Problem 24 - Chemistry The Central Science

Question

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. $$ \begin{array}{l}{\text { (a) } \mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s) \text { (acidic solution) }} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text { (acidic solution) }} \\ {\text { (c) } \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)(\text { acidic solution })} \\ {\text { (d) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (acidic solution) }} \\ {\text { (e) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (basic solution) }} \end{array} \\\ {\text { (f) } \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s) \text { (basic solution) }} \\ {\text { (g) } \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) \text { (basic solution) }} $$

Step-by-Step Solution

Answer

Here are the balanced reactions: a) \( \textrm{Mo}^{3+}(aq) + 3 \, \textrm{e}^- \rightarrow \textrm{Mo}(s) \) - Reduction b) \( \textrm{H}_{2}\textrm{SO}_{3}(aq) + \textrm{H}_{2}\textrm{O} \rightarrow \textrm{SO}_{4}^{2-}(aq) + 4 \, \textrm{H}^+ + 2 \, \textrm{e}^- \) - Oxidation c) \( \textrm{NO}_{3}^{-}(aq) + 2 \, \textrm{H}^+ + \textrm{e}^- \rightarrow \textrm{NO}(g) + \textrm{H}_{2}\textrm{O} \) - Reduction d) \( \textrm{O}_{2}(g) + 4 \, \textrm{H}^+ + 4 \, \textrm{e}^- \rightarrow 2 \, \textrm{H}_{2}\textrm{O}(l) \) - Reduction e) \( 2 \, \textrm{O}_{2}(g) + 4 \, \textrm{H}_{2}\textrm{O}(l) + 4 \, \textrm{e}^- \rightarrow 4 \, \textrm{OH}^{−}(aq) \) - Reduction f) \( \textrm{Mn}^{2+}(aq) + 2 \, \textrm{OH}^-(aq) + 2 \, \textrm{e}^- \rightarrow \textrm{MnO}_{2}(s) + \, \textrm{H}_{2}\textrm{O}(l) \) - Reduction g) \( \textrm{Cr}(\textrm{OH})_{3}(s) + 5 \, \textrm{e}^- + 6 \, \textrm{OH}^-(aq) \rightarrow \textrm{CrO}_{4}^{2-}(aq) + 3 \, \textrm{H}_{2}\textrm{O}(l) \) - Reduction

Step 1: Balance elements other than O and H

The molybdenum is already balanced; we have Mo^3+ on the left and Mo on the right.

Step 2: Balance O atoms

There are no O atoms in this reaction, so we can skip this step.

Step 3: Balance H atoms

There are no H atoms in this reaction, so we can skip this step as well.

Step 4: Balance charge

There is a charge of +3 (Mo^3+) on the left, and 0 on the right. To balance the charges, we add 3 electrons (e-) on the left side. \[ \textrm{Mo}^{3+}(aq) + 3 \, \textrm{e}^- \rightarrow \textrm{Mo}(s) \]

Step 5: Oxidation or reduction?

Since 3 electrons are gained, it is a reduction half-reaction. #b) H2SO3 (aq) -> SO4^2- (aq) (acidic solution)#

Step 1: Balance elements other than O and H

Balance S atoms: We have one S atom on both sides, so it is already balanced.

Step 2: Balance O atoms

The left side has 3 O atoms, while the right side has 4. Adding 1 H2O on the left side to balance the O atoms: \[ \textrm{H}_{2}\textrm{SO}_{3}(aq) + \textrm{H}_{2}\textrm{O} \rightarrow \textrm{SO}_{4}^{2-}(aq) \]

Step 3: Balance H atoms

The left side has 2 + 2 = 4 H atoms, and there are no H atoms on the right side. To balance, add 4 H+ ions on the right side: \( \textrm{H}_{2}\textrm{SO}_{3}(aq) + \textrm{H}_{2}\textrm{O} \rightarrow \textrm{SO}_{4}^{2-}(aq) + 4 \, \textrm{H}^+ \)

Step 4: Balance charge

The left side: 0 charge, right side: -2 + (4 * +1) = +2 charge. To balance the charges, add 2 electrons (e-) on the right side: \[ \textrm{H}_{2}\textrm{SO}_{3}(aq) + \textrm{H}_{2}\textrm{O} \rightarrow \textrm{SO}_{4}^{2-}(aq) + 4 \, \textrm{H}^+ + 2 \, \textrm{e}^- \]

Step 5: Oxidation or reduction?

Since 2 electrons are produced, it is an oxidation half-reaction. For the rest of the half-reactions, we will apply the same steps mentioned above. #c) NO3^-(aq) -> NO(g) (acidic solution)#

Step 11: Balanced reaction

\[ \textrm{NO}_{3}^{-}(aq) + 2 \, \textrm{H}^+ + \textrm{e}^- \rightarrow \textrm{NO}(g) + \textrm{H}_{2}\textrm{O} \] Reduction half-reaction. #d) O2(g) -> H2O(l) (acidic solution)#

Step 12: Balanced reaction

\[ \textrm{O}_{2}(g) + 4 \, \textrm{H}^+ + 4 \, \textrm{e}^- \rightarrow 2 \, \textrm{H}_{2}\textrm{O}(l) \] Reduction half-reaction. #e) O2(g) -> H2O(l) (basic solution)#

Step 13: Balanced reaction

\[ 2 \, \textrm{O}_{2}(g) + 4 \, \textrm{H}_{2}\textrm{O}(l) + 4 \, \textrm{e}^- \rightarrow 4 \, \textrm{OH}^{−}(aq) \] Reduction half-reaction. #f) Mn^2+(aq) -> MnO2(s) (basic solution)#

Step 14: Balanced reaction

\[ \textrm{Mn}^{2+}(aq) + 2 \, \textrm{OH}^-(aq) + 2 \, \textrm{e}^- \rightarrow \textrm{MnO}_{2}(s) + \, \textrm{H}_{2}\textrm{O}(l) \] Reduction half-reaction. #g) Cr(OH)3(s) -> CrO4^2-(aq) (basic solution)#

Step 15: Balanced reaction

\[ \textrm{Cr}(\textrm{OH})_{3}(s) + 5 \, \textrm{e}^- + 6 \, \textrm{OH}^-(aq) \rightarrow \textrm{CrO}_{4}^{2-}(aq) + 3 \, \textrm{H}_{2}\textrm{O}(l) \] Reduction half-reaction.

Key Concepts

Oxidation

Oxidation is one of the main types of chemical reactions crucial in various processes, such as metabolism and industrial applications. In oxidation, a substance loses electrons. To identify an oxidation reaction, look for the transfer of electrons or an increase in the oxidation state of the removing element.
For example, in the half-reaction involving \(\mathrm{H}_{2}\mathrm{SO}_{3}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq)\) in an acidic solution, \(\mathrm{H}_{2}\mathrm{SO}_{3}\) loses electrons to become \(\mathrm{SO}_{4}^{2-}.\) This indicates oxidation as the half-reaction releases electrons.

• Electrons are transferred from one element to another.
• Oxidation state of the element usually increases.

Understanding oxidation is essential not only for chemical reactions but also for broader applications involving energy generation and transfer.

Reduction

Reduction is the complementary process to oxidation, integral in redox reactions. A reduction half-reaction involves the gain of electrons by a molecule, atom, or ion. This gain typically results in a decrease in the oxidation state of the element receiving electrons.
For instance, when \(\mathrm{Mo}^{3+}(aq) \rightarrow \mathrm{Mo}(s)\), we find that the \(\mathrm{Mo}^{3+}\) ion gains electrons to form neutral \(\mathrm{Mo}\). This indicates a reduction process.

• Electrons are added to the element or compound.
• Oxidation state of the element decreases.

This concept is crucial not only for chemistry but also in biology, especially in cellular respiration and photosynthesis where energy transfer occurs through redox reactions.

Acidic Solution

An acidic solution is one where the concentration of hydrogen ions (\(\mathrm{H}^+\)) is greater than that of hydroxide ions (\(\mathrm{OH}^-\)). These ions play a critical role in balancing half-reactions by providing necessary ions to balance hydrogen or oxygen atoms.
In many redox reactions performed in acidic solutions, ions are needed to correctly balance the equation. For example, in the oxidation of \(\mathrm{H}_{2}\mathrm{SO}_{3} \rightarrow \mathrm{SO}_{4}^{2-}\) in an acidic medium, additional \(\mathrm{H}^+\) ions are required to balance the reaction.
Key aspects:

• \(\mathrm{H}^+\) ions are often used to balance reactions.
• Acidic solutions are pivotal in hydrogen atom transfers.

These solutions are common in chemical processes and metabolic pathways where maintaining a low pH is essential for reaction progress.

Basic Solution

A basic solution is characterized by a higher concentration of hydroxide ions (\(\mathrm{OH}^-\)) compared to hydrogen ions. Basic mediums are commonly involved in a variety of chemical processes, notably saponification and titrations.
When balancing half-reactions in a basic solution, \(\mathrm{OH}^-\) ions are used instead of \(\mathrm{H}^+\) ions to balance the hydrogen in the equation. For example, during the reduction of \(\mathrm{Mn}^{2+} \rightarrow \mathrm{MnO}_{2}\) in a basic solution, \(\mathrm{OH}^-\) ions help achieve attention balance.

• \(\mathrm{OH}^-\) ions contribute to reaction balancing.
• Creates an alkaline environment favorable for specific reactions.

Fundamentally, basic solutions are important in both organic and inorganic chemistry applications and are also crucial in environmental chemistry processes.

Electroneutrality

Electroneutrality principle states that, in a chemical reaction, the total charge must be conserved across reactants and products.
This principle ensures that the sum of positive and negative charges is zero, leading to stable and balanced equations.
To achieve electroneutrality in redox reactions:

• Balance both ions and molecules to ensure neutral charge.
• Use added electrons, \(\mathrm{H}^+\), or \(\mathrm{OH}^-\) ions when necessary.

For instance, when balancing the half-reaction \(\mathrm{Mo}^{3+} \to \mathrm{Mo}\) in acidic solution, three electrons are added to the left to achieve charge balance (since \(\mathrm{Mo}^{3+}\) has a charge of +3 and \(\mathrm{Mo}\) is neutral). Achieving electroneutrality is crucial to ensure that reactions are accurately represented and predict their behavior under different conditions.