Question

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0 \mathrm{~min}\). If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Short answer

The charge on the ruthenium ions in the given aqueous solution is approximately +3.

Step-by-step solution

Identify the known values

In this problem, we are given: - Current (I) = 2.50 A - Time (t) = 50.0 minutes - Mass of Ru produced (m) = 2.618 g

Convert time to seconds

First, let's convert the time given in minutes to seconds, as it will make further calculations more straightforward. \(t (in~seconds) = 50.0 ~min \times 60~ \frac{seconds}{min} = 3000~ s\)

Calculate the total charge

Now we need to calculate the total charge passed during electrolysis using the formula: \(Q = I \times t\) where Q is the total charge in Coulombs, I is the current in Amperes, and t is the time in seconds. \(Q = 2.50~A \times 3000~s = 7500~C\)

Determine the moles of ruthenium

Using the molar mass of ruthenium (Ru), we can determine the moles of ruthenium produced at the cathode: Molar mass of Ru = 101.1 g/mol \(\text{moles of Ru} = \frac{\text{mass of Ru}}{\text{molar mass of Ru}} = \frac{2.618~g}{101.1~\frac{g}{mol}} = 0.02588~mol\)

Apply Faraday's Law of Electrolysis

According to Faraday's Law of Electrolysis, the total charge (Q) is equal to the number of moles of ions (n), the charge per ion (z), and Faraday's constant (F): \(Q = n \times z \times F\) Faraday's constant (F) = 96,485 C/mol We already calculated Q (total charge) and n (number of moles). Now, we will solve for z, the charge on the ruthenium ions. Rearranging the equation to solve for z: \(z = \frac{Q}{n \times F}\) Calculate z: \(z = \frac{7500~C}{0.02588~mol \times 96485~\frac{C}{mol}} = 2.998 \approx 3\) Since z must be a whole number, we can round it to the nearest integer.

Determine the charge on ruthenium ions

The calculated value of z is approximately 3, which corresponds to a charge of +3 on the ruthenium ions in solution. Therefore, the charge on the ruthenium ions is +3.

Key concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis connects the flow of electric current to the chemical change at electrodes. It establishes a direct relationship between the amount of substance produced or consumed at an electrode and the total charge passed through the solution. This law is essential in understanding electrochemical reactions as it quantifies the proportionality factor, known as Faraday's constant.
Faraday's constant (\( F \)) is a key component in this law and is equal to approximately \(96,485 \ \text{C/mol} \). It represents the charge of one mole of electrons. Thus, knowing the number of moles of electrons and the total charge can help determine the charge on ions or evolve gases in electrochemical reactions.
In our exercise, we have used Faraday's Law to find out the charge number of Ruthenium by relating the total charge passed during the electrolysis to the moles of Ruthenium deposited and the Faraday's constant.

Charge Calculation

Charge calculation is an essential process when dealing with electrochemical cells, like the one in the given exercise. To find out how much charge has passed, you need to know the current and the amount of time the current was flowing.

• Current (\(I\)): Measured in Amperes; it is the rate of flow of electric charge.
• Time (\(t\)): Measured in seconds; length of time the current is applied must be converted from minutes if needed.

The formula for calculating total charge (\( Q \)) is:
\[ Q = I \times t \]
Example: If a current of \( 2.50 \text{A} \) flows for \( 50.0 \text{ minutes} \), you convert the time to seconds (\( 3000 \text{ s} \)) and then calculate: \[ Q = 2.50 \times 3000 = 7500 \text{ C} \].
This is the total charge that passed through the solution, and it can be applied to determine other aspects of the electrochemical cell, such as the moles of ions involved.

Molar Mass

Molar mass is a concept that is crucial when converting between the mass of a substance and the number of moles. It represents the mass of a given substance (in grams) divided by the amount of substance (in moles).
In our exercise, we use the molar mass of Ruthenium (Ru) which is approximately \( 101.1 \text{ g/mol} \).

• To find the number of moles of Ruthenium, divide the mass by its molar mass.

For example, given that \( 2.618 \text{ g} \) of Ruthenium is deposited, we calculate the moles of Ruthenium as follows:
\[ \text{Moles of Ru} = \frac{2.618}{101.1} \approx 0.02588 \text{ mol} \]
Molar mass provides a bridge between the macroscopic scales of chemistry (grams) and the microscopic scales (moles), facilitating precise calculations in various chemical reactions, including electrolysis.

Ruthenium Ions

Ruthenium is a transition metal that can exist in several oxidation states, the most common ones being +2, +3, and +4. In electrochemical cells, knowing the charge of the cationic form is essential for accurate calculations and predictions.
In this exercise, Ruthenium ions are produced through electrolysis, and we aim to determine the charge using Faraday's Law.

• By finding the moles of Ruthenium and using the total charge, we deduced the charge number (\( z \)) of the ruthenium ions.
• The calculation showed \( z \approx 3 \), indicating that the ions formed were \( Ru^{3+} \).

This illustrates how you can use electrochemical processes to infer the oxidation state of ions in a solution, a concept that supports the identification and manipulation of chemical species in laboratory and industrial settings.