Problem 24 - Chemistry Matter and Change

Question

Use the half-reaction method to balance the redox equations. Begin by writing the oxidation and reduction half-reactions. Leave the balanced equation in ionic form. \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Bi}^{2+}(\mathrm{aq})\) (in acid solution)

Step-by-Step Solution

Answer

Here's the simplified answer: The balanced redox equation in ionic form is: \(2\mathrm{Mn}^{2+}(\mathrm{aq})+5\mathrm{BiO}_{3}^{-}(\mathrm{aq})+16\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{MnO}_{4}^{-}(\mathrm{aq})+5\mathrm{Bi}^{2+}(\mathrm{aq})+22\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\)

Step 1: Identify Oxidation and Reduction Half-Reactions

In the given reaction, $\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Bi}^{2+}(\mathrm{aq})$ \(\mathrm{Mn}^{2+}\) is being oxidized to \(\mathrm{MnO}_{4}^{-}\), and \(\mathrm{BiO}_{3}^{-}\) is being reduced to \(\mathrm{Bi}^{2+}\). Oxidation half-reaction: \(\mathrm{Mn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})\) Reduction half-reaction: \(\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Bi}^{2+}(\mathrm{aq})\)

Step 2: Balance Atoms in the Half-Reactions

Balance the atoms other than O and H in each half-reaction. Oxidation half-reaction: \(\mathrm{Mn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})\) (Mn atoms are already balanced) Reduction half-reaction: \(\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Bi}^{2+}(\mathrm{aq})\) (Bi atoms are already balanced) Now, balance O atoms using H2O: Oxidation half-reaction: \(\mathrm{Mn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4 \mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\) Reduction half-reaction: \(\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Bi}^{2+}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\) Next, balance H atoms using H+ ions since it is in an acidic solution: Oxidation half-reaction: \(\mathrm{Mn}^{2+}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4 \mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\) Reduction half-reaction: \(\mathrm{BiO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Bi}^{2+}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\)

Step 3: Balance Charges in the Half-Reactions

Balance the charges in each half-reaction by adding electrons (e-) to the side with a higher positive charge. Oxidation half-reaction: \(\mathrm{Mn}^{2+}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4 \mathrm{H}_{2}\mathrm{O} (\mathrm{aq}) + 5e^-\) Reduction half-reaction: \(\mathrm{BiO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{H}^{+}(\mathrm{aq}) +2e^- \rightarrow \mathrm{Bi}^{2+}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\)

Step 4: Equalize the Electrons in Both Half-Reactions

Multiply each half-reaction by a suitable integer so that the total number of electrons in both half-reactions becomes equal. For oxidation half-reaction, multiply by 2: \(2[\mathrm{Mn}^{2+}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq})] \rightarrow 2[\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4 \mathrm{H}_{2}\mathrm{O} (\mathrm{aq}) + 5e^-]\) For reduction half-reaction, multiply by 5: \(5[\mathrm{BiO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{H}^{+}(\mathrm{aq}) +2e^-] \rightarrow 5[\mathrm{Bi}^{2+}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})]\)

Step 5: Combine Half-Reactions and Cancel Electrons

Add the two half-reactions together and cancel the common terms (electrons), and then simplify. \(2[\mathrm{Mn}^{2+}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq})] + 5[\mathrm{BiO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{H}^{+}(\mathrm{aq}) +2e^-] \rightarrow 2[\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4 \mathrm{H}_{2}\mathrm{O} (\mathrm{aq}) + 5e^-] + 5[\mathrm{Bi}^{2+}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})]\)

Step 6: Simplify and Write the Balanced Equation in Ionic Form

Simplify the equation and write it in ionic form: \(2\mathrm{Mn}^{2+}(\mathrm{aq})+5\mathrm{BiO}_{3}^{-}(\mathrm{aq})+16\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{MnO}_{4}^{-}(\mathrm{aq})+5\mathrm{Bi}^{2+}(\mathrm{aq})+22\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\) The balanced redox equation in ionic form is: \(2\mathrm{Mn}^{2+}(\mathrm{aq})+5\mathrm{BiO}_{3}^{-}(\mathrm{aq})+16\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{MnO}_{4}^{-}(\mathrm{aq})+5\mathrm{Bi}^{2+}(\mathrm{aq})+22\mathrm{H}_{2}\mathrm{O} (\mathrm{aq})\)

Key Concepts

Understanding the Half-Reaction Method

The half-reaction method is a popular technique for balancing redox reactions. It involves splitting the overall reaction into two separate equations—one for the oxidation process and one for the reduction process.
Each half-reaction is balanced separately and then combined to give a balanced overall equation.
This method is especially useful for complex reactions involving multiple elements that change their oxidation states.

• Oxidation Half-Reaction: In this half, focus on the species that lose electrons (become oxidized). Identify which atoms are oxidized, balance them, and add electrons to the product side.
• Reduction Half-Reaction: Similarly, pay attention to the species that gain electrons (are reduced). Balance these atoms and add electrons to the reactant side.

Using the half-reaction method ensures that the conservation of mass and charge is maintained, keeping our chemical equations precise and accurate. It offers a structured approach to tackling redox processes, essential for learning about future chemical analysis and applications.

Oxidation-Reduction Reactions Simplified

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between chemical species. They are fundamental processes representing how substances get transformed in their chemical structure.
To comprehend these reactions, you need to understand the concepts of oxidation and reduction individually:

• Oxidation: This is the process where a substance loses electrons. In the context of our exercise, \( \mathrm{Mn}^{2+} \) loses electrons to become \( \mathrm{MnO}_{4}^{-} \).
• Reduction: In contrast, reduction involves the gain of electrons. Here, \( \mathrm{BiO}_{3}^{-} \) gains electrons to become \( \mathrm{Bi}^{2+} \).

Combining these two processes forms a redox reaction. It's vital to note that the number of electrons lost in oxidation must equal the number gained in reduction. This equality ensures both sides of the reaction are balanced in terms of charge, not just atoms.

Balancing in Acidic Solutions

When balancing redox reactions, it's common to encounter reactions occurring in acidic solutions. Acidic conditions can influence the balancing method. Here’s how this environment affects the process:

• Use of \( \mathrm{H}^{+} \) ions: In acidic solutions, you'll often add \( \mathrm{H}^{+} \) ions to balance the hydrogen atoms after initially balancing other elements and oxygen with water molecules \( \mathrm{H}_{2}\mathrm{O} \).
• Final Charge Balancing: After balancing for atoms, ensure the overall charge of each half-reaction is equalized by adding electrons. In acidic solutions, this might mean having a net positive charge on one side that is counteracted with electrons.

Balancing in an acidic solution often requires extra attention to detail, since the medium significantly impacts the stoichiometry and often requires additional steps like adding \( \mathrm{H}_{2}\mathrm{O} \) or \( \mathrm{H}^{+} \). This systematic approach to balancing ensures our final equation is not only balanced atomically but also retains proper charge balance. It reflects the conditions of acid solutions, offering a realistic representation of these reactions in real-world chemical processes.