Problem 33 - Chemistry Principles and Practice

Question

Balance each of the following redox reactions in basic solution. (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{ClO}^{-}(\mathrm{aq}) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{OH})_{4}^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Al} + \mathrm{ClO}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Al(OH)}_{4}^{-} + \mathrm{Cl}^{-}\);\(b) \(2\mathrm{MnO}_{4}^{-} + 3\mathrm{SO}_{3}^{2-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{MnO}_{2} + 3\mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\);\(c) \(4\mathrm{Zn} + \mathrm{NO}_{3}^{-} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{Zn(OH)}_{4}^{2-} + \mathrm{NH}_{3}\).

Step 1: Separate into Half-Reactions (a)

Identify the oxidation and reduction processes. - Al changes from Al to Al(OH)₄⁻, oxidation occurs here.- ClO⁻ changes to Cl⁻, reduction occurs here. Separate these into two half-reactions:1. Oxidation: \(\mathrm{Al} \rightarrow \mathrm{Al(OH)}_{4}^{-}\)2. Reduction: \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\)

Step 2: Balance Elements and Oxygen (a)

For each half-reaction:- **Oxidation:** Balance Al and O by adding \(4\) OH⁻:\[\mathrm{Al} + 4\mathrm{OH}^{-} \rightarrow \mathrm{Al(OH)}_{4}^{-}\]- **Reduction:** Balance Cl and O by adding \(2\) H₂O:\[\mathrm{ClO}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Cl}^{-} + 4\mathrm{OH}^{-}\]

Step 3: Balance Charges with Electrons (a)

Add electrons to balance charges:- **Oxidation:** Add \(3e^{-}\) to the right:\[\mathrm{Al} + 4\mathrm{OH}^{-} \rightarrow \mathrm{Al(OH)}_{4}^{-} + 3e^{-}\]- **Reduction:** Add \(3e^{-}\) to the left:\[\mathrm{ClO}^{-} + 2\mathrm{H}_{2}\mathrm{O} + 3e^{-} \rightarrow \mathrm{Cl}^{-} + 4\mathrm{OH}^{-}\]

Step 4: Combine and Simplify (a)

Add the balanced half-reactions and cancel common terms:\[\mathrm{Al} + \mathrm{ClO}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Al(OH)}_{4}^{-} + \mathrm{Cl}^{-}\]This is the balanced equation in basic solution.

Step 1: Separate into Half-Reactions (b)

Identify the oxidation and reduction processes.- MnO₄⁻ reduced to MnO₂.- SO₃²⁻ oxidized to SO₄²⁻.Separate into half-reactions:1. Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{MnO}_{2}\)2. Oxidation: \(\mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\)

Step 2: Balance Elements and Oxygen (b)

Balance elements except for H and O.- **Reduction:** Mn is balanced; add \(2\) H₂O to right:\[\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{MnO}_{2} + 2 \mathrm{H}_{2}\mathrm{O}\]- **Oxidation:** Add one \mathrm{OH}^{-} to the right to balance O:\[\mathrm{SO}_{3}^{2-} + \mathrm{OH}^{-} \rightarrow \mathrm{SO}_{4}^{2-}\]- Add water to balance H:\[\mathrm{SO}_{3}^{2-} + \mathrm{OH}^{-} \rightarrow \mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\]

Step 3: Balance Charges with Electrons (b)

Add electrons to balance charges:- **Reduction:** Add \(3e^{-}\) to Mn side:\[\mathrm{MnO}_{4}^{-} + 3\mathrm{e}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{OH}^{-}\]- **Oxidation:** Add \(2e^{-}\) to left:\[2 \mathrm{SO}_{3}^{2-} + 2\mathrm{OH}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{e}^{-}\]

Step 4: Equalize Electrons and Combine (b)

Multiply half-reactions to equalize electrons:- Reduction: \[2\left(\mathrm{MnO}_{4}^{-} + 3\mathrm{e}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{OH}^{-}\right)\]- Oxidation:\[3\left(2 \mathrm{SO}_{3}^{2-} + 2\mathrm{OH}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{e}^{-}\right)\]Combine, cancel common terms and simplify:\[2\mathrm{MnO}_{4}^{-} + 3\mathrm{SO}_{3}^{2-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{MnO}_{2} + 3\mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\]

Step 1: Separate into Half-Reactions (c)

Identify the oxidation and reduction processes.- Zn is oxidized to Zn(OH)₄²⁻.- NO₃⁻ is reduced to NH₃.Separate into half-reactions:1. Oxidation: \(\mathrm{Zn} \rightarrow \mathrm{Zn(OH)}_{4}^{2-}\)2. Reduction: \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NH}_{3}\)

Step 2: Balance Elements and Oxygen (c)

Balance elements except for H and O.- **Oxidation:** Zn is balanced; add \(4\) OH⁻ to balance O:\[\mathrm{Zn} + 4\mathrm{OH}^{-} \rightarrow \mathrm{Zn(OH)}_{4}^{2-}\]- **Reduction:** Balance O on \(\mathrm{NH}_{3}\) side using additional compounds:\[\mathrm{NO}_{3}^{-} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NH}_{3} + 9\mathrm{OH}^{-}\]

Step 3: Balance Charges with Electrons (c)

Add electrons to balance charges:- **Oxidation:** Add \(2e^{-}\) to Zn side:\[\mathrm{Zn} + 4\mathrm{OH}^{-} \rightarrow \mathrm{Zn(OH)}_{4}^{2-} + 2\mathrm{e}^{-}\]- **Reduction:** Add \(8e^{-}\) to left:\[\mathrm{NO}_{3}^{-} + 6\mathrm{H}_{2}\mathrm{O} + 8 \mathrm{e}^{-} \rightarrow \mathrm{NH}_{3} + 9\mathrm{OH}^{-}\]

Step 4: Equalize Electrons and Combine (c)

Multiply half-reactions to equalize the number of electrons:- Oxidation: Multiply by \(4\)- Reduction: Use as is.Combine:\[4\mathrm{Zn} + \mathrm{NO}_{3}^{-} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{Zn(OH)}_{4}^{2-} + \mathrm{NH}_{3}\] and cancel common terms to simplify.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that involves making sure the number of atoms for each element is the same on both sides of the equation. This ensures the conservation of mass. When balancing, you adjust the coefficients (the numbers in front of the compounds) to balance the atoms. For complex reactions like redox reactions, make sure both the atoms and the charges are balanced. To achieve this, you may need to add water molecules (H₂O), hydroxide ions (OH⁻), or electrons (e⁻) to various parts of the equation. By doing so, you ensure that both mass and charge are conserved.
- Start by writing the unbalanced equation.
- Identify the different elements present and count their atoms on both sides.
- Adjust coefficients methodically to balance each element one at a time, updating your counts as you go.
Balancing equations is an iterative process that may require you to make several adjustments before achieving the correct balance.

Basic Solution Chemistry

In basic solution chemistry, reactions take place in an alkaline environment, meaning they occur in the presence of hydroxide ions (OH⁻). Understanding how the hydroxide ions interact in the reaction is key to solving redox equations in such solutions. In basic solutions, any hydrogen ions (H⁺) produced in reactions need to be neutralized by OH⁻ to form water. This is crucial in balancing redox reactions, as they often involve different charges and elements that require careful adjustment.
- If H⁺ ions are present, pair them with OH⁻ to form water:
\[ \mathrm{H}^{+} + \mathrm{OH}^{-} \rightarrow \mathrm{H}_{2}\mathrm{O} \]
- Ensure that the final equation has a net neutral or desired charge, characteristic of the solution's pH.
Understanding these interactions helps you predict the outcomes of reactions and maintain the correct ionic balance.

Half-Reaction Method

The half-reaction method is a systematic way to balance redox reactions, often used when reactions involve electron transfer. By separating the oxidation and reduction processes into two individual half-reactions, you can individually balance the atoms and charges. This method simplifies the balancing of complex equations by focusing on one process at a time.
Steps involved:

• Identify and separate the oxidation and reduction reactions.
• Balance each half-reaction for atoms other than hydrogen and oxygen.
• Balance oxygen atoms by adding H₂O molecules, then balance hydrogen atoms by adding OH⁻ ions.
• Finally, balance the charges by adding electrons to one side of each half-reaction.

Once both half-reactions are balanced, you combine them by making sure the electrons transferred in oxidation equal those in reduction, thus ensuring the whole reaction is balanced.

Oxidation and Reduction

Oxidation and reduction are two key concepts in redox reactions, referring to changes in the oxidation state of atoms. "LEO says GER" is a helpful mnemonic where LEO stands for Loss of Electrons is Oxidation, and GER stands for Gain of Electrons is Reduction. Oxidation involves an increase in oxidation number due to electron loss, while reduction involves a decrease in oxidation number due to electron gain.
In redox reactions:

• Identify which reactants lose electrons (they are oxidized) and which gain electrons (they are reduced).
• Assign oxidation numbers to track changes in electron density around individual atoms.
• Balance the electrons lost and gained between the oxidation and reduction half-reactions.

Understanding these electron transfers is vital in determining the outcome and direction of chemical reactions, particularly those occurring in aqueous environments.