Problem 83 - Chemistry The Central Science In Si Units

Question

Hydrogen peroxide is capable of oxidizing (a) hydrazine to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O},(\mathbf{b}) \mathrm{SO}_{2}\) to \(\mathrm{SO}_{4}^{2-},(\mathbf{c}) \mathrm{NO}_{2}^{-}\) to \(\mathrm{NO}_{3}^{-},(\mathbf{d}) \mathrm{H}_{2} \mathrm{~S}(g)\) to \(\mathrm{S}(s),(\mathbf{e}) \mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\). Write a balanced net ionic equation for each of these redox reactions.

Step-by-Step Solution

Answer

The balanced net ionic equations are: 1. \( \mathrm{N}_2\mathrm{H}_4 + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} \) 2. \( \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 4\mathrm{H}^+ \) 3. \( \mathrm{NO}_2^{-} + \mathrm{H}_2\mathrm{O}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NO}_3^{-} + 2\mathrm{H}_2\mathrm{O} \) 4. \( \mathrm{H}_2\mathrm{S} + \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{S} + 2\mathrm{H}_2\mathrm{O} \) 5. \( 2\mathrm{Fe}^{2+} + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_2\mathrm{O} \)

Step 1: Write the Half-Reactions for Oxidation and Reduction (Hydrazine)

For the oxidation of hydrazine (N_2H_4):- Oxidation: \[ \mathrm{N}_{2}\mathrm{H}_4 \rightarrow \mathrm{N}_{2} + 4\mathrm{H}^{+} + 4e^- \]- Reduction: \[ \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^{+} + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \]

Step 2: Balance the Equation and Combine (Hydrazine)

Multiply the reduction half-reaction by 2 so that electrons are equal: \[ 2(\mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^{+} + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O}) \]Combine:\[ \mathrm{N}_{2}\mathrm{H}_4 + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{N}_{2} + 4\mathrm{H}_2\mathrm{O} \]

Step 3: Write the Half-Reactions for SO2 Conversion

For the oxidation of \(\mathrm{SO}_2\):- Oxidation: \[ \mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 4\mathrm{H}^{+} + 2e^- \]- Reduction: \[ \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^{+} + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \]

Step 4: Balance and Combine (SO2)

The reduction reaction matches the oxidation in electrons:Combine:\[ \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 4\mathrm{H}^+ \]

Step 5: Write Half-Reactions for NO2- to NO3-

For the oxidation of \(\mathrm{NO}_2^-\):- Oxidation: \[ \mathrm{NO}_2^- + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NO}_3^- + 2\mathrm{H}^+ + 2e^- \]- Reduction: \[ \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \]

Step 6: Balance and Combine (NO2-)

The reduction reaction matches the number of electrons:Combine:\[ \mathrm{NO}_2^- + \mathrm{H}_2\mathrm{O}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NO}_3^- + 2\mathrm{H}_2\mathrm{O} \]

Step 7: Write Half-Reactions for H2S to Sulfur

For the oxidation of \(\mathrm{H}_2\mathrm{S}\):- Oxidation: \[ \mathrm{H}_2\mathrm{S} \rightarrow \mathrm{S} + 2\mathrm{H}^+ + 2e^- \]- Reduction: \[ \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \]

Step 8: Balance and Combine (H2S)

The reduction reaction matches the oxidation:Combine:\[ \mathrm{H}_2\mathrm{S} + \mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{S} + 2\mathrm{H}_2\mathrm{O} \]

Step 9: Write Half-Reactions for Fe2+ to Fe3+

For the oxidation of \(\mathrm{Fe}^{2+}\):- Oxidation: \[ \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^- \]- Reduction: \[ \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \]

Step 10: Balance and Combine (Fe2+)

Multiply the oxidation half-reaction by 2: \[ 2(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-) \]Combine:\[ 2\mathrm{Fe}^{2+} + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_2\mathrm{O} \]

Key Concepts

Oxidation and Reduction

In a redox reaction, which stands for reduction-oxidation reaction, two key processes occur. These are oxidation and reduction. They always happen in tandem. A substance getting oxidized will lose electrons, whereas a substance getting reduced will gain them. This electron exchange is what defines these reactions.
For example, when hydrogen peroxide (H_2O_2) oxidizes hydrazine (N_2H_4) to nitrogen gas (N_2), hydrazine loses electrons (oxidation), and hydrogen peroxide gains electrons (reduction).

• Oxidation: The species losing electrons increases its oxidation state.
• Reduction: The species gaining electrons decreases its oxidation state.

When learning about redox reactions, it's important to remember the mnemonic 'OIL RIG'. This stands for "Oxidation Is Loss, Reduction Is Gain," referring to the loss and gain of electrons, respectively.

Balancing Chemical Equations

Balancing chemical equations is crucial in chemistry to ensure the conservation of mass. In redox reactions, it’s even more pivotal because it involves balancing atoms and the transfer of electrons among molecules.
The procedure involves writing each half-reaction for both oxidation and reduction separately. For example, in the hydrazine oxidation, the half-reaction for oxidation produces four electrons, while the hydrogen peroxide reduction requires only two. Thus, the reduction half-reaction needs to be multiplied by two to balance the number of electrons on both sides.

• Step 1: Write separate half-reactions for oxidation and reduction.
• Step 2: Balance all atoms except for H and O.
• Step 3: Balance oxygen atoms by adding water (H_2O).
• Step 4: Balance hydrogen atoms by adding protons (H^+ or use OH^- for basic solutions).
• Step 5: Balance the charge by adding electrons.
• Step 6: Ensure electron transfer numbers are equal and combine the half-reactions.

The aim is to have an equal number of each type of atom on both sides of the equation and the same net charge on both sides, reflecting true mass and charge conservation.

Half-reactions

Half-reactions split redox reactions into two distinct parts: one showing oxidation and the other showing reduction. This breakdown aids in systematically balancing redox equations. By separately analyzing these two processes, it becomes manageable to keep track of electron transfer.
For instance, in the redox reaction of H_2S converting to sulfur while reacting with hydrogen peroxide, the oxidation half-reaction involves the conversion of H_2S to solid sulfur, producing electrons. Concurrently, the reduction half-reaction shows hydrogen peroxide receiving these electrons to form water.

• Oxidation Half-Reaction: Describes the electron-donor process.
• Reduction Half-Reaction: Describes the electron-acceptor process.
• Advantages: Clearly differentiates the flow of electrons, enabling precise equation balancing.

Balancing reactions by half-reactions allows for clarity. It isolates the electron shifts, letting us address each separately. Once each half-reaction is balanced, combine them to yield the balanced overall equation.