Electrolysis Of Cadmium Sulfate: Anode Substance Calculation

Hey guys! Let's dive into a chemistry problem involving electrolysis! We're going to figure out how much stuff gets released at the anode (that's the positive electrode, in case you forgot!) when we pass an electric current through a solution of cadmium sulfate (CdSO₄). This is a pretty common type of problem you might encounter, and understanding it will definitely boost your chemistry game. So, grab your notebooks and let's break it down! The core concept here revolves around Faraday's laws of electrolysis, which link the amount of substance produced to the amount of electric charge passed through the solution. We'll be using this fundamental principle to solve the question. Let’s get started.

We are given the following information: The current is 3 Faradays (3 Ф), and the solution contains 2 moles of CdSO₄. The question asks us to find the mass of the substance deposited at the anode. Since we have an inert electrode, the substance deposited at the anode will be the substance from the electrolyte. We know that the cadmium sulfate solution will undergo electrolysis, which causes the cadmium ions (Cd²⁺) to migrate toward the cathode (negative electrode) where they will be reduced, while sulfate ions (SO₄²⁻) will migrate toward the anode (positive electrode) where they will be oxidized. So, we need to know what happens at the anode.

At the anode, the sulfate ions (SO₄²⁻) will be oxidized. However, since we have an inert electrode, it's more likely that the water molecules (H₂O) will be oxidized instead of the sulfate ions. The oxidation of water will produce oxygen gas (O₂) and hydrogen ions (H⁺). The reaction is 2H₂O → O₂ + 4H⁺ + 4e⁻. This reaction means that for every 4 moles of electrons (4F), 1 mole of oxygen gas is produced. Let's consider the number of Faradays passed through the solution. We have 3 Faradays. This means 3 moles of electrons are involved. Based on the equation, to produce one mole of oxygen, we need 4 Faradays. With 3 Faradays, we produce a fraction of a mole of oxygen. We can use the following calculation: (3 Faradays / 4 Faradays/mole O₂) = 0.75 moles of O₂. Since the question asks about the quantity of the substance released, we now need to determine the mass of oxygen released. The molar mass of O₂ is 32 g/mol. Therefore, the mass of oxygen released is 0.75 moles * 32 g/mol = 24 g. The problem provides answer choices with no specific units (A) 8 (B) 16 (C) 32 (D) 24. Therefore, the answer is D) 24.

Understanding the Basics: Electrolysis and Faraday's Laws

Alright, before we get deeper, let's refresh our memories on the key concepts here. Electrolysis is the process of using electricity to drive a non-spontaneous chemical reaction. Think of it as using electricity to force a reaction to happen that wouldn't normally occur on its own. In our case, we're using electricity to break down the cadmium sulfate solution. The Faraday's laws of electrolysis are super important here. They tell us the relationship between the amount of electricity passed through the solution and the amount of substance produced at the electrodes. The first law states that the mass of a substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The second law states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are in the ratio of their chemical equivalents.

Specifically, the amount of substance produced is directly proportional to the amount of charge (in Coulombs) passed through the solution. The constant of proportionality is related to the electrochemical equivalent of the substance. One Faraday (1 F) of charge is equivalent to the charge of one mole of electrons (approximately 96,485 Coulombs). This is a crucial number to remember because it links the amount of electricity to the number of moles of electrons involved in the reaction. In our problem, we were given the current in Faradays (3 F), which simplifies our calculations. If we were given the current in Amperes and the time in seconds, we'd need to convert that into Coulombs using the formula: Charge (Coulombs) = Current (Amperes) * Time (seconds). Understanding Faraday’s laws is key to solving electrolysis problems, and we’ll use these principles to calculate the mass of the substance formed at the anode.

Electrolysis happens in an electrochemical cell, which has two electrodes: the anode (where oxidation happens) and the cathode (where reduction happens). Oxidation is the loss of electrons, and reduction is the gain of electrons. Remember the mnemonic OIL RIG: Oxidation Is Loss, Reduction Is Gain. The substance deposited at the anode is the substance from the electrolyte or the electrode itself if it is not inert. The substance deposited at the cathode is the substance from the electrolyte. In our specific problem, we're interested in the anode reaction, where oxidation of water occurs, leading to the formation of oxygen gas. This is a crucial point in understanding how the problem works and how to approach the solution.

Step-by-Step Solution: Unveiling the Anode Reaction

Let's break down the problem step-by-step. First, let's consider what's happening at the anode. Remember, the anode is where oxidation takes place. In the electrolysis of a CdSO₄ solution with an inert electrode (like platinum or graphite), the water molecules will undergo oxidation instead of the sulfate ions, since water is more easily oxidized. The oxidation reaction at the anode is: 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e⁻. This equation tells us that for every 2 molecules of water that are oxidized, one molecule of oxygen gas is produced, and four electrons are released. Now, we are given that we have 3 Faradays of current. One Faraday represents one mole of electrons. So, 3 Faradays mean 3 moles of electrons are involved. From the equation, we can see that 4 moles of electrons are needed to produce 1 mole of O₂. We can set up a proportion: (4 moles of e⁻ / 1 mole O₂) = (3 moles of e⁻ / x moles of O₂). Solving for x, we get x = 0.75 moles of O₂. This means that 0.75 moles of oxygen gas will be produced at the anode. This is how we use the balanced chemical equation to figure out the mole ratio between electrons and the product at the anode. Next, to get the mass of oxygen, we need the molar mass of oxygen gas (O₂). The molar mass of O₂ is 32 g/mol. Finally, we can calculate the mass of oxygen produced using the following formula: mass = moles × molar mass. Mass of O₂ = 0.75 moles × 32 g/mol = 24 g. Therefore, 24 g of oxygen will be released at the anode.

This calculation helps us determine which answer choice is correct. This is the whole process of solving this kind of problem. This method provides a clear and organized approach to tackle electrolysis questions, making them less intimidating and easier to understand.

Decoding the Answer Choices and Choosing Wisely

Now that we've crunched the numbers, let's look at the answer choices. The problem gives us these options:

A) 8 B) 16 C) 32 D) 24

We calculated that 24 grams of oxygen gas is released at the anode. So, the correct answer is D) 24. Remember, the key is to understand the chemical reactions at the electrodes, use Faraday's laws, and do the calculations systematically. Always make sure to check the units and that they are consistent. In these types of problems, the units can be a trap. Ensure you are using the correct units throughout your calculations to avoid mistakes. Double-check your calculations to ensure you didn't make any errors in your arithmetic. Carefully read the question and make sure you are answering what is asked. For example, the question asked about the mass of the substance released at the anode, not the number of moles. That's why we had to convert moles to grams.

Also, keep in mind that the amount of cadmium sulfate (2 moles) is extra information. The amount of cadmium sulfate in the solution is not relevant to this calculation. The amount of the current (3 Ф) is what's important here, as it directly relates to the amount of substance produced. Also, note that if the electrode was not inert, then the anode would be oxidized, and the metal of the anode would dissolve. The problem will be solved in a different way in this case. It is useful to practice multiple problems. The more problems you solve, the more comfortable you'll become with this type of problem. Good luck, and keep practicing!