Problem 65 - Chemistry: Principles and Reactions

Question

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{SO}_{2}(g)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{SO}_{3}(g)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{Zn}^{2+}(a q)\) (c) \(\mathrm{ClO}^{-}(a q)+\mathrm{CrO}_{2}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{CrO}_{4}^{2-}(a q)\) (d) \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{H}_{2}(g)\)

Step-by-Step Solution

Answer

In the given redox reactions, the balanced equations are: 1. \(\mathrm{SO}_{2} + \mathrm{I}_{2} \longrightarrow \mathrm{SO}_{3} + 2 \mathrm{I}^{-}\) 2. \(5 \mathrm{Zn} + 10 \mathrm{NO}_{3}^{-} + 12 \mathrm{H}_{2}\mathrm{O} \longrightarrow 5 \mathrm{Zn}^{2+} + 2 \mathrm{NH}_{3} + 6 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{OH}^{-}\) 3. \(2 \mathrm{ClO}^{-} + \mathrm{CrO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow 2 \mathrm{Cl}^{-} + \mathrm{O}_{2} + 2 \mathrm{OH}^{-} + \mathrm{CrO}_{4}^{2-} + 4 \mathrm{OH}^{-}\) 4. \(2 \mathrm{H}_{2}\mathrm{O} + 2\mathrm{K} \longrightarrow 2\mathrm{OH}^{-} + \mathrm{H}_{2} + 2\mathrm{K}^{+}\)

Step 1: Reaction (a) - Identify Oxidation and Reduction Atoms

In this reaction, sulfur in \(\mathrm{SO}_{2}\) is oxidized to \(\mathrm{SO}_{3}\) and iodine in \(\mathrm{I}_{2}\) is reduced to \(\mathrm{I}^{-}\).

Step 2: Reaction (a) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{SO}_{2} \longrightarrow \mathrm{SO}_{3}\) Reduction half-reaction: \(\mathrm{I}_{2} \longrightarrow 2 \mathrm{I}^{-}\)

Step 3: Reaction (a) - Balance Atoms and Charges in Half-reactions

Oxidation half-reaction: \(\mathrm{SO}_{2} \longrightarrow \mathrm{SO}_{3} + 2e^{-}\) Reduction half-reaction: \(\mathrm{I}_{2} + 2e^{-} \longrightarrow 2 \mathrm{I}^{-}\)

Step 4: Reaction (a) - Combine and Simplify Half-reaction

\(\mathrm{SO}_{2} + \mathrm{I}_{2} + 2e^{-} \longrightarrow \mathrm{SO}_{3} + 2 \mathrm{I}^{-} + 2e^{-}\) Remove 2 electrons from both sides: \(\mathrm{SO}_{2} + \mathrm{I}_{2} \longrightarrow \mathrm{SO}_{3} + 2 \mathrm{I}^{-}\) (Balanced)

Step 5: Reaction (b) - Identify Oxidation and Reduction Atoms

In this reaction, zinc in \(\mathrm{Zn}\) is oxidized to \(\mathrm{Zn}^{2+}\) and nitrogen in \(\mathrm{NO}_{3}^{-}\) is reduced to \(\mathrm{NH}_{3}\).

Step 6: Reaction (b) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}\) Reduction half-reaction: \(\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NH}_{3}\)

Step 7: Reaction (b) - Balance Atoms and Charges in Half-reactions

Oxidation half-reaction: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+} + 2e^{-}\) Reduction half-reaction: \(2 \mathrm{NO}_{3}^{-} + 10e^{-} + 12 \mathrm{H}_{2}\mathrm{O} \longrightarrow 2 \mathrm{NH}_{3} + 6 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{OH}^{-}\)

Step 8: Reaction (b) - Balance Electrons and Combine Half-reactions

Multiply the oxidation half-reaction by 5 and add to reduction half-reaction: \(5 \mathrm{Zn} + 10 \mathrm{NO}_{3}^{-} + 10e^{-} + 12 \mathrm{H}_{2}\mathrm{O} \longrightarrow 5 \mathrm{Zn}^{2+} + 2 \mathrm{NH}_{3} + 6 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{OH}^{-} + 10e^{-}\) Remove 10 electrons from both sides: \(5 \mathrm{Zn} + 10 \mathrm{NO}_{3}^{-} + 12 \mathrm{H}_{2}\mathrm{O} \longrightarrow 5 \mathrm{Zn}^{2+} + 2 \mathrm{NH}_{3} + 6 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{OH}^{-}\) (Balanced)

Step 9: Reaction (c) - Identify Oxidation and Reduction Atoms

In this reaction, chlorine in \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\) and chromium in \(\mathrm{CrO}_{2}^{-}\) is oxidized to \(\mathrm{CrO}_{4}^{2-}\).

Step 10: Reaction (c) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{CrO}_{2}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}\) Reduction half-reaction: \(\mathrm{ClO}^{-} \longrightarrow \mathrm{Cl}^{-}\)

Step 11: Reaction (c) - Balance Atoms and Charges in Half-reactions

Oxidation half-reaction: \(\mathrm{CrO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{CrO}_{4}^{2-} + 2e^{-} + 4 \mathrm{OH}^{-}\) Reduction half-reaction: \(2 \mathrm{ClO}^{-} + 2e^{-} \longrightarrow 2 \mathrm{Cl}^{-} + \mathrm{O}_{2} + 2 \mathrm{OH}^{-}\)

Step 12: Reaction (c) - Combine and Simplify Half-reaction

\(2 \mathrm{ClO}^{-} + 2e^{-} + \mathrm{CrO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow 2 \mathrm{Cl}^{-} + \mathrm{O}_{2} + 2 \mathrm{OH}^{-} + \mathrm{CrO}_{4}^{2-} + 2e^{-} + 4 \mathrm{OH}^{-}\) Remove 2 electrons from both sides: \(2 \mathrm{ClO}^{-} + \mathrm{CrO}_{2}^{-} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow 2 \mathrm{Cl}^{-} + \mathrm{O}_{2} + 2 \mathrm{OH}^{-} + \mathrm{CrO}_{4}^{2-} + 4 \mathrm{OH}^{-}\) (Balanced)

Step 13: Reaction (d) - Identify Oxidation and Reduction Atoms

In this reaction, potassium in \(\mathrm{K}\) is oxidized to \(\mathrm{K}^{+}\) and hydrogen in \(\mathrm{H}_{2}\mathrm{O}\) is reduced to \(\mathrm{H}_{2}\).

Step 14: Reaction (d) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{K} \longrightarrow \mathrm{K}^{+}\) Reduction half-reaction: \(\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{H}_{2}\)

Step 15: Reaction (d) - Balance Atoms and Charges in Half-reactions

Oxidation half-reaction: \(\mathrm{K} \longrightarrow \mathrm{K}^{+} + e^{-}\) Reduction half-reaction: \(2 \mathrm{H}_{2}\mathrm{O} + 2e^{-} \longrightarrow 2 \mathrm{OH}^{-} + \mathrm{H}_{2}\)

Step 16: Reaction (d) - Combine and Simplify Half-reaction

\(2 \mathrm{H}_{2}\mathrm{O} + 2e^{-} + 2\mathrm{K} \longrightarrow 2\mathrm{OH}^{-} + \mathrm{H}_{2} + 2\mathrm{K}^{+} + 2e^{-}\) Remove 2 electrons from both sides: \(2 \mathrm{H}_{2}\mathrm{O} + 2\mathrm{K} \longrightarrow 2\mathrm{OH}^{-} + \mathrm{H}_{2} + 2\mathrm{K}^{+}\) (Balanced)

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, involve the transfer of electrons between two substances. One substance loses electrons, while the other gains them. The substance that loses electrons undergoes oxidation. On the other hand, the substance that gains electrons undergoes reduction.
In redox reactions, identifying which elements are oxidized and which are reduced is crucial. Consider reaction (a): \(\mathrm{SO}_{2}+\mathrm{I}_{2} \rightarrow \mathrm{SO}_{3}+2\mathrm{I}^{-}\). Here, sulfur in \(\mathrm{SO}_{2}\) is oxidized to form \(\mathrm{SO}_{3}\), shedding electron \(e^{-} ightarrow e^{-}\). Meanwhile, iodine in \(\mathrm{I}_{2}\) is reduced by gaining electrons to form \(2\mathrm{I}^{-}\).
Examining these reactions helps students comprehend the dynamic electron transfers that drive these chemical reactions. Mastering concepts of oxidation and reduction not only clarifies the balance of equations but aids in predicting the behavior of substances under varying conditions.

Half-Reactions

Half-reactions split the overall redox reaction into two parts: one for oxidation and one for reduction. Each half-reaction details the process of oxidation or reduction with a deep focus on electron transfer.
Taking reaction (b) as an example: \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}\) represents the oxidation and \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NH}_{3}\) the reduction. Here, zinc loses two electrons, noted as \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^{-}\), implying its oxidation. Similarly, nitrogen in \(\mathrm{NO}_{3}^{-}\) gains electrons in converting to \(\mathrm{NH}_{3}\), detailed by showing electron absorption.
Expressing reactions in such a manner gives clearer insights into electron flow, ensuring each component plays its role in balance. Students can then manipulate reactions, ensuring they consider conservation of mass and charge throughout.

Basic Solution Chemistry

Basic solutions, characterized by higher concentrations of hydroxide ions \(\mathrm{OH}^{-}\), often play a pivotal role in balancing redox reactions, specifically in basic mediums. When dealing with basic solutions, it's essential to balance additional elements like water and \(\mathrm{OH}^{-}\) ions to ensure the reaction's atoms and charges are fully balanced.
Take reaction (c): \(\mathrm{ClO}^{-} + \mathrm{CrO}_{2}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{CrO}_{4}^{2-}\). Here, balancing involves aligning \(\mathrm{OH}^{-}\) ions on both sides to maintain neutrality. In a basic solution setting, you ensure the reaction accommodates every water molecule and hydroxide ion affecting the transformation.
Balancing complex reactions in basic solutions demands attentiveness to all participants, including seemingly minor ones, accentuating the integrative nature of chemistry in everyday contexts.