Mastering Redox Reactions: Practice Problems & Solutions

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Hey chemistry enthusiasts! Ever found yourself tangled in the web of redox reactions? Balancing these equations can seem tricky at first, but trust me, with the right approach – like the oxidation number method – you'll be acing these problems in no time. Today, we're diving deep into some practice problems to sharpen your skills. We'll be working through two examples, one in an acidic environment and the other in a basic environment. So, grab your pencils, get ready to crunch some numbers, and let's conquer those redox reactions together!

Understanding the Oxidation Number Method: A Quick Refresher

Before we jump into the problems, let's quickly recap the oxidation number method. This method is a systematic way to balance redox reactions by tracking the changes in oxidation numbers of the atoms involved. Here's a simplified version of the steps:

  1. Assign Oxidation Numbers: Determine the oxidation numbers for each atom in the reactants and products. Remember, the oxidation number represents the charge an atom would have if all the bonds were ionic.
  2. Identify Oxidation and Reduction: Identify which atoms are being oxidized (losing electrons, oxidation number increases) and which are being reduced (gaining electrons, oxidation number decreases).
  3. Calculate the Change in Oxidation Number: Determine the total change in oxidation number for both oxidation and reduction.
  4. Balance the Change: Multiply the reactants and products by appropriate coefficients to make the total increase in oxidation number equal to the total decrease.
  5. Balance Other Atoms: Balance the remaining atoms (other than those undergoing oxidation or reduction) using the trial-and-error method.
  6. Balance Oxygen and Hydrogen (for Acidic and Basic Conditions):
    • Acidic: Add water (H₂O) to the side deficient in oxygen atoms and add hydrogen ions (H⁺) to the side deficient in hydrogen atoms.
    • Basic: Add water (H₂O) to the side deficient in oxygen atoms. Then, add the same number of hydroxide ions (OH⁻) to the opposite side of the water molecules, and add hydrogen ions (H⁺) to balance the hydrogen.
  7. Check the Charges: Ensure that the total charges on both sides of the equation are equal.

Sounds like a lot, right? But with practice, this method becomes second nature. Ready to put it to the test?

Problem 1: Balancing in an Acidic Environment

Let's get started with our first problem. We have the following reaction happening in an acidic environment:

Cr₂O₇²⁻ + C₂O₄²⁻ → 2Cr³⁺ + 2CO₂

Alright, guys, let's break this down step-by-step to balance this redox reaction using the oxidation number method. Remember, we're in an acidic environment, so we'll need to account for those H⁺ ions and H₂O molecules!

  1. Assign Oxidation Numbers:

    • Cr in Cr₂O₇²⁻: Oxygen has an oxidation number of -2. Since there are seven oxygen atoms, the total negative charge from oxygen is -14. The overall charge of the ion is -2. Therefore, 2Cr + (-14) = -2, which means 2Cr = +12, and Cr = +6.
    • C in C₂O₄²⁻: Oxygen has an oxidation number of -2. There are four oxygen atoms, so the total negative charge from oxygen is -8. The overall charge is -2. Therefore, 2C + (-8) = -2, which means 2C = +6, and C = +3.
    • Cr in Cr³⁺: +3 (given).
    • C in CO₂: Oxygen has an oxidation number of -2. There are two oxygen atoms, so the total negative charge from oxygen is -4. Therefore, C + (-4) = 0, which means C = +4.

  2. Identify Oxidation and Reduction:

    • Cr goes from +6 to +3 (reduction – gains electrons).
    • C goes from +3 to +4 (oxidation – loses electrons).

  3. Calculate the Change in Oxidation Number:

    • Cr: The oxidation number changes by 3. Since there are two Cr atoms in Cr₂O₇²⁻, the total change for Cr is 2 × 3 = 6.
    • C: The oxidation number changes by 1. Since there are two C atoms in C₂O₄²⁻, the total change for C is 2 × 1 = 2.

  4. Balance the Change: We need to balance the electron transfer. The least common multiple of 6 and 2 is 6. We will multiply the carbon-containing species by a factor of 3 to balance the total change in oxidation numbers, so the ratio of Cr to C should be 1:3

    • Multiply C₂O₄²⁻ and CO₂ by 3: Cr₂O₇²⁻ + 3C₂O₄²⁻ → 2Cr³⁺ + 6CO₂

  5. Balance Other Atoms:

    • Chromium is already balanced (2 Cr on both sides).
    • Carbon is now balanced (6 C on both sides).
    • Oxygen and hydrogen will be taken care of in the next step, as we are in an acidic environment.

  6. Balance Oxygen and Hydrogen (Acidic Conditions):

    • We have 7 oxygen atoms on the left side (from dichromate) and 12 on the right side (from carbon dioxide). We need 5 more oxygen atoms on the left. We add 7 H₂O to the right to balance the oxygen: Cr₂O₇²⁻ + 3C₂O₄²⁻ → 2Cr³⁺ + 6CO₂ + 7H₂O
    • Now, we need to balance the hydrogen. There are 14 hydrogen atoms on the right. We add 14 H⁺ ions to the left to balance: Cr₂O₇²⁻ + 3C₂O₄²⁻ + 14H⁺ → 2Cr³⁺ + 6CO₂ + 7H₂O

  7. Check the Charges:

    • Left side: -2 (from dichromate) + -6 (from oxalate) + 14 (from H⁺) = +6
    • Right side: 2 × (+3) = +6
    • The charges are balanced!

The balanced equation is: Cr₂O₇²⁻ + 3C₂O₄²⁻ + 14H⁺ → 2Cr³⁺ + 6CO₂ + 7H₂O

Problem 2: Balancing in a Basic Environment

Now, let's tackle a problem in a basic environment. Here's our reaction:

2CrO₄²⁻ + Fe(OH)₂ → Cr₂O₃ + Fe(OH)₃

Alright, buckle up, guys! We're doing another redox reaction, this time in a basic solution. This means we have to incorporate those OH⁻ ions and balance our equation accordingly. Let's walk through it:

  1. Assign Oxidation Numbers:

    • Cr in CrO₄²⁻: Oxygen has an oxidation number of -2. There are four oxygen atoms, so the total negative charge from oxygen is -8. The overall charge is -2. Therefore, Cr + (-8) = -2, which means Cr = +6.
    • Fe in Fe(OH)₂: The hydroxide ion (OH) has a charge of -1. Therefore, Fe + 2 × (-1) = 0, which means Fe = +2.
    • Cr in Cr₂O₃: Oxygen has an oxidation number of -2. There are three oxygen atoms, so the total negative charge from oxygen is -6. Therefore, 2Cr + (-6) = 0, which means 2Cr = +6, and Cr = +3.
    • Fe in Fe(OH)₃: The hydroxide ion (OH) has a charge of -1. Therefore, Fe + 3 × (-1) = 0, which means Fe = +3.

  2. Identify Oxidation and Reduction:

    • Cr goes from +6 to +3 (reduction – gains electrons).
    • Fe goes from +2 to +3 (oxidation – loses electrons).

  3. Calculate the Change in Oxidation Number:

    • Cr: The oxidation number changes by 3. Since there are two Cr atoms, the total change for Cr is 2 × 3 = 6.
    • Fe: The oxidation number changes by 1.

  4. Balance the Change:

    • Multiply Fe(OH)₂ and Fe(OH)₃ by 6 to balance the total change. We will multiply Cr species by 1: 2CrO₄²⁻ + 6Fe(OH)₂ → Cr₂O₃ + 6Fe(OH)₃

  5. Balance Other Atoms:

    • Chromium is already balanced (2 Cr on both sides).
    • Iron is now balanced (6 Fe on both sides).

  6. Balance Oxygen and Hydrogen (Basic Conditions):

    • We have 8 oxygen atoms on the left (from chromate) and 3 on the right side (from Cr₂O₃) and 18 from Fe(OH)₃, for a total of 21. We have 12 oxygen atoms from Fe(OH)₂. We have to balance the water by adding 8H₂O to the left side: 2CrO₄²⁻ + 6Fe(OH)₂ + 8H₂O → Cr₂O₃ + 6Fe(OH)₃
    • We have 16 hydrogen atoms on the left, and 18 on the right side. Add 2 OH⁻ to the right side to balance: 2CrO₄²⁻ + 6Fe(OH)₂ + 8H₂O → Cr₂O₃ + 6Fe(OH)₃ + 2OH⁻
    • Add 2 OH⁻ to the left side and 2 H₂O to the right to balance 2CrO₄²⁻ + 6Fe(OH)₂ + 8H₂O + 2OH⁻ → Cr₂O₃ + 6Fe(OH)₃ + 2OH⁻

  7. Check the Charges:

    • Left side: -2 (from chromate) - 2 (from hydroxide) = -4
    • Right side: -2 (from hydroxide) = -2
    • So, it is 2CrO₄²⁻ + 6Fe(OH)₂ + 4H₂O → Cr₂O₃ + 6Fe(OH)₃ + 2OH⁻ is the correct answer

The balanced equation is: 2CrO₄²⁻ + 6Fe(OH)₂ + 4H₂O → 2Cr₂O₃ + 6Fe(OH)₃ + 2OH⁻

Tips for Success

Alright, you've seen how to work through these problems, but here are some extra tips to help you master redox reactions:

  • Practice, practice, practice! The more you work through these problems, the easier they'll become. Try different examples from your textbook or online resources.
  • Don't be afraid to make mistakes. Everyone makes mistakes when learning. Learn from them and try again.
  • Break down the problem into smaller steps. Don't try to do everything at once. Focus on one step at a time.
  • Double-check your work. Make sure your oxidation numbers are correct and that you've balanced the charges.
  • Understand the concepts. Don't just memorize the steps. Understand the underlying principles of oxidation and reduction.

Conclusion

And there you have it, guys! We've successfully navigated through two redox reactions, one in an acidic environment and one in a basic environment. The oxidation number method is your friend – it just takes a bit of practice. Keep at it, and you'll be balancing those equations like a pro in no time! Remember to stay curious, keep practicing, and don't hesitate to ask for help if you get stuck. Happy chemistry-ing!