Problem 31 - Chemistry Principles and Practice

Question

Balance each of the following redox reactions in acid solution. (a) \(\mathrm{Fe}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{I}_{2}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})\) (c) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Mn}^{2+}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) \( \mathrm{Fe} + 2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag} + \mathrm{Fe}^{2+} \); (b) \( \mathrm{I}_{2} + 2 \mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-} + \mathrm{S}_{4}\mathrm{O}_{6}^{2-} \); (c) \( \mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 \mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H}_{2}\mathrm{O} \).

Step 1: Separate and Identify Half-Reactions (a)

For the reaction \( \mathrm{Fe} + \mathrm{Ag}^{+} \rightarrow \mathrm{Ag} + \mathrm{Fe}^{2+} \), identify the oxidation and reduction half-reactions. - Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^{-} \)- Reduction: \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \)

Step 2: Balance Electrons in Half-Reactions (a)

The number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. Therefore, multiply the entire reduction half-reaction by 2:- Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^{-} \)- Reduction: \( 2 \mathrm{Ag}^{+} + 2e^{-} \rightarrow 2 \mathrm{Ag} \)

Step 3: Combine and Balance Overall Reaction (a)

Combine the balanced half-reactions:\[ \mathrm{Fe} + 2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag} + \mathrm{Fe}^{2+} \] Verify that atoms and charges are balanced. The equation is now balanced.

Step 4: Separate and Identify Half-Reactions (b)

For the reaction \( \mathrm{I}_{2} + \mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow \mathrm{I}^{-} + \mathrm{S}_{4}\mathrm{O}_{6}^{2-} \), identify the half-reactions.- Reduction: \( \mathrm{I}_{2} + 2e^{-} \rightarrow 2 \mathrm{I}^{-} \)- Oxidation: \( 2 \mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow \mathrm{S}_{4}\mathrm{O}_{6}^{2-} + 2e^{-} \)

Step 5: Combine and Balance Overall Reaction (b)

Since the electrons in both half-reactions are already balanced, combine them:\[ \mathrm{I}_{2} + 2 \mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-} + \mathrm{S}_{4}\mathrm{O}_{6}^{2-} \] Verify that atoms and charges are balanced. The equation is balanced.

Step 6: Separate and Identify Half-Reactions (c)

For the reaction \( \mathrm{MnO}_{4}^{-} + \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} \), identify the half-reactions.- Reduction: \( \mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O} \)- Oxidation: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^{-} \)

Step 7: Balance Electrons in Half-Reactions (c)

To balance electrons, multiply the oxidation half-reaction by 5:- Reduction: \( \mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O} \)- Oxidation: \( 5 \mathrm{Fe}^{2+} \rightarrow 5 \mathrm{Fe}^{3+} + 5e^{-} \)

Step 8: Combine and Balance Overall Reaction (c)

Combine the balanced half-reactions:\[ \mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 \mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H}_{2}\mathrm{O} \] Ensure atoms and charges are balanced. The equation is balanced.

Key Concepts

Oxidation and Reduction

In chemistry, oxidation and reduction are two fundamental processes that involve the transfer of electrons. These reactions are often paired, as they occur simultaneously. This is why they are commonly referred to as redox reactions.

• Oxidation is defined as the loss of electrons by a molecule, atom, or ion. During this process, the oxidation state of the substance increases.
• Reduction involves the gain of electrons, leading to a decrease in oxidation state.

In the context of the original exercise, we can see oxidation happening as iron (Fe) loses electrons to form Fe²⁺, and reduction as silver ions (Ag⁺) gain electrons to form silver (Ag).
In any redox reaction, the total number of electrons lost in oxidation must equal the number of electrons gained in reduction. This balance is crucial for maintaining charge neutrality and stabilizing the reaction.

Balancing Chemical Equations

Balancing chemical equations is a critical skill in chemistry, particularly when dealing with redox reactions. The goal is to have the same number of each type of atom on both sides of the equation, ensuring mass and charge conservation.
To balance a redox reaction, following these steps can be helpful:

• First, separate the reaction into half-reactions that involve oxidation and reduction.
• Next, balance all atoms in each half-reaction, starting with elements other than hydrogen and oxygen.
• Then, balance oxygen atoms by adding water molecules.
• Balance hydrogen atoms by adding hydrogen ions (H⁺).
• Finally, balance charges by adding electrons to one side of the half-reaction.

The charge and atoms in the reactions must be equal both before and after the reaction occurs. For example, in the exercise's reaction involving MnO₄^- and Fe²⁺, we balanced the electrons by ensuring both half-reactions shared an equal electron count, leading to a balanced chemical equation.

Half-Reaction Method

The half-reaction method is an effective technique for balancing redox reactions, especially in complex equations. This method involves breaking down a redox reaction into its two distinct processes: oxidation and reduction.
Here's how to apply the half-reaction method:

• Identify and write out the two half-reactions.
• Each half-reaction should have elements and charges balanced independently first.
• Balance the electrons in each half-reaction so that they cancel out in sum.
• Combine the balanced half-reactions, ensuring that all electrons are canceled out, resulting in a charge-balanced full reaction.

This method is particularly useful because it focuses on smaller, more manageable parts of the equation first.
In the initial exercise, each reaction was split into its oxidation and reduction components before balancing. This systematic approach helps in visualizing the changes that occur during the reaction and ensures a smooth pathway to a balanced equation.