Problem 71 - Chemistry Matter and Change

Question

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

Step-by-Step Solution

Answer

a. Oxidation: \(2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 6\mathrm{H}^+ + 6e^-\); Reduction: \(\mathrm{PbO} + 2e^- + 2\mathrm{H}^+ \rightarrow \mathrm{Pb} + \mathrm{H_2O}\). b. Red.: \(\mathrm{I}_2 + 2e^- \rightarrow 2\mathrm{I}^-\); Ox.: \(2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} + 2e^-\). c. Ox.: \(\mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^-\); Red.: \(2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2\).

Step 1: Assign Oxidation States for Equation (a)

For each element in the equation \( \mathrm{PbO}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{Pb}(\mathrm{s}) \):- \( \mathrm{Pb} \) in \( \mathrm{PbO} \) is +2 and 0 in \( \mathrm{Pb} \).- \( \mathrm{N} \) in \( \mathrm{NH}_3 \) is -3 and 0 in \( \mathrm{N}_2 \).

Step 2: Determine Oxidation and Reduction for Equation (a)

- Oxidation: \( \mathrm{N} \) goes from -3 in \( \mathrm{NH}_3 \) to 0 in \( \mathrm{N}_2 \).- Reduction: \( \mathrm{Pb} \) goes from +2 in \( \mathrm{PbO} \) to 0 in \( \mathrm{Pb} \).

Step 3: Write Half-Reactions for Equation (a)

- Oxidation half-reaction: \( \mathrm{2NH}_3 \rightarrow \mathrm{N}_2 + 6\mathrm{H}^+ + 6e^- \).- Reduction half-reaction: \( \mathrm{PbO} + 2e^- + 2\mathrm{H}^+ \rightarrow \mathrm{Pb} + \mathrm{H_2O} \).

Step 4: Assign Oxidation States for Equation (b)

For each element in \( \mathrm{I}_{2}(\mathrm{~s})+\mathrm{NaS}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq}) \):- \( \mathrm{I} \) in \( \mathrm{I}_2 \) is 0 and -1 in \( \mathrm{NaI} \).- \( \mathrm{S} \) changes from +2 in \( \mathrm{NaS}_2\mathrm{O}_3 \) to +2.5 in \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_4 \).

Step 5: Determine Oxidation and Reduction for Equation (b)

- Reduction: \( \mathrm{I} \) goes from 0 in \( \mathrm{I}_2 \) to -1 in \( \mathrm{NaI} \).- Oxidation: \( \mathrm{S} \) goes from +2 to +2.5 in the conversion of \( \mathrm{NaS}_2\mathrm{O}_3 \) to \( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_4 \).

Step 6: Write Half-Reactions for Equation (b)

- Reduction half-reaction: \( \mathrm{I}_2 + 2e^- \rightarrow 2\mathrm{I}^- \).- Oxidation half-reaction: \( 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} + 2e^- \).

Step 7: Assign Oxidation States for Equation (c)

For each element in \( \mathrm{Sn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{SnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \):- \( \mathrm{Sn} \) is 0 in elemental form and +2 in \( \mathrm{SnCl}_2 \).- \( \mathrm{H} \) is +1 in \( \mathrm{HCl} \) and 0 in \( \mathrm{H}_2 \).

Step 8: Determine Oxidation and Reduction for Equation (c)

- Oxidation: \( \mathrm{Sn} \) goes from 0 to +2.- Reduction: \( \mathrm{H} \) goes from +1 in \( \mathrm{HCl} \) to 0 in \( \mathrm{H}_2 \).

Step 9: Write Half-Reactions for Equation (c)

- Oxidation half-reaction: \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \).- Reduction half-reaction: \( 2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2 \).

Key Concepts

Oxidation

In redox reactions, oxidation refers to the loss of electrons by a molecule, atom, or ion. When a species undergoes oxidation, its electron count decreases, and its oxidation state increases. This concept is fundamental to redox chemistry, where oxidation and reduction occur simultaneously.

For example, in the reaction where ammonia (\(\mathrm{NH}_3\)) is converted into nitrogen gas (\(\mathrm{N}_2\)), nitrogen is oxidized. The oxidation state of nitrogen in \(\mathrm{NH}_3\) is -3. During the reaction, it increases to 0 in \(\mathrm{N}_2\). This change indicates an oxidation process, as electrons have been lost by nitrogen.

Key points about oxidation include:

• An increase in oxidation state.
• Loss of electrons (commonly expressed in half-reactions).
• Often involves adding oxygen or removing hydrogen in reactions.

It's essential to understand that oxidation cannot occur without a corresponding reduction, forming a balanced redox pair.

Reduction

Reduction is a chemical process where a molecule, atom, or ion gains electrons, leading to a decrease in its oxidation state. It typically accompanies oxidation in redox reactions. While one species loses electrons and is oxidized, another species gains those electrons and is reduced.

Consider the example from equation (a) in the original exercise. Lead in lead oxide (\(\mathrm{PbO}\)) is reduced to lead (\(\mathrm{Pb}\)) during the reaction. The oxidation state changes from +2 in \(\mathrm{PbO}\) to 0 in \(\mathrm{Pb}\), showcasing a gain of electrons or reduction.

Some characteristics of reduction include:

• A decrease in oxidation state.
• Gain of electrons (also shown in half-reactions).
• Often involves addition of hydrogen or removal of oxygen in reactions.

This process is the yin to oxidation's yang, creating a balanced chemical scenario where electrons are transferred from one element to another.

Oxidation States

The concept of oxidation states, or oxidation numbers, helps us track the transfer of electrons in redox reactions. An oxidation state is a theoretical charge an atom would have if all bonds to atoms of different elements were entirely ionic. These states are crucial for identifying which species get oxidized and reduced.

In the exercise, different elements have been assigned specific oxidation states based on standard rules, such as:

• The oxidation state of a pure element is always 0. For instance, the oxidation state of \(\mathrm{Sn}\) in its elemental form is 0.
• The sum of oxidation states for all atoms in a molecule or ion equals the charge on the molecule or ion.
• In \(\mathrm{HCl}\), \(\mathrm{H}\) is +1, and \(\mathrm{Cl}\) is -1, resulting in a neutral molecule.

Accurate determination of these states allows us to dissect changes during chemical reactions, sorting out which atoms are oxidized and which are reduced.

Half-Reactions

Half-reactions are an essential part of understanding redox processes. They split a redox reaction into two separate parts. One represents oxidation and the other reduction. By doing this, we can see clearly how electrons are transferred between the reactants.

For instance, in the original problems, each redox equation is broken down into an oxidation half-reaction and a reduction half-reaction. For example, the oxidation half-reaction involved \(\mathrm{Sn}\) changing to \(\mathrm{Sn}^{2+}\) in Eq.(c), indicating electron loss. Simultaneously, the reduction half-reaction depicts \(2\mathrm{H}^+\) ions gaining electrons to form \(\mathrm{H}_2\).

The process to write half-reactions include:

• Determining the correct oxidation states of each element involved.
• Writing separate reactions for loss and gain of electrons.
• Balancing the number of electrons transferred in oxidation and reduction.

Breaking a redox reaction into half-reactions is vital for balancing complex redox processes and understanding electron flow.