Question

Write balanced equations for the following half-reactions. (a) \(\mathrm{UO}_{2}^{+}(\mathrm{aq}) \rightarrow \mathrm{U}^{4+}(\mathrm{aq})\) (acid solution) (b) \(\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})\) (acid solution) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq}) \rightarrow \mathrm{N}_{2}(\mathrm{g})\) (basic solution) (d) \(\mathrm{ClO}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})\) (basic solution)

Short answer

The balanced equations for the half-reactions in acid and basic solutions are provided in Step 4.

Step-by-step solution

Balance Atoms Other Than O and H

For each half-reaction, begin by balancing any elements other than oxygen and hydrogen.**(a)** For \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+}\), uranium (U) is already balanced.**(b)** For \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\), chlorine (Cl) is already balanced.**(c)** For \(\mathrm{N}_{2}\mathrm{H}_{4} \rightarrow \mathrm{N}_{2}\), nitrogen (N) is already balanced.**(d)** For \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\), chlorine (Cl) is already balanced.

Balance Oxygen Atoms

Balance the oxygen atoms by adding water molecules (\(\mathrm{H}_2\mathrm{O}\)) to the appropriate side of each equation.**(a)** \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+}\): Add 2 \(\mathrm{H}_2\mathrm{O}\) to the right. **(b)** \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\): Add 3 \(\mathrm{H}_2\mathrm{O}\) to the right.**(c)** No action needed, as there are no O atoms in reactants or products.**(d)** \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\): Add 1 \(\mathrm{H}_2\mathrm{O}\) to the left.

Balance Hydrogen Atoms

Balance the hydrogen atoms by adding \(\mathrm{H}^{+}\) ions. In basic solutions, subsequently neutralize the \(\mathrm{H}^{+}\) by adding \(\mathrm{OH}^{-}\) ions to both sides of the equation.**(a)** \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): Add 4 \(\mathrm{H}^{+}\) to the left.**(b)** \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): Add 6 \(\mathrm{H}^{+}\) to the left.**(c)** \(\mathrm{N}_{2}\mathrm{H}_{4} \rightarrow \mathrm{N}_{2}\): Add 4 \(\mathrm{OH}^{-}\) to the right to balance the 4 \(\mathrm{H}\), rearrange for basic: \(\mathrm{N}_{2}\mathrm{H}_{4} + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_{2} + 4\mathrm{H}_2\mathrm{O}\).**(d)** \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cl}^{-}\): Add 2 \(\mathrm{H}^{+}\) to the right, then neutralize by adding 2 \(\mathrm{OH}^{-}\) to both sides: \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\).

Balance Charges

Balance the charge by adding electrons (\(\mathrm{e}^{-}\)) to the side with the greater positive charge.**(a)** \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): Add 3 electrons to the left: \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\).**(b)** \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): Add 6 electrons to the right: \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\).**(c)** \(\mathrm{N}_2\mathrm{H}_{4} + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O}\): Add 4 electrons to the right: \(\mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^{-}\).**(d)** \(\mathrm{ClO}^{-} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{OH}^{-}\): Add 2 electrons to the right: \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\).

Verify Balance in Atoms and Charges

Ensure that the number of atoms and the charges on both sides of each half-reaction are equal.**(a)** \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): 1 U, 2 O, 4 H, total charge = +2 both sides.**(b)** \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): 1 Cl, 3 O, 6 H, total charge = 0 both sides.**(c)** \(\mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^{-}\): 2 N, 8 H, 4 O, total charge = -4 both sides.**(d)** \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\): 1 Cl, 1 O, 2 H, total charge = -3 both sides.

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between chemical species. In these reactions, one species experiences oxidation (loses electrons), and the other one undergoes reduction (gains electrons). Whether a substance is oxidized or reduced defines the nature and direction of the electron transfer.
Within a redox reaction, the substance that loses electrons (is oxidized) acts as a reducing agent, while the substance that gains electrons (is reduced) serves as an oxidizing agent. Recognizing these transfers is crucial for balancing redox reactions in both acidic and basic solutions.

• Oxidation: Loss of electrons, increase in oxidation state.
• Reduction: Gain of electrons, decrease in oxidation state.
• Reducing agent: Donates electrons, is oxidized.
• Oxidizing agent: Accepts electrons, is reduced.

To effectively balance redox reactions, one must often write separate half-reactions that clearly illustrate the oxidation and reduction processes, balance them for mass and charge, and then combine them to form the full redox equation.

Acidic Solution

In the context of redox reactions, an acidic solution contains excess hydrogen ions ( \(\mathrm{H}^{+} \)), which are crucial for balancing certain chemical equations.
When balancing redox reactions in acidic conditions, the additional \(\mathrm{H}^{+} \) ions are added to balance hydrogen atoms after using water (\(\mathrm{H}_2\mathrm{O}\)) molecules to balance oxygen atoms, if required.

• Add water molecules to balance oxygen atoms where needed.
• Add \(\mathrm{H}^{+} \) ions to balance hydrogen atoms after the oxygen atoms are balanced.
• Add electrons to balance the charges and complete the half-reaction.

By using this systematic approach, equations are balanced for mass and charge, ensuring the chemical reaction obeys the law of conservation of matter.

Basic Solution

Balancing redox reactions in a basic solution involves additional steps compared to acidic solutions due to the presence of hydroxide ions ( \(\mathrm{OH}^{-} \)). These basic conditions require that any hydrogen ions added during balancing must be neutralized to form water, ensuring the reaction conditions reflect basic pH.
The process typically involves:

• Adding \(\mathrm{OH}^{-} \) ions to both sides of the equation to neutralize any \(\mathrm{H}^{+} \) ions, forming water.
• Balancing oxygen using water molecules, and then hydrogen with additional \(\mathrm{OH}^{-} \) ions. This ensures both mass and charge are balanced.
• Finally, balancing the charges by adding electrons to either side of the equation.

This method maintains the balance of atoms while adapting the reaction to the basic environment, complying with both physical chemistry principles and the unique conditions of basic solutions.

Electron Transfer

Understanding electron transfer is vital to grasping redox reactions as it involves the movement of electrons from one reactant to another. Electrons flow from the species being oxidized to the one being reduced, directly affecting the balancing of charges.
The steps to ensure balanced electron transfer include:

• Identifying and writing separate half-reactions for oxidation and reduction.
• Balancing each half-reaction for both atoms and charges.
• Adding the same number of electrons to both sides to ensure that the electrons lost in oxidation equal the electrons gained in reduction.

Balancing the electrons is a key step, ensuring that the total charge is neutral when the complete redox equation is constructed. This balance reflects the conservation of charge principle in chemical reactions, which is a cornerstone in chemistry.