Question

When 12,000 coulombs of electricity is passed through the electrolyte, \(3.0 \mathrm{~g}\) of a metal of atomic mass \(96.5 \mathrm{~g} / \mathrm{mol}\) is deposited. The electro-valency of the metal cation in the electrolyte is (a) \(+4\) (b) \(+3\) (c) \(+2\) (d) \(-4\)

Short answer

The electro-valency of the metal cation in the electrolyte is \(+4\).

Step-by-step solution

Understand the Problem and Determine the Data Given

The problem provides the quantity of charge passed through an electrolyte (12,000 coulombs), the mass of metal deposited (3.0 g), and the atomic mass of the metal (96.5 g/mol). The goal is to find the electro-valency of the metal cation.

Use Faraday's First Law of Electrolysis

According to Faraday's first law of electrolysis, the mass of a substance deposited at an electrode is directly proportional to the quantity of electricity that is passed through the electrolyte. This is given by the formula: \( m = (MIt) / (nF) \) where \( m \) is the mass of the substance deposited (in grams), \( M \) is the atomic mass (in grams per mole), \( I \) is the current (in amperes), \( t \) is the time in seconds during which the current flows, \( n \) is the number of electrons or the electro-valency of the metal cation, and \( F \) is the Faraday constant, approximately \( 96,485 \) coulombs per mole.

Isolate the Unknown (Electro-valency, n) and Solve

Rearrange the formula to solve for \( n \): \( n = (MIt) / (mF) \). We are given that \( m = 3.0 \) g, \( M = 96.5 \) g/mol, and the total charge (\( It \)) is \( 12,000 \) coulombs. Using Faraday's constant \( F = 96,485 \) C/mol, we can calculate the electro-valency.

Calculate the Electro-valency

Substitute the known values into the formula and calculate \( n \): \( n = (96.5 \text{ g/mol} \times 12,000 \text{ C}) / (3.0 \text{ g} \times 96,485 \text{ C/mol}) \).

Simplify the Calculation

Cancel out the units and simplify to find \( n \): \( n = (96.5 \times 12,000) / (3.0 \times 96,485) \).

Complete the Calculation and Determine the Electro-valency

After calculating, we find that \( n = 4 \), which means the electro-valency of the metal cation in the electrolyte is \(+4\). Therefore, the correct answer is (a) \(+4\).

Key concepts

Faraday’s First Law of Electrolysis

Understanding Faraday’s First Law of Electrolysis is critical for anyone delving into electrochemistry. This law constitutes the backbone of electrolysis problem solving, as it connects the amount of substance deposited at an electrode to the quantity of electricity that flows through the electrolyte. Simply put, it states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the amount of electric charge passed through the electric circuit.

For instance, if you double the electric charge passing through the electrolyte, you'll double the mass of the substance deposited. This relationship is mathematically represented by the equation, \( m = (MIt) / (nF) \), where \( m \) is the mass of the substance deposited, \( M \) is its molar mass, \( I \) is the current, \( t \) is the time, \( n \) is the electro-valency, and \( F \) is the Faraday constant. The Faraday constant, approximately 96,485 coulombs per mole, is a fundamental value in electrochemistry and signifies the charge of one mole of electrons.

To apply this law in problem-solving, first, identify the given variables, then use the law’s formula to find the unknown. This methodical approach simplifies complex electrolysis problems and makes the calculations more manageable. Recognizing the direct proportion allows us to isolate the desired variable and solve for it, as seen in the exercise presented.

Electrolysis Problem Solving

Tackling electrolysis problems effectively involves a clear understanding of the principles behind electrolysis and Faraday’s laws. The process often breaks down into several key steps that guide us to the correct answer. Firstly, get a firm grip on the problem by identifying all the known quantities and the unknown variable. This aligns with the strategy we see in the initial step of the given exercise.

Next, utilize the relevant equations and constants such as Faraday’s laws and the Faraday constant \( F \) to establish a relationship between the knowns and the unknown. Often, this involves manipulating the formula to isolate the variable you’re solving for, such as the electro-valency \( n \) in our exercise. With all variables in the correct place, insert the known values and carry out the calculation, taking care to keep track of your units. Whenever possible, cancel out units algebraically to simplify the calculations before arriving at the numerical answer.

Remember, paying meticulous attention to the units and conversion factors is just as important as understanding the theory behind the equations. Missteps here can lead to incorrect results despite a correct application of the theory. Thus, mastering problem-solving in electrolysis not only requires mathematical skill but also a detailed awareness of the underlying chemical principles.

Mole Concept

The mole concept is a fundamental pillar in chemistry, providing a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and observe. At its core, the mole is a unit of measurement for the amount of substance, similar to how a 'dozen' refers to the quantity twelve. One mole is defined as exactly 6.02214076 x 10²³ (Avogadro's number) of particles, whether they're atoms, ions, or molecules.

Understanding the mole concept allows us to relate the mass of a substance to the number of its entities. In electrochemistry and chemical calculations, we commonly use it in relation to molar mass - the mass of one mole of a substance, expressed in grams per mole (g/mol). This measure is vital to solving electrolysis problems as it links the macroscopic grams of a sample to the microscopic moles needed for stoichiometric calculations.

In the provided exercise, the molar mass \( M \) of the metal and the mass \( m \) of metal deposited are used in conjunction with Faraday's First Law to determine the electro-valency. A clear grasp of how moles, molar mass, and mass are related is essential to successfully solving such problems. Always remember, the molar mass serves as a conversion factor between a substance's mass and the number of moles, paving the way to explore deeper relationships in chemistry.