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Chapter 4: Problem 65

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) \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{H}_{2}(g)\)

Short Answer

Expert verified
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 by step solution

01

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 \(\mathrm{I}^{-}\).
02

Reaction (a) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{SO}_{2} \longrightarrow \mathrm{SO}_{3}\) Reduction half-reaction: \(\mathrm{I}_{2} \longrightarrow 2 \mathrm{I}^{-}\)
03

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

Oxidation half-reaction: \(\mathrm{SO}_{2} \longrightarrow \mathrm{SO}_{3} + 2e^{-}\) Reduction half-reaction: \(\mathrm{I}_{2} + 2e^{-} \longrightarrow 2 \mathrm{I}^{-}\)
04

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)
05

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 \(\mathrm{NH}_{3}\).
06

Reaction (b) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}\) Reduction half-reaction: \(\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NH}_{3}\)
07

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

Oxidation half-reaction: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+} + 2e^{-}\) Reduction half-reaction: \(2 \mathrm{NO}_{3}^{-} + 10e^{-} + 12 \mathrm{H}_{2}\mathrm{O} \longrightarrow 2 \mathrm{NH}_{3} + 6 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{OH}^{-}\)
08

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)
09

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 \(\mathrm{CrO}_{4}^{2-}\).
10

Reaction (c) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{CrO}_{2}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}\) Reduction half-reaction: \(\mathrm{ClO}^{-} \longrightarrow \mathrm{Cl}^{-}\)
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: \(2 \mathrm{ClO}^{-} + 2e^{-} \longrightarrow 2 \mathrm{Cl}^{-} + \mathrm{O}_{2} + 2 \mathrm{OH}^{-}\)
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)
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 \(\mathrm{H}_{2}\).
14

Reaction (d) - Write Oxidation and Reduction Half-reactions

Oxidation half-reaction: \(\mathrm{K} \longrightarrow \mathrm{K}^{+}\) Reduction half-reaction: \(\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{H}_{2}\)
15

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

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

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