Question

Balance these reactions: (a) \(\mathrm{Mo-O}_{u}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MoO}_{4}(s)+\mathrm{Mn}^{2}(a q)\) (acidic solution) (b) \(\mathrm{BrO}^{-}(a q)+\mathrm{Cr}\left(\mathrm{OH}_{4}^{-}(a q) \longrightarrow\right.\) \(\mathrm{Br}^{-}\left(\mathrm{aq}_{)}\right)+\mathrm{CrO}_{4}^{-}(\mathrm{aq})\) (basic solution) (c) \(\mathrm{S}_{2} \mathrm{O}^{2}-(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow\) \(\mathrm{SO}_{4}^{2}(a q)+\mathrm{MnO}_{2}(s)\) (basic solution)

Short answer

For part (a): 8H⁺ + MnO₄⁻ + Mo-Oₓ → Mn²⁺ + MoO₄ + 4H₂O. For part (b): 4Cr(OH)₄⁻ + 3BrO⁻ → 2CrO₄²⁻ + 3Br⁻ + 4OH⁻. For part (c): 2MnO₄⁻ + 3S₂O₃²⁻ → 2MnO₂ + 3SO₄²⁻.

Step-by-step solution

- Identify oxidation and reduction half-reactions (for part a)

Identify the oxidation states of each element to determine which species are being oxidized and reduced. For the given reaction:Mo-Oₓ(s) + MnO₄⁻(aq) → MoO₄(s) + Mn²⁺(aq)Molybdenum (Mo) is oxidized, changing from a lower oxidation state to +6 in MoO₄, and manganese (Mn) is reduced from +7 in MnO₄⁻ to +2 in Mn²⁺.

- Write the half-reactions (for part a)

Write separate half-reactions for oxidation and reduction:Oxidation half-reaction: Mo-Oₓ → MoO₄Reduction half-reaction: MnO₄⁻ → Mn²⁺

- Balance oxygen atoms (for part a)

Balance oxygen atoms in the half-reactions by adding H₂O:Oxidation: Mo-Oₓ + 4H₂O → MoO₄Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O

- Balance hydrogen atoms (for part a)

Balance hydrogen by adding H⁺ ions:Oxidation: Mo-Oₓ + 4H₂O → MoO₄ + 8H⁺Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

- Balance the charges (for part a)

Balance the charge by adding electrons (e⁻):Oxidation: Mo-Oₓ + 4H₂O → MoO₄ + 8H⁺ + 6e⁻Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

- Combine and balance the overall equation (for part a)

Multiply the half-reactions to equalize the number of electrons and combine them:10H⁺ + MnO₄⁻ + Mo-Oₓ → Mn²⁺ + MoO₄ + 4H₂OBalanced equation in acidic solution: 10H⁺ + 2MnO₄⁻ + 2Mo-Oₓ → 2Mn²⁺ + 2MoO₄ + 8H₂O

- Identify oxidation and reduction half-reactions (for part b)

For the reaction:BrO⁻(aq) + Cr(OH)₄⁻(aq) → Br⁻(aq) + CrO₄²⁻(aq)Bromine (Br) is reduced from +1 in BrO⁻ to -1 in Br⁻, and chromium (Cr) is oxidized from +3 in Cr(OH)₄⁻ to +6 in CrO₄²⁻.

- Write the half-reactions (for part b)

Oxidation half-reaction: Cr(OH)₄⁻ → CrO₄²⁻Reduction half-reaction: BrO⁻ → Br⁻

- Balance oxygen and hydrogen atoms (part b)

Oxidation: Cr(OH)₄⁻ → CrO₄²⁻ + 2H₂OReduction: BrO⁻ + 2H₂O → Br⁻ + 2OH⁻

- Balance the charges (for part b)

Oxidation: Cr(OH)₄⁻ → CrO₄²⁻ + 4e⁻Reduction: BrO⁻ + 6e⁻ → Br⁻ + 6H₂O

- Combine and balance the overall equation (for part b)

Combine the reactions:2Cr(OH)₄⁻ + 3BrO⁻ → 2CrO₄²⁻ + 3Br⁻ + 4OH⁻

- Identify oxidation and reduction half-reactions (for part c)

For the reaction:S₂O₃²⁻(aq) + MnO₄⁻(aq) → SO₄²⁻(aq) + MnO₂(s)Sulfur (S) in S₂O₃²⁻ is oxidized to SO₄²⁻, and manganese (Mn) is reduced from +7 in MnO₄⁻ to +4 in MnO₂.

- Write the half-reactions (for part c)

Oxidation half-reaction: S₂O₃²⁻ → SO₄²⁻Reduction half-reaction: MnO₄⁻ → MnO₂

- Balance oxygen and hydrogen atoms (for part c)

Oxidation: S₂O₃²⁻ + 11H₂O → 2SO₄²⁻ + 20H⁺Reduction: MnO₄⁻ + 4H₂O → MnO₂ + 8H⁺

- Balance the charges (for part c)

Oxidation: S₂O₃²⁻ → 2SO₄²⁻ + 2e⁻Reduction: MnO₄⁻ + 4e⁻ → MnO₂

- Combine and balance the overall equation (for part c)

Multiply and combine the reactions:2MnO₄⁻ + 3S₂O₃²⁻ → 2MnO₂ + 6SO₄²⁻

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, are chemical reactions that involve the transfer of electrons between two substances. These reactions are fundamental in various biological, industrial, and environmental processes. In a redox reaction, one substance loses electrons (oxidation) while the other gains electrons (reduction). Since these reactions are coupled, the loss and gain of electrons happen simultaneously.

Here’s an easy way to remember: OIL RIG.

• OIL: Oxidation Is Loss (of electrons)
• RIG: Reduction Is Gain (of electrons)

Let's break it down with an example: In the reaction \[\mathrm{Mo-O}_{u}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MoO}_{4}(s)+\mathrm{Mn}^{2}(a q)\]

The Molybdenum (Mo) is oxidized as it loses electrons, while the Manganese (Mn) is reduced because it gains electrons. Identifying and understanding these reactions are key to mastering topics like energy production, metabolism, and even corrosion.

Half-Reactions

Half-reactions are a method of breaking down redox reactions into two simpler parts to simplify balancing. Each half-reaction either represents oxidation or reduction. By separating them, you isolate the transfer of electrons, making it easier to balance the equation.

For instance, in the redox reaction: \[\mathrm{Mo-O}_{u}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MoO}_{4}(s)+\mathrm{Mn}^{2}(a q)\]

You can split it into:

• Oxidation half-reaction: \[\mathrm{Mo-O}_{u} \rightarrow \mathrm{MoO}_{4} + 4H_{2}O\]
• Reduction half-reaction: \[\mathrm{MnO}_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4H_{2}O\]

Steps to balance half-reactions involve:

• Balancing atoms other than oxygen and hydrogen first
• Balancing oxygen atoms by adding \(H_{2}O\)
• Balancing hydrogen atoms by adding \(H^{+}\) ions
• Balancing charges by adding electrons \(e^{-}\)

Finally, you combine the balanced half-reactions, ensuring the same number of electrons are lost in oxidation and gained in reduction. This method makes it a lot easier to systematically approach more complex redox reactions.

Acidic and Basic Solutions

Balancing chemical equations often depend on whether the solution is acidic or basic. The environment changes how you address hydrogen and oxygen atoms in half-reactions.

In **acidic solutions**, focus on using \(H^{+}\) ions to balance hydrogen atoms. For example:

• Oxidation half-reaction: \[\mathrm{Mo-O}_{u} + 4H_{2}O \rightarrow \mathrm{MoO}_{4} + 8H^{+} + 6e^{-}\]
• Reduction half-reaction: \[\mathrm{MnO}_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4H_{2}O\]

**Basic solutions** require using \(OH^{-}\) ions to balance hydrogen atoms. For instance:

• Oxidation half-reaction: \[\mathrm{Cr(OH)_{4}^{-} \rightarrow CrO_{4}^{-} + 2H_{2}O + 4e^{-}}\]
• Reduction half-reaction: \[\mathrm{BrO^{-} + 2H_{2}O + 6e^{-} \rightarrow Br^{-} + 6OH^{-}}\]

When balancing equations in basic solutions, you sometimes end up with extra \(H_{2}O\) or \(OH^{-}\) ions that you'll need to adjust by adding more water or hydroxide ions to both sides of the equation. Recognizing the solution type is essential because it dictates your approach to balancing redox reactions and helps ensure accuracy.