Problem 5 - Chemistry and Chemical Reactivity

Question

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{CrO}_{4}^{2-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s}) \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) (d) \(\mathrm{HS}^{-}(\mathrm{aq})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

Balance each redox equation step-by-step using half-reaction methods.

Step 1: Assign Oxidation Numbers - Part a

For \(\text{Al} + \text{OH}^- \rightarrow \text{Al(OH)}_4^- + \text{H}_2\), assign oxidation numbers: \(\text{Al}\) goes from 0 to +3, and \(\text{H}_2\) goes from +1 (in \(\text{OH}^-\)) to 0. The oxidation change occurs from 0 to +3 for \(\text{Al}\) and from +1 to 0 for \(\text{H}_2\).

Step 2: Split the Equation into Half-Reactions - Part a

Split the reaction into oxidation and reduction half-reactions: \(\text{Al} \rightarrow \text{Al(OH)}_4^-\) and \(\text{OH}^- \rightarrow \text{H}_2\).

Step 3: Balance Elements Except O and H - Part a

Balance the atoms other than O and H. Here, Al is already balanced in the oxidation half-reaction. In the reduction half-reaction, the equation for \(\text{H}_2\) formation is balanced concerning \(\text{H}\) as is.

Step 4: Balance Oxygen Using Water - Part a

For the oxidation half-reaction: \(\text{Al} + 4\text{OH}^- \rightarrow \text{Al(OH)}_4^-\). Oxygen is balanced by the set-up of \(\text{Al(OH)}_4^-\) already.

Step 5: Balance Hydrogen Using Water - Part a

For the reduction half-reaction: Each \(\text{OH}^-\) provides a source of hydrogen, and given \(\text{OH}^- \rightarrow \text{H}_2\), the hydrogen balance is intrinsic. No adjustments are necessary.

Step 6: Balance Charge Using Electrons - Part a

Balance the charge by adding electrons: In the oxidation half-reaction, \(\text{Al} \rightarrow \text{Al(OH)}_4^- + 3e^-\). In the reduction, \(4\text{e}^- + 4\text{OH}^- \rightarrow \text{2H}_2O + \text{2H}_2\).

Step 7: Balance and Combine Half-Reactions - Part a

Balance electrons transferred: Multiply half-reactions to equalize electron transfer (8 total electrons on each side), then combine the equations.

Step 8: Repeat Steps for Parts b, c, and d

Use the same steps (1-7) to separate, balance elements, balance \(\text{O}/\text{H}\) with water, balance charges using electrons, and finally, recombine for equations b, c, and d as indicated in part (a). Repeat the process to ensure proper balanced equations for all parts.

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is essential in mastering redox reactions. Each atom in a chemical reaction is assigned an oxidation number, which helps in determining how electrons are transferred. For instance, in a reaction like \( \text{Al} + \text{OH}^- \rightarrow \text{Al(OH)}_4^- + \text{H}_2 \), aluminum's oxidation number changes from 0 to +3, indicating it loses electrons and is oxidized. Meanwhile, the hydrogen in \( \text{OH}^- \) goes from +1 to 0, signaling a gain of electrons, hence it's reduced.
These changes indicate which atoms undergo oxidation and reduction, setting the stage for balancing redox equations. Understanding how to assign and interpret these numbers is the first step in solving redox problems.

• If the oxidation number increases, it’s an oxidation process.
• If the oxidation number decreases, it involves reduction.

This foundational concept is the key to recognizing electron flow within a redox reaction.

Half-Reactions

Redox reactions can be intimidating due to their complexity. A useful strategy to simplify them is by breaking them into half-reactions, separating the oxidation process from the reduction. Take, for example, the reaction \( \text{Al} \rightarrow \text{Al(OH)}_4^- \) and \( \text{OH}^- \rightarrow \text{H}_2 \); these are the half-reactions.

• Oxidation half-reaction: Describes the species losing electrons. Here, \( \text{Al} \rightarrow \text{Al(OH)}_4^- \).
• Reduction half-reaction: Describes the species gaining electrons, like \( \text{OH}^- \rightarrow \text{H}_2 \).

By analyzing each half-reaction independently, you can ensure individual processes are balanced before recombining them for a full picture of the redox equation. It simplifies the electron transfer involved, making balancing easier.

Balancing Equations

Balancing redox equations maintains the concept of conservation of mass and charge. Each element must have the same number of atoms on both the reactant and product sides, and the total charge should also be equal.
Initially, leave oxygen and hydrogen for later adjustment; focus first on atoms other than O and H.

• For example, in balancing aluminum: \( \text{Al} \rightarrow \text{Al(OH)}_4^- \), aluminum atoms are equal from the start.
• For charge balancing, equate the number of electrons lost and gained across reactions.

Working through sequential steps ensures that both mass and charge are balanced, essential criteria for a correct redox equation. Doing this systematically across all processes in the reaction ensures overall balance.

Basic Solution

Redox reactions may occur in neutral, acidic, or basic solutions. Each environment affects the method used to balance the equation. For reactions in a basic solution, water \( (\text{H}_2\text{O}) \) and hydroxide ions \( (\text{OH}^-) \) are crucial for balancing certain elements and charges.

• Generally, after balancing for neutral charge and mass, add \( \text{OH}^- \) ions to balance any \( \text{H}^+ \) ions present, thereby ensuring the equation remains consistent with the basic solution.
• This might involve adding water molecules as necessary to the opposite side to maintain both balance and basic conditions.

This method ensures the chemical equation accurately reflects the environment it happens in.

Electron Transfer

At the heart of redox reactions, electron transfer describes how electrons move between atoms. Oxidation is associated with a loss of electrons, and reduction with a gain, defining how compounds interact chemically.
Converting the flow of electrons into an idea of exchange helps understand each reactant's role.

• Often balanced by ensuring the number of electrons lost in the oxidation reaction equals those gained in the reduction reaction.
• Sometimes requires multiplying entire half-reactions to achieve equal electron transfer.

Grasping the electron transfer process is crucial as it gives insight into the very nature of chemical bonding and rearrangement. This understanding is key to mastering better reaction mechanisms and predicting product formation.