Problem 21 - Chemistry The Molecular Science

Question

Balance this redox reaction (in acidic solution). $$ \mathrm{Fe}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{NO}_{2}(\mathrm{~g}) $$

Step-by-Step Solution

Answer

\( \mathrm{Fe} + 3 \mathrm{NO}_{3}^{-} + 6 \mathrm{H}^{+} \rightarrow \mathrm{Fe}^{3+} + 3\mathrm{NO}_2 + 3\mathrm{H}_2\mathrm{O} \)

Step 1: Separate into Half-Reactions

Identify the two half-reactions in the overall equation. The oxidation half-reaction is \( \mathrm{Fe} \longrightarrow \mathrm{Fe}^{3+} \), and the reduction half-reaction is \( \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2} \).

Step 2: Balance Atoms Other than O and H

For the first half-reaction, Fe is already balanced as there is one Fe on each side. For the second half-reaction, N is balanced with one \( \mathrm{NO}_3^- \) reducing to one \( \mathrm{NO}_2 \).

Step 3: Balance Oxygen Atoms with Water

In the reduction half-reaction \( \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2} \), balance the oxygen by adding one \( \mathrm{H}_2\mathrm{O} \) to the right side: \( \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2} + \mathrm{H}_{2}\mathrm{O} \).

Step 4: Balance Hydrogen Atoms with Hydrogen Ions

To balance the hydrogens in the second half-reaction, add two \( \mathrm{H}^{+} \) ions to the left side: \( \mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} \longrightarrow \mathrm{NO}_{2} + \mathrm{H}_{2}\mathrm{O} \).

Step 5: Balance Charges with Electrons

For the oxidation half-reaction, add 3 electrons to the right side: \( \mathrm{Fe} \longrightarrow \mathrm{Fe}^{3+} + 3\mathrm{e}^{-} \). For the reduction half-reaction, add one electron to the left side: \( \mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \).

Step 6: Equalize Electrons between Half-Reactions

Multiply the entire reduction half-reaction by 3 to equalize the number of electrons. The revised half-reaction is: \( 3(\mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}) \).

Step 7: Combine the Half-Reactions

Combine the two balanced half-reactions, ensuring electrons cancel out:\[ \mathrm{Fe} + 3\mathrm{NO}_{3}^{-} + 6 \mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+} + 3\mathrm{NO}_2 + 3\mathrm{H}_2\mathrm{O} \].

Step 8: Verify Balance

Check that the number of atoms of each kind and the charges are equal on both sides:- Iron: 1 on each side- Nitrogen: 3 on each side- Oxygen: 9 (3 from \(\mathrm{NO}_3^-\), balanced by 3 from \(\mathrm{H}_2\mathrm{O}\))- Hydrogen: 6 on each side- Total charge: +6 on each side with 3 \( \mathrm{Fe}^{3+} \) balancing 6 \( \mathrm{H}^{+} \).

Key Concepts

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. In this method, reactions are split into two halves labeled as **oxidation** and **reduction** half-reactions. This helps in focusing on changes occurring to specific elements.

For example, in the reaction given, you have: - **Oxidation half-reaction**: \( \mathrm{Fe} \longrightarrow \mathrm{Fe}^{3+} \) - **Reduction half-reaction**: \( \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2} \)
Each half-reaction deals only with either oxidation or reduction at a time. This allows chemists to independently balance these reactions more easily and focus separately on atoms and charges involved. Once each is balanced individually, they are recombined to give the fully balanced redox equation.

Electron Transfer

**Electron transfer** is the heart of redox reactions. It involves electrons moving from one specie to another. The specie losing electrons is said to undergo **oxidation**, and the specie gaining electrons undergoes **reduction**.

In the given problem:

• **Iron (Fe)** is oxidizing as it loses 3 electrons to become \( \mathrm{Fe}^{3+} \).
• **Nitrate ion (\( \mathrm{NO}_{3}^{-} \))** is reducing as it gains electrons to form \( \mathrm{NO}_{2} \).

The number of electrons lost in oxidation must equal those gained in reduction to uphold the law of conservation of charge. This is why the calculation of electron transfer is critical. You adjust half-reactions to equalize electron count before combining them.

Balancing Chemical Equations

Balancing chemical equations ensures that matter is conserved during a reaction, meaning atoms are neither created nor destroyed. The task is to have the same number of each kind of atom on both sides of the arrow.

In a redox reaction, besides atom count, charges must also be balanced on either side. Here's a stepwise approach:

• Identify atoms to balance excluding hydrogen and oxygen initially.
• Balance oxygen using water molecules \( (\mathrm{H}_2\mathrm{O}) \).
• Balance hydrogen using hydrogen ions \((\mathrm{H}^{+}) \) for reactions in acidic conditions.
• Balance the charges by adding electrons appropriately to equalize the charges of both sides.

Each of these steps ensures compliance with conservation principles which is fundamental in chemical processes.

Acidic Solution Chemistry

Acidic solution chemistry comes into play when redox reactions occur in solutions rich in hydrogen ions \((\mathrm{H}^{+}) \). This is typically represented by the presence of an acid, indicating the medium where balancing must consider hydrogen ions for full accuracy.

In this case:

• Balance hydrogen by adding \( \mathrm{H}^{+} \) ions as needed. For the reduction reaction, \( 2 \mathrm{H}^{+} \) ions were added to the reactants to balance hydrogen associated with additional water molecules.

Acidic environment adjustments ensure not only the stability of your reaction but also confirm true and complete balance is achieved, reflecting how the reaction progresses naturally in such conditions.