Problem 51 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

Balanced half-reactions: (a) \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^+ + e^- \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\) (b) \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^+ + 6e^- \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\) (c) \(\mathrm{N}_2\mathrm{H}_4 + 2\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O} + 2e^-\) (d) \(\mathrm{ClO}^{-} + 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\).

Step 1: Identify Elements and Charges

First, identify the oxidation state of the elements in each reaction. For (a), uranium in \(\mathrm{UO}_{2}^{+}\) is in the +5 state, and \(\mathrm{U}^{4+}\) is in +4. (b) Chlorine in \(\mathrm{ClO}_{3}^{-}\) is in +5, and in \(\mathrm{Cl}^{-}\) is -1. (c) Nitrogen in \(\mathrm{N}_{2}\mathrm{H}_4\) is -2, and in \(\mathrm{N}_{2}\) is 0. (d) Chlorine in \(\mathrm{ClO}^{-}\) is in +1, and in \(\mathrm{Cl}^{-}\) is -1.

Step 2: Balance Non-Oxygen and Non-Hydrogen Elements

Ensure that the main element for reduction/oxidation is balanced. In each of the cases (a-d), this step is already balanced with the same number of corresponding atoms on both sides.

Step 3: Balance Oxygen Atoms using Water

Balance oxygen atoms by adding \(\text{H}_2\text{O}\) molecules to the side lacking oxygen. For (a), add 2 \(\text{H}_2\text{O}\) to the right. In (b), add 3 \(\text{H}_2\text{O}\) to the right. In (c), no need to add water as there are no oxygens to balance. In (d), add 1 \(\text{H}_2\text{O}\) to the right.

Step 4: Balance Hydrogen Atoms using H+ (Acidic) or OH- (Basic)

For acidic solutions in (a) and (b): Add \(\text{H}^+\) ions to balance hydrogen. For basic solutions in (c) and (d): Add \(\text{OH}^-\) ions on both sides to balance and neutralize added hydrogen from water. Thus, for (a) add 4 \(\text{H}^+\) to the left. For (b) add 6 \(\text{H}^+\) to the left. For (c) add 2 \(\text{OH}^-\) to both sides. For (d) add 2 \(\text{OH}^-\) to both sides.

Step 5: Balance the Charge using Electrons

Add electrons to balance the charges on both sides of the equation.- (a): Add 1e\(^-\) to the left side.- (b): Add 6e\(^-\) to the left side.- (c): Add 2e\(^-\) to the left side.- (d): Add 2e\(^-\) to the left side.

Step 6: Write the Balanced Equations

(a) \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} + 1e^- \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\)(b) \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 6e^- \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\)(c) \(\mathrm{N}_2\mathrm{H}_4 + 2\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 2\mathrm{H}_2\mathrm{O} + 2e^-\)(d) \(\mathrm{ClO}^{-} + 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is like a puzzle. The goal is to make sure that the number of each type of atom is the same on both sides of the equation. Start by identifying the elements and their oxidation states. Then move on to ensuring the same count of atoms for the main element involved in the reaction on both sides. This is usually straightforward for the main element. Next, balance any oxygen atoms by adding water (\(\text{H}_2\text{O}\)) molecules to the equation.

• For instance, if there are more oxygen atoms on one side, add \(\text{H}_2\text{O}\) to the opposite side.

Finally, balance hydrogen atoms by either adding hydrogen ions (\(\text{H}^+\)) in acidic solutions or hydroxide ions (\(\text{OH}^-\)) in basic solutions. This brings balance to all atoms.
Once atoms are balanced, focus on balancing charges by adding electrons \(e^-\) to the side with a higher positive charge. This ensures that the overall charge is the same on both sides. Mastering these steps leads to a balanced equation and a deeper understanding of chemical reactions.

Oxidation and Reduction

Oxidation and reduction are twin processes known as redox reactions. In any redox reaction, one substance loses electrons (oxidation), while another gains electrons (reduction). To identify these processes, look at the change in oxidation states.

• "Oxidation" involves an increase in oxidation state.
• "Reduction" involves a decrease in oxidation state.

For example, when uranium goes from \(\mathrm{UO}_{2}^{+}\) to \(\mathrm{U}^{4+}\), it undergoes reduction by gaining electrons.
Reduction and oxidation have some common terms:- The substance that gets oxidized is called the "reducing agent."- The substance that gets reduced is called the "oxidizing agent."Understanding these terms helps in identifying the roles each substance plays in a chemical reaction. Recognizing these patterns takes practice but aids in simplifying complex reactions.

Electrochemistry

Electrochemistry ties together chemical reactions and electricity. In redox reactions, electrons are transferred, and this electron flow can generate electricity. This is at the heart of batteries and galvanic cells.

• An "electrochemical cell" converts chemical energy into electrical energy.
• "Galvanic cells," on the other hand, use spontaneous redox reactions to produce electricity.

These cells consist of two electrodes called the anode and cathode. The anode is where oxidation occurs, and electrons flow from the anode to the cathode.
Understanding electrochemistry reveals how batteries work and helps us harness electrical energy from chemical processes. This knowledge is essential for creating efficient energy solutions and technologies.