Balancing Redox Reactions Determining The Coefficient Of Sn(OH)₃⁻

by Mei Lin 66 views

Hey guys! Ever get tangled up in balancing redox reactions, especially those happening in basic solutions? It can feel like deciphering a secret code, right? But don't worry, we're going to break down a specific example today and make it crystal clear. We'll focus on the reaction:

Bi(OH)₃(s) + Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq) + Bi(s)

Our mission? To balance this equation using the smallest whole number coefficients and, most importantly, figure out the coefficient for Sn(OH)₃⁻. So, grab your lab coats (or just your thinking caps!) and let's dive in!

Understanding Redox Reactions: Oxidation and Reduction

Before we jump into balancing, let's quickly recap what redox reactions are all about. The term "redox" is short for reduction-oxidation, and it describes reactions where electrons are transferred between chemical species. Think of it as a chemical tug-of-war where electrons are the rope.

  • Oxidation is the loss of electrons. A species that loses electrons is said to be oxidized, and its oxidation state increases. Remember the handy mnemonic OIL RIG: Oxidation Is Loss (of electrons).
  • Reduction is the gain of electrons. A species that gains electrons is said to be reduced, and its oxidation state decreases. Following OIL RIG, Reduction Is Gain (of electrons).

In any redox reaction, oxidation and reduction always occur together. One species can't lose electrons unless another species is there to accept them, and vice versa. The species that causes another species to be oxidized is called the oxidizing agent, and it itself is reduced. Conversely, the species that causes another species to be reduced is called the reducing agent, and it itself is oxidized. In our reaction, we need to identify which species are being oxidized and reduced.

Let's assign oxidation states to each element in our reaction:

  • Bi(OH)₃(s): Oxygen usually has an oxidation state of -2, and hydrogen is +1. Therefore, the three hydroxide ions (OH⁻) contribute a total charge of 3 * (-1) = -3. Since the compound is neutral, bismuth (Bi) must have an oxidation state of +3.
  • Sn(OH)₃⁻(aq): Again, oxygen is -2 and hydrogen is +1, so the three hydroxide ions contribute -3. The overall ion has a charge of -1, so tin (Sn) must have an oxidation state of +2 to balance the charges (+2 - 3 = -1).
  • Sn(OH)₆²⁻(aq): Six hydroxide ions contribute 6 * (-1) = -6. The overall ion has a charge of -2, so tin (Sn) must have an oxidation state of +4 (+4 - 6 = -2).
  • Bi(s): Elements in their elemental form have an oxidation state of 0.

Now we can clearly see what's happening:

  • Bismuth (Bi) goes from +3 in Bi(OH)₃ to 0 in Bi(s). This is a decrease in oxidation state, meaning bismuth is being reduced.
  • Tin (Sn) goes from +2 in Sn(OH)₃⁻ to +4 in Sn(OH)₆²⁻. This is an increase in oxidation state, meaning tin is being oxidized.

With this knowledge, we can now move on to the balancing process, keeping in mind that we're dealing with a basic solution.

The Half-Reaction Method: A Step-by-Step Guide for Balancing Redox Reactions

The half-reaction method is a powerful technique for balancing redox reactions, especially when dealing with acidic or basic solutions. This method involves breaking down the overall reaction into two half-reactions: one for oxidation and one for reduction. We then balance each half-reaction separately before combining them to obtain the balanced overall equation.

Here's how it works, step-by-step, for our reaction in basic solution:

1. Write the Unbalanced Half-Reactions:

Based on our oxidation state analysis, we can write the two half-reactions:

  • Reduction half-reaction: Bi(OH)₃(s) → Bi(s)
  • Oxidation half-reaction: Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq)

2. Balance the Atoms (Except Oxygen and Hydrogen):

In both half-reactions, the elements other than oxygen and hydrogen (Bi and Sn) are already balanced. We have one bismuth atom on each side of the reduction half-reaction and one tin atom on each side of the oxidation half-reaction.

3. Balance Oxygen Atoms by Adding H₂O:

  • Reduction half-reaction: Bi(OH)₃(s) → Bi(s) (3 O on the left, 0 on the right. Add 3 H₂O to the right) Bi(OH)₃(s) → Bi(s) + 3H₂O(l)
  • Oxidation half-reaction: Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq) (3 O on the left, 6 on the right. Add 3 H₂O to the left) 3H₂O(l) + Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq)

4. Balance Hydrogen Atoms by Adding H⁺:

  • Reduction half-reaction: Bi(OH)₃(s) → Bi(s) + 3H₂O(l) (3 H on the left, 6 H on the right. Add 3 H⁺ to the left) 3H⁺(aq) + Bi(OH)₃(s) → Bi(s) + 3H₂O(l)
  • Oxidation half-reaction: 3H₂O(l) + Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq) (6 H on the left, 6 H on the right. Add 3 H⁺ to the right) 3H₂O(l) + Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq) + 3H⁺(aq)

5. Neutralize H⁺ with OH⁻ (Since the Reaction is in Basic Solution):

This is a crucial step for reactions in basic solutions. For every H⁺ ion, we add an OH⁻ ion to both sides of the half-reaction. The H⁺ and OH⁻ ions on the same side will combine to form H₂O.

  • Reduction half-reaction: 3H⁺(aq) + Bi(OH)₃(s) → Bi(s) + 3H₂O(l) (Add 3 OH⁻ to both sides) 3H⁺(aq) + 3OH⁻(aq) + Bi(OH)₃(s) → Bi(s) + 3H₂O(l) + 3OH⁻(aq) 3H₂O(l) + Bi(OH)₃(s) → Bi(s) + 3H₂O(l) + 3OH⁻(aq)
  • Oxidation half-reaction: 3H₂O(l) + Sn(OH)₃⁻(aq) → Sn(OH)₆²⁻(aq) + 3H⁺(aq) (Add 3 OH⁻ to both sides) 3H₂O(l) + Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) + 3H⁺(aq) + 3OH⁻(aq) 3H₂O(l) + Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) + 3H₂O(l)

6. Simplify by Canceling Out Water Molecules (If Any):

  • Reduction half-reaction: 3H₂O(l) + Bi(OH)₃(s) → Bi(s) + 3H₂O(l) + 3OH⁻(aq) (Cancel 3 H₂O from both sides) Bi(OH)₃(s) → Bi(s) + 3OH⁻(aq)
  • Oxidation half-reaction: 3H₂O(l) + Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) + 3H₂O(l) (Cancel 3 H₂O from both sides) Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq)

7. Balance the Charge by Adding Electrons (e⁻):

  • Reduction half-reaction: Bi(OH)₃(s) → Bi(s) + 3OH⁻(aq) (0 charge on the left, -3 on the right. Add 3 e⁻ to the left) 3e⁻ + Bi(OH)₃(s) → Bi(s) + 3OH⁻(aq)
  • Oxidation half-reaction: Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) (-4 charge on the left, -2 on the right. Add 2 e⁻ to the right) Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) + 2e⁻

8. Make the Number of Electrons Equal in Both Half-Reactions:

To do this, we multiply each half-reaction by the appropriate factor so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. The least common multiple of 3 and 2 is 6.

  • Multiply the reduction half-reaction by 2: 2 * [3e⁻ + Bi(OH)₃(s) → Bi(s) + 3OH⁻(aq)] => 6e⁻ + 2Bi(OH)₃(s) → 2Bi(s) + 6OH⁻(aq)
  • Multiply the oxidation half-reaction by 3: 3 * [Sn(OH)₃⁻(aq) + 3OH⁻(aq) → Sn(OH)₆²⁻(aq) + 2e⁻] => 3Sn(OH)₃⁻(aq) + 9OH⁻(aq) → 3Sn(OH)₆²⁻(aq) + 6e⁻

9. Add the Balanced Half-Reactions Together:

Now we add the two balanced half-reactions, canceling out anything that appears on both sides of the equation (in this case, the electrons).

6e⁻ + 2Bi(OH)₃(s) → 2Bi(s) + 6OH⁻(aq) 3Sn(OH)₃⁻(aq) + 9OH⁻(aq) → 3Sn(OH)₆²⁻(aq) + 6e⁻

Adding these together, we get:

2Bi(OH)₃(s) + 3Sn(OH)₃⁻(aq) + 9OH⁻(aq) → 2Bi(s) + 3Sn(OH)₆²⁻(aq) + 6OH⁻(aq)

10. Simplify the Equation (If Possible):

We can simplify this equation by canceling out common terms. We have 9 OH⁻ ions on the left and 6 OH⁻ ions on the right, so we can cancel out 6 OH⁻ ions from both sides, leaving 3 OH⁻ ions on the left:

2Bi(OH)₃(s) + 3Sn(OH)₃⁻(aq) + 3OH⁻(aq) → 2Bi(s) + 3Sn(OH)₆²⁻(aq)

The Final Balanced Equation and the Coefficient of Sn(OH)₃⁻

So, after all that hard work, we have our balanced redox equation in basic solution:

2Bi(OH)₃(s) + 3Sn(OH)₃⁻(aq) + 3OH⁻(aq) → 2Bi(s) + 3Sn(OH)₆²⁻(aq)

And the answer to our original question? The coefficient of Sn(OH)₃⁻ is 3. Woohoo! We did it!

Why Balancing Redox Reactions Matters

You might be thinking, "Okay, that was a lot of steps. But why do we even need to balance these equations?" Well, balancing redox reactions is essential for a few key reasons:

  • Conservation of Mass: A balanced equation ensures that the number of atoms of each element is the same on both sides of the equation. This reflects the fundamental principle of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
  • Stoichiometry: Balanced equations provide the quantitative relationships between reactants and products. This allows us to calculate how much of a reactant is needed to produce a certain amount of product, which is crucial in many chemical applications.
  • Electrochemistry: Redox reactions are the foundation of electrochemistry, the study of the relationship between chemical reactions and electrical energy. Balanced equations are necessary for understanding and designing electrochemical devices like batteries and fuel cells.

Practice Makes Perfect: Tips for Mastering Redox Balancing

Balancing redox reactions can seem daunting at first, but with practice, it becomes much easier. Here are a few tips to help you master the art:

  • Understand Oxidation States: A solid grasp of oxidation state rules is essential for identifying which species are being oxidized and reduced. Review these rules regularly.
  • Break It Down: The half-reaction method is your friend. Breaking the reaction into smaller, manageable parts makes the balancing process much less overwhelming.
  • Practice, Practice, Practice: The more you practice balancing redox reactions, the more comfortable you'll become with the steps involved. Work through various examples, and don't be afraid to make mistakes – that's how we learn!
  • Check Your Work: Always double-check that your final equation is balanced in terms of both atoms and charge. This will help you catch any errors.

Conclusion: Redox Reactions Demystified

Balancing redox reactions in basic solutions might seem like a challenge, but hopefully, this step-by-step guide has made the process clearer. We've seen how the half-reaction method can be used to systematically balance even complex equations, and we've learned why balancing these reactions is so important in chemistry. Remember, the coefficient of Sn(OH)₃⁻ in our example was 3. Now you're armed with the knowledge and skills to tackle any redox balancing problem that comes your way. Keep practicing, and you'll be a redox balancing pro in no time!