Problem 3 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in acid solution. (a) \(\mathrm{Ag}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{Ag}^{+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\operatorname{Cr}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Ag} + \mathrm{NO}_{3}^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO}_{2} + \mathrm{Ag}^{+} + \mathrm{H}_2\mathrm{O}\). (b) \(2\mathrm{MnO}_4^{-} + 5\mathrm{HSO}_3^{-} + 6\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{SO}_4^{2-} + 3\mathrm{H}_2\mathrm{O}\). (c) \(4\mathrm{Zn} + 2\mathrm{NO}_3^{-} + 10\mathrm{H}^+ \rightarrow 4\mathrm{Zn}^{2+} + \mathrm{N}_2\mathrm{O} + 5\mathrm{H}_2\mathrm{O}\). (d) \(\operatorname{Cr} + \mathrm{NO}_3^{-} + 4\mathrm{H}^+ \rightarrow \mathrm{Cr}^{3+} + \mathrm{NO} + 2\mathrm{H}_2\mathrm{O}\).

Step 1: Identify Oxidation and Reduction Half-Reactions for (a)

In the reaction \(\mathrm{Ag}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{Ag}^{+}(\mathrm{aq})\), silver is oxidized from \(\mathrm{Ag}^0\) to \(\mathrm{Ag}^+\), and nitrate \(\mathrm{NO}_{3}^{-}\) is reduced to \(\mathrm{NO}_2\). Separate the reaction into two half-reactions:- **Oxidation:** \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^{-}\)- **Reduction:** \(\mathrm{NO}_{3}^{-} + e^{-} \rightarrow \mathrm{NO}_{2}\).

Step 2: Balance Each Half-Reaction for (a)

For the oxidation half-reaction \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^{-}\), the equation is already balanced. For the reduction half-reaction, balance as follows: - Add \(\mathrm{H}_2\mathrm{O}\) to balance oxygen: \(\mathrm{NO}_{3}^{-} + e^{-} \rightarrow \mathrm{NO}_{2} + \mathrm{H}_2\mathrm{O}\) - Add \(\mathrm{H}^{+}\) to balance hydrogen: \(\mathrm{NO}_{3}^{-} + e^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO}_{2} + \mathrm{H}_2\mathrm{O}\).

Step 3: Combine and Balance Overall Reaction for (a)

Add the balanced half-reactions together making sure electrons cancel out: \[\begin{align*}\text{Oxidation: } & \mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^{-} \\text{Reduction: } & \mathrm{NO}_{3}^{-} + e^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO}_{2} + \mathrm{H}_2\mathrm{O} \\text{Overall:} & \mathrm{Ag} + \mathrm{NO}_{3}^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{NO}_{2} + \mathrm{Ag}^{+} + \mathrm{H}_2\mathrm{O}\end{align*}\]

Step 4: Balance Reaction (b) - Identify and Separate Half-Reactions

In the reaction \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\):- **Reduction:** \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)- **Oxidation:** \(\mathrm{HSO}_{3}^{-} \rightarrow \mathrm{SO}_{4}^{2-}\).

Step 5: Balance Each Half-Reaction for (b)

**Reduction half-reaction:**- Balance Mn and O: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\).**Oxidation half-reaction:**- Balance S and O: \(\mathrm{HSO}_{3}^{-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 2\mathrm{H}^{+} + 2e^{-}\).

Step 6: Combine and Balance Overall Reaction for (b)

To combine half-reactions, adjust coefficients to balance electrons: \[\begin{align*}\text{Reduction (multiplied by 2): } & 2(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}) \\text{Oxidation (multiplied by 5): } & 5(\mathrm{HSO}_{3}^{-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 2\mathrm{H}^{+} + 2e^{-}) \\text{Overall:} & 2\mathrm{MnO}_{4}^{-} + 5\mathrm{HSO}_{3}^{-} + 6\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{SO}_{4}^{2-} + 3\mathrm{H}_2\mathrm{O}\end{align*}\]

Step 7: Balance Reaction (c) - Identify and Separate Half-Reactions

In the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\):- **Oxidation:** \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}\)- **Reduction:** \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{N}_{2}\mathrm{O}\).

Step 8: Balance Each Half-Reaction for (c)

**Oxidation half-reaction:**- \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^{-}\).**Reduction half-reaction:**- Balance N and O: \(2\mathrm{NO}_{3}^{-} + 10\mathrm{H}^{+} + 8e^{-} \rightarrow \mathrm{N}_{2}\mathrm{O} + 5\mathrm{H}_2\mathrm{O}\).

Step 9: Combine and Balance Overall Reaction for (c)

Adjust half-reaction coefficients to balance electrons before combining: \[\begin{align*}\text{Oxidation (multiplied by 4): } & 4(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^{-}) \\text{Reduction: } & 2\mathrm{NO}_{3}^{-} + 10\mathrm{H}^{+} + 8e^{-} \rightarrow \mathrm{N}_{2}\mathrm{O} + 5\mathrm{H}_2\mathrm{O} \\text{Overall:} & 4\mathrm{Zn} + 2\mathrm{NO}_{3}^{-} + 10\mathrm{H}^{+} \rightarrow 4\mathrm{Zn}^{2+} + \mathrm{N}_{2}\mathrm{O} + 5\mathrm{H}_2\mathrm{O}\end{align*}\]

Step 10: Balance Reaction (d) - Identify and Separate Half-Reactions

In the reaction \(\operatorname{Cr}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\):- **Oxidation:** \(\operatorname{Cr} \rightarrow \mathrm{Cr}^{3+}\)- **Reduction:** \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NO}\).

Step 11: Balance Each Half-Reaction for (d)

**Oxidation half-reaction:**- \(\operatorname{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^{-}\).**Reduction half-reaction:**- Balance N and O: \(\mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} + 3e^{-} \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O}\).

Step 12: Combine and Balance Overall Reaction for (d)

Electrons balance naturally, so add half-reactions directly: \[\begin{align*}\text{Oxidation: } & \operatorname{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^{-} \\text{Reduction: } & \mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} + 3e^{-} \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O} \\text{Overall:} & \operatorname{Cr} + \mathrm{NO}_{3}^{-} + 4\mathrm{H}^{+} \rightarrow \mathrm{Cr}^{3+} + \mathrm{NO} + 2\mathrm{H}_2\mathrm{O}\end{align*}\]

Key Concepts

Half-Reaction Method

One of the most effective techniques for balancing redox reactions is the Half-Reaction Method. It simplifies the balancing process by separating the overall reaction into two half-reactions, one for oxidation and one for reduction.
In each half-reaction, we focus on the transfer of electrons. This is crucial because redox reactions involve the transfer of electrons from one substance to another. Here's how it works:

• First, identify the species that undergo oxidation (losing electrons) and reduction (gaining electrons) in the equation.
• Next, write the two half-reactions. One will show oxidation and electrons on the product side, the other shows reduction with electrons on the reactant side.
• Balance each half-reaction separately, ensuring that the number of atoms and the total charge are the same on both sides.

Combining these balanced half-reactions will give you the overall balanced redox reaction. The key is ensuring electrons are conserved, meaning the electrons lost in oxidation equal the electrons gained in reduction.

Balancing Chemical Equations

Balancing Chemical Equations is an essential skill in chemistry. The goal is to ensure that there are equal numbers of each type of atom on both sides of the equation. This respects the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.
When balancing equations, particularly in redox reactions:

• Start by balancing atoms that appear in only one reactant and one product.
• Next, balance the remaining atoms using coefficients. These are numbers placed before compounds to indicate how many times they appear in the reaction.
• Finally, ensure the charges are balanced as well, which is particularly important in ionic equations.

Using this methodological approach helps maintain balance in both mass and charge across the equation. It's often helpful to balance more complex reactions using systematic methods like the Half-Reaction Method in redox scenarios.

Acidic Solutions

In chemistry, redox reactions often occur in specific environments, one of which is Acidic Solutions. An acidic solution implies a high concentration of \( H^+ \) ions. This happens when the reaction environment provides excess hydrogen ions, contributing to the ability to balance hydrogen atoms in redox equations.
When balancing reactions in acidic solutions:

• Carefully add water molecules \( (H_2O) \) to balance oxygen atoms as needed.
• Add \( H^+ \) ions to balance the hydrogen atoms that may have been introduced with water.
• Finally, ensure that the total number of electrons is equal between the oxidation and reduction half-reactions, and that the overall charge is balanced on both sides of the equation.

This approach of using \( H^+ \) in acidic solutions helps you achieve a balanced redox equation, harmonizing the number of atoms and charges across the entire reaction.