Calculating Copper Mass Deposited In Electrolysis

Hey guys! Ever wondered how much copper gets deposited during electrolysis? It's a fascinating concept in chemistry, and today, we're going to break it down step by step. We'll tackle a common problem: calculating the mass of copper that deposits at the cathode when an electric current flows through a copper sulfate ($CuSO_4$) solution. So, grab your calculators, and let's dive in!

Understanding Electrolysis and Faraday's Laws

Before we jump into the calculation, let's quickly recap the basics of electrolysis. Electrolysis is the process of using electricity to drive a non-spontaneous chemical reaction. In simpler terms, we're using electrical energy to make a chemical change happen. Think of it like using a battery to plate a metal object – that's electrolysis in action!

Key Concepts in Electrolysis:

• Electrolytic Cell: The setup where electrolysis occurs, consisting of electrodes (cathode and anode) immersed in an electrolyte solution.

• Cathode: The negatively charged electrode where reduction (gain of electrons) occurs. In our case, copper ions ($Cu^{2+}$) will gain electrons and deposit as solid copper (Cu).

• Anode: The positively charged electrode where oxidation (loss of electrons) occurs.

• Electrolyte: A solution containing ions that can conduct electricity. Here, it's copper sulfate ($CuSO_4$), which dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions in water.

• Faraday's Laws of Electrolysis: These laws are the cornerstone of quantitative electrolysis. They relate the amount of substance produced or consumed at an electrode to the quantity of electricity passed through the electrolytic cell. There are two main laws, but we'll primarily focus on the first one for this problem.

  • Faraday's First Law: The mass of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, this is represented as:

    $m \propto Q$

    Where:

    • m is the mass of the substance deposited or liberated
    • Q is the quantity of electricity passed

  • Faraday's Second Law: The masses of different substances liberated or dissolved by the same quantity of electricity are proportional to their equivalent weights.

Faraday's Constant and Its Significance

To convert the proportionality in Faraday's First Law into an equation, we introduce Faraday's constant (F). This constant represents the amount of electric charge carried by one mole of electrons, and its value is approximately 96,485 Coulombs per mole (C/mol). For practical purposes, it's often rounded to 96,500 C/mol.

Faraday's constant is crucial because it links the electrical world (charge) to the chemical world (moles of substance). It allows us to calculate how much of a substance will be produced or consumed in an electrolytic reaction based on the amount of electricity passed.

Now, let's express the quantity of electricity (Q) in terms of current (I) and time (t). Current is the rate of flow of electric charge, so:

$Q = I \times t$

Where:

• Q is the quantity of electricity in Coulombs (C)
• I is the current in Amperes (A)
• t is the time in seconds (s)

Combining this with Faraday's First Law and incorporating Faraday's constant, we get the fundamental equation for calculating the mass of a substance deposited during electrolysis:

$m = \frac{M \times I \times t}{n \times F}$

Where:

• m is the mass of the substance deposited (in grams)
• M is the molar mass of the substance (in grams per mole)
• I is the current (in Amperes)
• t is the time (in seconds)
• n is the number of moles of electrons required to deposit one mole of the substance (this is determined from the balanced half-reaction)
• F is Faraday's constant (approximately 96,500 C/mol)

With this equation in our toolkit, we're ready to tackle the copper deposition problem!

Problem Breakdown: Calculating the Mass of Deposited Copper

Okay, let's get to the heart of the problem. We need to figure out the mass of copper deposited at the cathode when a 10-ampere current flows through a copper sulfate ($CuSO_4$) solution for 193 seconds. We're also given the atomic mass of copper ($Ar ext{ }Cu=63.5$).

Here’s how we'll approach this:

1. Identify the Given Information:

   • Current (I) = 10 Amperes
   • Time (t) = 193 seconds
   • Atomic mass of Copper (Ar Cu) = 63.5 g/mol (This is essentially the molar mass, M)
   • Faraday's constant (F) = 96,500 C/mol (This is a constant value)

2. Determine the Number of Electrons Transferred (n):

   This is a crucial step. We need to look at the half-reaction that's happening at the cathode. Copper ions ($Cu^{2+}$) are being reduced to solid copper (Cu). The balanced half-reaction is:

   $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

   Notice that each copper ion gains 2 electrons to become a neutral copper atom. So, n = 2.

3. Apply Faraday's Law Equation:

   Now we have all the pieces of the puzzle! We'll use the equation we derived earlier:

   $m = \frac{M \times I \times t}{n \times F}$

   Plug in the values:

   $m = \frac{63.5 ext{ g/mol} \times 10 ext{ A} \times 193 ext{ s}}{2 \times 96,500 ext{ C/mol}}$

4. Calculate the Mass (m):

   Time to crunch the numbers! Using your calculator (or doing it the old-fashioned way!), you'll find:

   $m \approx 0.635 ext{ g}$

Step-by-Step Solution with Detailed Explanation

Let's walk through the entire calculation again, but this time, we'll really break down each step to make sure everyone's on the same page.

Step 1: Gather the Known Information

First, we need to clearly identify what we already know from the problem statement. This is like gathering your ingredients before you start cooking. We have:

• Current (I): 10 Amperes. This tells us how much electric charge is flowing per unit of time.
• Time (t): 193 seconds. This is the duration of the electrolysis process.
• Molar mass of Copper (M): 63.5 g/mol. This is the mass of one mole of copper atoms. We get this from the problem statement, which gives us the atomic mass (Ar) of copper.
• Faraday's Constant (F): 96,500 C/mol. This is a fundamental constant in electrochemistry, representing the charge of one mole of electrons.

Step 2: Determine the Number of Electrons (n) Involved

This step is about understanding the chemistry of what's happening at the electrode. We need to know how many electrons are required to deposit one copper atom. For that, we look at the half-reaction:

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

This equation tells us that a copper ion ($Cu^{2+}$) in solution gains two electrons ($2e^-$) to become solid copper (Cu). Therefore, the number of electrons transferred per copper ion is 2. So, n = 2.

Why is this important? The number of electrons is crucial because it directly relates to the amount of charge needed to deposit a certain amount of copper. If it took a different number of electrons, the mass of copper deposited would be different.

Step 3: Apply Faraday's Law Formula

Now comes the main equation that ties everything together. We'll use the formula we discussed earlier:

$m = \frac{M \times I \times t}{n \times F}$

This formula tells us that the mass of copper deposited (m) is directly proportional to the molar mass (M), current (I), and time (t), and inversely proportional to the number of electrons (n) and Faraday's constant (F).

Step 4: Plug in the Values

This is where we substitute the values we gathered in Step 1 and Step 2 into the formula:

$m = \frac{63.5 ext{ g/mol} \times 10 ext{ A} \times 193 ext{ s}}{2 \times 96,500 ext{ C/mol}}$

Make sure you include the units! This helps you keep track of what you're calculating and ensures your final answer has the correct units (grams in this case).

Step 5: Calculate the Result

Now it's calculator time! We perform the multiplication and division:

$m = \frac{122555 ext{ g A s/mol}}{193000 ext{ C/mol}}$

Remember that 1 Ampere (A) is equal to 1 Coulomb per second (C/s). So, the units A s in the numerator effectively become Coulombs (C).

$m \approx 0.635 ext{ g}$

Step 6: State the Answer

Finally, we state our answer clearly:

The mass of copper deposited at the cathode is approximately 0.635 grams.

Key Takeaways from the Step-by-Step Solution

• Organization is Key: Breaking the problem down into steps makes it much easier to manage.
• Units Matter: Keeping track of units helps prevent errors and ensures your final answer makes sense.
• Understand the Chemistry: Knowing the half-reaction and the number of electrons transferred is crucial for using Faraday's Law correctly.

Common Mistakes and How to Avoid Them

Electrolysis calculations can sometimes be tricky, so let's look at some common mistakes students make and how to avoid them.

1. Forgetting to Balance the Half-Reaction: The most frequent error is not correctly determining the number of electrons (n) involved in the electrode reaction. Always write out the balanced half-reaction to ensure you get the right value. For example, if you incorrectly assumed that copper deposited by gaining only one electron, your calculation would be way off.

2. Using the Wrong Units: Make sure your units are consistent. Time should be in seconds, current in Amperes, and molar mass in grams per mole. If time is given in minutes, for example, you'll need to convert it to seconds before plugging it into the formula.

3. Confusing Atomic Mass and Molar Mass: Atomic mass (Ar) is the mass of a single atom, while molar mass (M) is the mass of one mole of atoms. They have the same numerical value but different units (amu vs. g/mol). In electrolysis calculations, we use molar mass.

4. Rounding Errors: Avoid rounding intermediate values during the calculation. Round only your final answer to the appropriate number of significant figures. This helps minimize errors in your result.

5. Misunderstanding Faraday's Constant: Remember that Faraday's constant (F) is the charge of one mole of electrons. Don't confuse it with other constants or use it in the wrong context.

Practice Problems to Master Electrolysis Calculations

Now that we've gone through the theory, the calculation, and common mistakes, it's time to put your knowledge to the test! Here are a few practice problems to help you master electrolysis calculations:

1. Silver Deposition: A solution of silver nitrate ($AgNO_3$) is electrolyzed using a current of 5 Amperes for 30 minutes. Calculate the mass of silver (Ag) deposited at the cathode. (Ar Ag = 107.87 g/mol)

2. Zinc Plating: What current is required to deposit 1.5 grams of zinc (Zn) from a zinc sulfate ($ZnSO_4$) solution in 45 minutes? (Ar Zn = 65.38 g/mol)

3. Electrolysis of Water: During the electrolysis of water, how many liters of oxygen gas ($O_2$) are produced at standard temperature and pressure (STP) when a current of 2 Amperes is passed for 1 hour? (Molar volume of gas at STP = 22.4 L/mol)

Tips for Solving Practice Problems:

• Read Carefully: Understand what the problem is asking before you start calculating.
• List the Givens: Identify the known quantities and their units.
• Write the Half-Reaction: Determine the number of electrons involved in the electrode process.
• Use the Formula: Apply Faraday's Law equation correctly.
• Check Your Units: Make sure your units are consistent throughout the calculation.
• Show Your Work: This helps you (and your instructor) track your steps and identify any errors.

Real-World Applications of Electrolysis

Electrolysis isn't just a theoretical concept you learn in chemistry class; it has numerous practical applications in our daily lives and in various industries. Understanding these applications can make the topic even more interesting and relevant.

1. Electroplating: This is one of the most common applications of electrolysis. Electroplating is the process of coating a metal object with a thin layer of another metal using electrolysis. This is done to improve the appearance, protect against corrosion, or enhance the surface properties of the object. For example, jewelry is often plated with gold or silver, and car bumpers are chrome-plated for a shiny finish and rust protection.

2. Extraction and Purification of Metals: Electrolysis is used to extract highly reactive metals like sodium (Na), potassium (K), and aluminum (Al) from their ores. It's also used to purify metals like copper. Impure copper is used as the anode in an electrolytic cell, and pure copper deposits at the cathode. The impurities settle at the bottom of the cell as