Problem 24 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

(a) Reduction, (b) Oxidation, (c) Reduction, (d) Reduction, (e) Reduction, (f) Oxidation, (g) Oxidation.

Step 1: Determine Oxidation States

For each half-reaction, determine the initial and final oxidation states of the species involved to identify changes and thus deduce whether it's an oxidation or reduction.

Step 2: Balance Atoms Other Than Oxygen and Hydrogen

Adjust the coefficients to ensure that all atoms other than oxygen and hydrogen are balanced on both sides of each half-reaction.

Step 3: Balance Oxygen Atoms

Add water (H₂O) molecules to the side deficient in oxygen to balance the oxygen atoms in both acidic and basic solutions. This step corrects any discrepancies in the numbers of oxygen atoms.

Step 4: Balance Hydrogen Atoms

In acidic solutions, add hydrogen ions (H⁺) to the side deficient in hydrogen. In basic solutions, add water molecules to the side deficient and an equal number of hydroxide ions (OH⁻) to the opposite side.

Step 5: Balance Charges

Introduce electrons (e⁻) to the side with higher positive charge or lower negative charge to balance the charges for each half-reaction.

Step 6: Verify the Redox Type

Check if the electrons are on the reduction (gain electrons) or oxidation (lose electrons) side to label the half-reaction accordingly.

Step 7: Completion of Balance for (a) Mo Half-Reaction

For (a) \[ \mathrm{Mo}^{3+} + 3e^- \rightarrow \mathrm{Mo}(s) \]This is a reduction since electrons are gained by Mo.

Step 8: Completion of Balance for (b) Sulfurous Acid Half-Reaction

For (b) \[ \mathrm{H}_2\mathrm{SO}_3(aq) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{SO}_4^{2-}(aq) + 4H^+ + 2e^- \]This is an oxidation since electrons are lost.

Step 9: Completion of Balance for (c) Nitrate Ion Half-Reaction

For (c) \[ \mathrm{NO}_3^-(aq) + 4H^+ + 3e^- \rightarrow \mathrm{NO}(g) + 2H_2O(l) \]This is a reduction since electrons are gained by NO.

Step 10: Completion of Balance for (d) Oxygen to Water (Acidic)

For (d) \[ \mathrm{O}_2(g) + 4H^+ + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}(l) \]This is a reduction since electrons are gained.

Step 11: Completion of Balance for (e) Oxygen to Water (Basic)

For (e) \[ \mathrm{O}_2(g) + 2H_2O(l) + 4e^- \rightarrow 4\mathrm{OH}^- \]This is a reduction as electrons are gained.

Step 12: Completion of Balance for (f) Manganese Half-Reaction

For (f) \[ \mathrm{Mn}^{2+}(aq) + 2H_2O(l) + 2e^- \rightarrow \mathrm{MnO}_2(s) + 4\mathrm{OH}^- \]This is an oxidation since electrons are lost.

Step 13: Completion of Balance for (g) Chromium Hydroxide Half-Reaction

For (g) \[ \mathrm{Cr(OH)}_3(s) + \mathrm{OH}^- \rightarrow \mathrm{CrO}_4^{2-}(aq) + 2\mathrm{H}_2O(l) + 3e^- \]This is an oxidation as electrons are lost.

Key Concepts

Oxidation

Understanding oxidation in redox reactions is key to grasping how electrons are transferred between substances. Oxidation involves the loss of electrons by a molecule, atom, or ion. This process results in an increase in oxidation state.

During oxidation, you might notice that the species becomes more positive. A classic example from our exercise is the transformation of \[\mathrm{Mn}^{2+} \rightarrow \mathrm{MnO}_2\] in solution where electrons are lost, indicating oxidation. Remember the mnemonic "OIL RIG"—Oxidation Is Loss, Reduction Is Gain—to help recall these concepts.

• Oxidation is marked by a loss of electrons.
• It results in an increase in oxidation state.
• The species becomes more positive or less negative.

Reduction

Reduction is the counterpart to oxidation and involves the gaining of electrons by a molecule, atom, or ion. This process leads to a decrease in oxidation state and is often seen when a substance becomes less positive.

In our examples, the conversion of \[\mathrm{NO}_3^- \rightarrow \mathrm{NO}\] demonstrates reduction, as the nitrate gains electrons. Reduction often occurs in tandem with oxidation, balancing the electron flow in a reaction.

• Reduction involves gaining electrons.
• It results in a decrease in oxidation state.
• The species becomes less positive or more negative.

Half-Reaction Balancing

Half-reaction balancing is a technique to simplify and solve redox reactions, where the reaction is split into two halves—one for oxidation and one for reduction. Each half-reaction is balanced separately for atoms and charges.

Let's consider a step-by-step example of balancing a half-reaction like the one with molybdenum:

• Determine oxidation states to identify changes.
• Balance atoms other than \(\mathrm{H}\) and \(\mathrm{O}\).
• Add \(\mathrm{H}_2\mathrm{O}\) to balance \(\mathrm{O}\) atoms.
• Balance \(\mathrm{H}\) with \(\mathrm{H}^+\) (acidic) or \(\mathrm{OH}^-\) (basic).
• Balance charge with electrons.

This detailed method ensures that the overall reaction remains correct and centered around electron balance.

Acidic Solutions

In redox reactions occurring in acidic solutions, special steps are involved to balance hydrogen ions. Since acidic solutions are abundant in \(\mathrm{H}^+\), they facilitate hydrogen balancing without affecting the rest of the reaction dynamics.

While balancing half-reactions in an acidic environment:

• Add \(\mathrm{H}_2\mathrm{O}\) to balance oxygen atoms.
• Use \(\mathrm{H}^+\) to balance hydrogen atoms.
• Adjust electrons to balance charges.

The use of \(\mathrm{H}^+\) ions seamlessly integrates with the reaction medium, making redox reactions manageable and straightforward in acidic conditions.

Basic Solutions

Redox reactions in basic solutions involve additional steps since such solutions are rich in \(\mathrm{OH}^-\) ions rather than \(\mathrm{H}^+\). This requires a different approach compared to acidic media.

For balancing in a basic environment:

• Add \(\mathrm{H}_2\mathrm{O}\) to the side deficient in hydrogen to balance hydrogens initially.
• Introduce \(\mathrm{OH}^-\) to counterbalance any added \(\mathrm{H}^+\), converting it to water.
• Balance remaining charges by adjusting electrons.

Understanding this technique lets us solve redox equations correctly in basic environments, using water and hydroxide ions efficiently.