Problem 55 - Chemistry: Principles and Reactions

Question

Wlassify each of the following half-equations as oxidation or reduction and balance. (a) (acidic) \(\quad \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\) (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)

Step-by-Step Solution

Answer

Question: Identify and balance the given half-equations as oxidation or reduction: (a) (acidic) ℳn^2+(aq) ➡ ℳnO₄^-(aq) (b) (basic) CrO₄^(2-)(aq) ➡ Cr^(3+)(aq) (c) (basic) PbO₂(s) ➡ Pb^(2+)(aq) (d) (acidic) ClO₂^-(aq) ➡ ClO^-(aq) Answer: (a) Oxidation: 5H+(aq) + Mn^2+(aq) + 5e^- ➡ MnO₄^-(aq) + 4H₂O(l) (b) Reduction: 2e^- + CrO₄^2-(aq) + 4H₂O(l) ➡ Cr^3+(aq) + 8OH^-(aq) (c) Reduction: 2e^- + PbO₂(s) ➡ Pb^2+(aq) + 2OH^-(aq) (d) Reduction: 2H+(aq) + ClO₂^-(aq) + e^- ➡ ClO^-(aq) + H₂O(l)

Step 1: Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Mn has an oxidation number of +2 on the left side and Mn in MnO4- has an oxidation number of +7.

Step 2: Identify oxidation or reduction

Since the oxidation number of Mn increases from +2 to +7, this half-equation represents an oxidation process.

Step 3: Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Mn is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen with H+: \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\) and \(5 \mathrm{H}^{+}(a q) + \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\). Finally, balance the charge by adding electrons: \(5 \mathrm{H}^{+}(a q) + \mathrm{Mn}^{2+}(a q) + 5 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l)\). (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\)

Step 1: Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Cr in CrO4^(2-) has an oxidation number of +6 and Cr^(3+) has an oxidation number of +3.

Step 2: Identify oxidation or reduction

Since the oxidation number of Cr decreases from +6 to +3, this half-equation represents a reduction process.

Step 3: Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Cr is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen by adding OH- ions: \(\mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q)\) and \(\mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q) + 8\mathrm{OH}^{-}(a q)\). Finally, balance the charge by adding electrons: \(2 \mathrm{e}^{-} + \mathrm{CrO}_{4}^{2-}(a q) + 4 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}^{3+}(a q) + 8\mathrm{OH}^{-}(a q)\). (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\)

Step 1: Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Pb in PbO2 has an oxidation number of +4 and Pb^(2+) has an oxidation number of +2.

Step 2: Identify oxidation or reduction

Since the oxidation number of Pb decreases from +4 to +2, this half-equation represents a reduction process.

Step 3: Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Pb is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen by adding OH- ions: \(\mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q) + 2\mathrm{OH}^{-}(a q)\). Finally, balance the charge by adding electrons: \(2 \mathrm{e}^{-} + \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q) + 2\mathrm{OH}^{-}(a q)\). (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)

Step 1: Determine oxidation numbers

Assign oxidation numbers to the elements on both sides of the equation: Cl in ClO2- has an oxidation number of +3 and Cl in ClO- has an oxidation number of +1.

Step 2: Identify oxidation or reduction

Since the oxidation number of Cl decreases from +3 to +1, this half-equation represents a reduction process.

Step 3: Balance the half-equation

Balance the atoms other than oxygen and hydrogen first. In this case, Cl is already balanced. Then, balance the oxygens by adding water molecules and balance hydrogen with H+: \(\mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) and \(\mathrm{ClO}_{2}^{-}(a q) + 2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\). Finally, balance the charge by adding electrons: \(2 \mathrm{H}^{+}(a q) + \mathrm{ClO}_{2}^{-}(a q) + \mathrm{e}^{-} \longrightarrow \mathrm{ClO}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\).

Key Concepts

Oxidation Numbers

Oxidation numbers are a crucial concept in understanding redox reactions. An oxidation number is a hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. These numbers help identify how electrons are being transferred in a reaction.

To assign oxidation numbers, keep the following rules in mind:

• Elements in their pure form have an oxidation number of 0. For example, \( ext{O}_2\) and \( ext{H}_2\) both have oxidation numbers of 0.
• The oxidation number of a monoatomic ion is equal to its charge. For example, \( ext{Na}^+\) has an oxidation number of +1.
• Oxygen typically has an oxidation number of -2 in most compounds, while hydrogen generally has +1.
• In a neutral molecule, the sum of oxidation numbers equals zero, while in a polyatomic ion, it equals the charge of the ion.

Using these guidelines helps determine the oxidation states of elements in a reaction, paving the way for understanding which atoms are oxidized and which are reduced.

Half-Equations Balancing

Half-equations represent one part of the chemical process involved in a redox reaction, specifically either oxidation or reduction. Balancing these equations is essential because it ensures the conservation of mass and charge in a reaction.

Here's a streamlined method to balance half-equations:

• Firstly, balance the atoms other than oxygen and hydrogen.
• For reactions in acidic solutions, add \( ext{H}_2 ext{O} \) to balance oxygen atoms, and \( ext{H}^+ \) to balance hydrogen atoms.
• In basic solutions, add \( ext{OH}^- \) to balance hydrogen after adding water for oxygen balance.
• Finally, balance the charge by adding electrons. You may need to multiply the entire equation to ensure electron balance when pairing oxidation and reduction reactions.

This method is essential for understanding the step-by-step changes each element undergoes in a reaction and, ultimately, solving the complete redox equation.

Oxidation and Reduction Identification

Identifying oxidation and reduction is a vital skill for analyzing redox reactions. Redox involves the transfer of electrons, evident through changes in oxidation states. Here’s how to identify oxidation and reduction:

• Oxidation occurs when an atom's oxidation number increases, signifying a loss of electrons.
• Reduction happens when an atom's oxidation number decreases, indicating a gain of electrons.
• The substance undergoing oxidation is the reducing agent—it donates electrons.
• The substance being reduced is the oxidizing agent—it accepts electrons.

By tracking these changes, you can uncover which elements undergo oxidation and reduction, providing insight into the mechanics of the chemical reaction.