lithium), the theoretical mass of metal (0.0135 g) is less than the actual deposited mass (0.109 g). This indicates that our assumption of a \(1+\) charge for the cations is incorrect, and it's not possible for the cations to have a charge of \(1+.\)

Step-by-step solution

Determine the total charge passed through the cell

We need to determine the total charge that passed through the cell in order to calculate the amount of metal deposited. We can calculate the total charge by using the following formula: Total charge (Q) = Current (I) × Time (t) Where I = 1.25 A (amperes) and t = 150 s (seconds).

Calculate the total charge

Plug the given values of current and time into the formula: Q = 1.25 A × 150 s Q = 187.5 C (coulombs)

Use Faraday's law of electrolysis

According to Faraday's law of electrolysis, the amount of metal deposited (in moles) is proportional to the total charge passed through the cell. The relation can be written as: Amount of metal deposited (in moles) = Total charge (Q) / (Charge of cations (z) × Faraday's constant (F)) Where z = charge of cations and F = Faraday's constant, approximately 96485 C/mol.

Calculate the number of moles for a \(1+\) cation

Now let's assume the charge of the cations is \(1+,\) which means z = 1. We can calculate the number of moles of metal deposited using the total charge obtained in step 2: Amount of metal deposited (in moles) = 187.5 C / (1 × 96485 C/mol) Amount of metal deposited (in moles) ≈ 0.00194 mol

Calculate the theoretical mass of the metal deposited

To find the theoretical mass of the metal deposited, we need to know the molar mass of the metal (M). However, we do not have this information. So, we can write a general equation for the theoretical mass: Theoretical mass of metal = Amount of metal deposited (in moles) × Molar mass of metal (M) Theoretical mass of metal ≈ 0.00194 mol × M

Analyze the results

We know that 0.109 g of metal was actually deposited. Since we don't have the molar mass of the metal, it is difficult to reach a definite conclusion. However, we can analyze the results by assuming that the metal is from Group 1 of the periodic table. If we assume that the metal is Lithium (the lightest Group 1 element) with a molar mass of approximately 6.94 g/mol: Theoretical mass of metal ≈ 0.00194 mol × 6.94 g/mol ≈ 0.0135 g This theoretical mass is less than the actual mass of metal deposited (0.109 g). This indicates that, if the metal is from Group 1, our assumption that the charge of the cations is \(1+\) is incorrect, since a charge of \(1+\) would have led to less mass being deposited. Therefore, it's not possible for the cations to have a charge of \(1+.\)

Key concepts

Electrochemistry Basics

Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical change.
In electrochemical cells, chemical reactions are driven by the movement of electrons through an external circuit.
There are two main types of electrochemical cells: galvanic cells, which generate electricity through spontaneous reactions, and electrolytic cells, where electrical energy is used to drive non-spontaneous reactions. In this exercise, we deal with electrolytic cells.
Here, a current is used to deposit a metal from a solution onto an electrode.

• The substance undergoing reduction is known as the "electrolyte" and contains ions that move to the electrodes.
• In the context of metal deposition, the electrodes involved are usually referred to as the cathode and anode.
• A real-world example of this process is electroplating, a method used to coat an object with a thin layer of metal.

Understanding electrochemistry allows us to harness electrical energy to perform chemical transformations.

Understanding Cations

Cations are positively charged ions that result from the loss of electrons.
In electrolysis, cations in the solution move towards the cathode (the negative electrode) where they gain electrons and get reduced.
These cations originate from the dissociation of compounds in the solution.

• Cations are designated by their charge, indicated by a superscript, such as the lithium ion \(Li^{+}\).
• The charge on a cation plays a critical role in determining how much of the substance can be deposited during electrolysis.
• For example, if the cation has a charge of \(1+\), one mole of electrons will reduce one mole of cations.

This principle comes into play in the given problem where we suspect a discrepancy in the expected charge of the deposited metal's cation.

The Process of Charge Calculation

Calculating the total charge passing through an electrolytic cell is essential for determining the amount of metal deposited.
The formula simply uses the current and the time duration, expressed as:\[Q = I \times t\]Where

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

Once we have the total charge, we can relate it to deposited moles of metal through Faraday’s Law of Electrolysis.
This law indicates that the amount of substance (in moles) transformed at an electrode is directly proportional to the quantity of electricity that passed through the cell.
Using this calculation, we can determine whether the assumed charge state of cations matches the obtained results.

Principles of Metal Deposition

Metal deposition through the process of electrolysis involves a sequence of controlled chemical reactions.
These reactions transform cations into their metallic form, allowing them to deposit onto a substrate, usually an electrode surface.
The amount of metal deposited is dependent on several key factors.

• The charge of the cations, which determines the stoichiometry of the reaction as each ion requires a specific number of electrons to be reduced.
• The total charge passed, calculated by the current and time, determining how much energy is available for the reduction process.
• The molar mass of the metal, which is needed to convert moles of metal into grams for practical applications.

In the problem, an assumed charge of \(1+\) for the cations did not provide the observed mass of the metal, hence suggesting a higher charge, \(2+\) or \(3+\), might be more appropriate, assuming all other factors remain constant.