Question

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. \(\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)\) c. \(\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow\) $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

Short answer

The short answer for the balanced reactions are: a. \(3 \mathrm{PO}_{3}^{3-} + 2\: \mathrm{MnO}_{4}^{-} + 14\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 2\: \mathrm{MnO}_{2} + 7\: \mathrm{H}_{2}\mathrm{O}\) b. \(2 \mathrm{Mg}(s) + 2 \mathrm{OCl}^{-}(a q) + 4 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{Mg}(\mathrm{OH})_{2}(s) + 2 \mathrm{Cl}^{-}(a q)\) c. \(3 \mathrm{H}_{2} \mathrm{CO}(a q) + 2 \mathrm{Ag}(\mathrm{NH}_{3})_{2}^{+}(a q) \rightarrow 3 \mathrm{HCO}_{3}^{-}(a q) + 2 \mathrm{Ag}(s) + 6 \mathrm{NH}_{3}(a q)\)

Step-by-step solution

Determine the oxidation states

Assign oxidation states for all elements in the reactants and products. Here: P: +3 (in \(\mathrm{PO}_{3}^{3-}\)), +5 (in \(\mathrm{PO}_{4}^{3-}\)) Mn: +7 (in \(\mathrm{MnO}_{4}^{-}\)), +4 (in \(\mathrm{MnO}_{2}\))

Write half-reactions

Write the oxidation half-reaction and reduction half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-}\) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2}\)

Balance atoms

Balance atoms other than oxygen and hydrogen in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-}\) (P atoms balanced) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2}\) (Mn atoms balanced)

Balance oxygen atoms

Add water molecules to balance the number of oxygen atoms in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O}\)

Balance hydrogen atoms

Add hydroxide ions (\(\mathrm{OH}^-\)) to balance the number of hydrogen atoms in each half-reaction. Oxidation: \(\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O}\)

Balance charge

Balance the charges in each half-reaction by adding electrons. Oxidation: \(\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-} + 2\: e^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O}\) Reduction: \(\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O} + 3\: e^{-}\)

Equalize electrons and add half-reactions

Equalize the number of electrons in both half-reactions and add the half-reactions. Oxidation: \(3\cdot (\mathrm{PO}_{3}^{3-} + 2\: \mathrm{OH}^{-} + 2\: e^{-}\rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2}\mathrm{O})\) Reduction: \(2\cdot (\mathrm{MnO}_{4}^{-} + 4\: \mathrm{OH}^{-}\rightarrow \mathrm{MnO}_{2} + 2\: \mathrm{H}_{2}\mathrm{O} + 3\: e^{-})\) Adding the equalized half-reactions: \(3 \mathrm{PO}_{3}^{3-} + 6\: \mathrm{OH}^{-} + 2\: \mathrm{MnO}_{4}^{-} + 8\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 3\: \mathrm{H}_{2}\mathrm{O} + 2\: \mathrm{MnO}_{2} + 4\: \mathrm{H}_{2}\mathrm{O}\)

Simplify and write balanced equation

Simplify the equation and write the final balanced equation. \(3 \mathrm{PO}_{3}^{3-} + 2\: \mathrm{MnO}_{4}^{-} + 14\: \mathrm{OH}^{-} \rightarrow 3 \mathrm{PO}_{4}^{3-} + 2\: \mathrm{MnO}_{2} + 7\: \mathrm{H}_{2}\mathrm{O}\) Similarly, analyze and balance reactions b. and c. using the same step-by-step procedure in base.

Key concepts

Half-Reaction Method

To balance oxidation-reduction reactions, we often use the half-reaction method, especially in aqueous solutions. This approach divides the overall reaction into two parts: one showcasing oxidation, where electrons are lost, and the other showing reduction, where electrons are gained. This separation highlights the electron transfer process, which is essential in redox reactions.

**Oxidation and Reduction**
Each half-reaction contains steps that depict the oxidation and reduction processes separately. To apply this method:

• Identify and write down the oxidation and reduction half-reactions.
• Balance these for all atoms except oxygen and hydrogen initially.
• Balance oxygen by adding water ( H_2O ight).
• Balance hydrogen using hydroxide ions ( OH^- ight) when in basic conditions.
• Add electrons to equalize the charge.

This breakdown simplifies the complex task of balancing the full chemical equation, making it more manageable and methodical.

Redox Balancing in Basic Solution

Balancing redox reactions in basic solutions requires a slightly adjusted approach from standard conditions because the presence of OH^- ight) ions instead of protons ( H^+ ight) can complicate matters. When you're dealing with basic solutions:

**Steps to Balance**

• First, follow the typical steps of the half-reaction method, including balancing oxygen and hydrogen.
• Once hydrogen atoms are adjusted in an acidic context, add OH^- ight ions equal to the H^+ ight ions on both sides to form water, keeping the solution basic.
• Simplify water molecules on both sides if necessary, which might involve canceling out equivalent amounts of H_2O ight.

This strategy ensures the balanced equation accurately reflects the basic medium, respecting the reaction environment specific to the problem.

Chemical Equations Balancing

Balancing chemical equations involves ensuring the number of atoms for each element is the same on both sides of the equation. For redox reactions, a systematic approach like the half-reaction method can ease this task, particularly under different environmental conditions (acidic or basic).

**Key Considerations**

• Identifying oxidation states is crucial – it allows you to determine which elements are oxidized and reduced.
• Focus on conservation of mass and charge; neither should be created nor destroyed.
• After completing half-reactions, ensure that electrons lost in oxidation equal those gained in reduction.
• Simplify your final equation by canceling common species, such as H_2O ight or OH^- ight, appearing on both sides of the reaction.

Mastering chemical equations by understanding the nuances involved not only builds foundational chemistry skills but also nurtures an appreciation for these vital chemical processes.