Problem 64 - Chemistry Matter and Change

Question

Write the oxidation and reduction half-reactions represented in each of these redox equations. Write the half-reactions in net ionic form if they occur in aqueous solution. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{PbO}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Pb}(\mathrm{s})} \\ {\text { b. } \mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq})} \\ {\text { c. } \mathrm{Sn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{SnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})}\end{array} \end{equation}

Step-by-Step Solution

Answer

(a) \( \mathrm{PbO} + 2 \mathrm{e}^- \rightarrow \mathrm{Pb} \); \( 2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 6 \mathrm{H}^+ + 6 \mathrm{e}^- \). (b) \( \mathrm{I}_2 + 2 \mathrm{e}^- \rightarrow 2 \mathrm{I}^- \); \( 2 \mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} + 2 \mathrm{e}^- \). (c) \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2 \mathrm{e}^- \); \( 2 \mathrm{H}^+ + 2 \mathrm{e}^- \rightarrow \mathrm{H}_2 \).

Step 1: Identifying Oxidation and Reduction Components in Equation (a)

In the reaction \[ \mathrm{PbO}(s) + \mathrm{NH}_3(g) \rightarrow \mathrm{N}_2(g) + \mathrm{H}_2\mathrm{O}(l) + \mathrm{Pb}(s) \]lead(IV) oxide \( \mathrm{PbO}(s) \) is reduced to lead \( \mathrm{Pb}(s) \), and ammonia \( \mathrm{NH}_3(g) \) is oxidized to nitrogen \( \mathrm{N}_2(g) \).First, write the reduction of \( \mathrm{PbO} \):1. Convert PbO to Pb: \[ \mathrm{PbO}(s) + 2 \mathrm{e}^- \rightarrow \mathrm{Pb}(s) + \mathrm{O}^{2-} \]Then, write the oxidation of \( \mathrm{NH}_3 \): 1. Convert \( \mathrm{NH}_3 \) to \( \mathrm{N}_2 \):\[ 2 \mathrm{NH}_3(g) \rightarrow \mathrm{N}_2(g) + 6 \mathrm{H}^+ + 6 \mathrm{e}^- \]

Step 2: Identifying Oxidation and Reduction Components in Equation (b)

The given reaction is:\[ \mathrm{I}_2(s) + \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3(aq) \rightarrow \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_4(aq) + \mathrm{NaI}(aq) \]Iodine \( \mathrm{I}_2 \) is reduced to iodide \( \mathrm{I}^- \), and thiosulfate \( \mathrm{S}_2\mathrm{O}_3^{2-} \) is oxidized to tetrathionate \( \mathrm{S}_2\mathrm{O}_4^{2-} \).1. Reduction half-reaction of iodine:\[ \mathrm{I}_2(s) + 2 \mathrm{e}^- \rightarrow 2 \mathrm{I}^-(aq) \]2. Oxidation half-reaction of thiosulfate:\[ 2 \mathrm{S}_2\mathrm{O}_3^{2-}(aq) \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-}(aq) + 2 \mathrm{e}^- \]

Step 3: Identifying Oxidation and Reduction Components in Equation (c)

For the reaction:\[ \mathrm{Sn}(s) + 2 \mathrm{HCl}(aq) \rightarrow \mathrm{SnCl}_2(aq) + \mathrm{H}_2(g) \]Tin \( \mathrm{Sn} \) is oxidized to \( \mathrm{Sn}^{2+} \), and hydrogen ions from \( \mathrm{HCl} \) are reduced to hydrogen gas \( \mathrm{H}_2 \).1. Oxidation half-reaction of tin:\[ \mathrm{Sn}(s) \rightarrow \mathrm{Sn}^{2+}(aq) + 2 \mathrm{e}^- \]2. Reduction half-reaction of hydrogen:\[ 2 \mathrm{H}^+(aq) + 2 \mathrm{e}^- \rightarrow \mathrm{H}_2(g) \]

Key Concepts

Oxidation Half-Reactions

In redox chemistry, the oxidation half-reaction is a critical component where a species loses electrons, increasing its oxidation state. For instance, in the reaction

• PbO(s) is reduced to Pb(s).
• The process involves ammonia, \( \text{NH}_3 \), which is oxidized to nitrogen, \( \text{N}_2 \).

Take Equation (b):

• Thiosulfate, \( \text{S}_2\text{O}_3^{2-} \), is oxidized in the reaction to produce tetrathionate, \( \text{S}_2\text{O}_4^{2-} \).
• The oxidation part can be written as: \[ 2 \text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_2\text{O}_4^{2-}(aq) + 2 \text{e}^- \]

For students, it's essential to remember that oxidation is always coupled with the loss of electrons.

Reduction Half-Reactions

Reduction half-reactions involve the gain of electrons by a substance, leading to a decrease in its oxidation state.

• In Equation (a), PbO is reduced as electrons are gained, converting \( \text{O}^{2-} \) to element Pb.
• The reduction reaction is: \[ \text{PbO}(s) + 2 \text{e}^- \rightarrow \text{Pb}(s) + \text{O}^{2-} \]

Moving on to Equation (b):

• Iodine, \( \text{I}_2 \), does not remain in elemental form; it reduces to iodide, \( \text{I}^- \).
• The reduction half-reaction can be shown as: \[ \text{I}_2(s) + 2 \text{e}^- \rightarrow 2 \text{I}^-(aq) \]

It is fundamental in these reactions to pair the oxidation and reduction sides, keeping electron transfer balanced.

Net Ionic Equations

Net ionic equations simplify reactions by only showing the species that undergo chemical change, omitting the spectator ions. These equations become particularly relevant in aqueous solution chemistry.For reaction (c):

• The net ionic form eliminates non-essential components, focusing solely on the reacting species \( \text{Sn} \) and \( \text{H}^+ \).
• This is illustrated as: \[ \text{Sn}(s) + 2 \text{H}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + \text{H}_2(g) \]

Net ionic equations confer clarity, making it easier to see the transformation during the redox processes.

Aqueous Solution Chemistry

Understanding reactions in the context of aqueous solution chemistry is critical, as many redox reactions occur in such conditions.Substances dissolve in water, which leads to separation into ions:

• Some equations, like that in Equation (b), involve aqueous solutions such as Na\(_2\)S\(_2\)O\(_3\)(aq).
• Knowing how ions interact and react in solution lets you focus on the fundamental chemistry involved.

In any redox reactions set in aqueous solutions, it's important to consider both solubility and the states of matter, as they affect reactions. The concentration of ions can also affect reaction rates and equilibria, ultimately impacting the redox processes.