Problem 25 - Chemistry Matter and Change

Question

Use the half-reaction method to balance the redox equations. Begin by writing the oxidation and reduction half-reactions. Leave the balanced equation in ionic form. Challenge \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{ClO}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) (in basic solution)

Step-by-Step Solution

Answer

The balanced redox equation in ionic form in basic solution is: \(2 N_2O(g) + 2 ClO^-(aq) \rightarrow 4 NO_2^-(aq) + 2 Cl^-(aq) + 3 H_2O(l) + 4 OH^-(aq)\)

Step 1: Identify the oxidation and reduction half-reactions

In this exercise, we are given the following redox equation: \(N_2O(g) + ClO^-(aq) \rightarrow NO_2^-(aq) + Cl^-(aq)\) We need to determine the oxidation and reduction half-reactions. To do this, we must first assign oxidation numbers to all elements in the reaction. N2O: N = +1; O = -2 ClO-: Cl = +1; O = -2 NO2-: N = +3; O = -2 Cl-: Cl = -1 The oxidation half-reaction is: \( N_2O(g) \rightarrow NO_2^-(aq)\) The reduction half-reaction is: \( ClO^-(aq) \rightarrow Cl^-(aq)\)

Step 2: Balance the oxidation and reduction half-reactions

Next, we balance each of the half-reactions by atoms and charges. Oxidation half-reaction (Balance Nitrogen atoms, add water to balance Oxygen atoms, and add hydroxide ions to balance charges): \(2 N_2O(g) \rightarrow 4 NO_2^-(aq) + 2 H_2O(l) + 4 OH^-(aq)\) Charge before: 0 Charge after: -4 + -4 = -8 Reduction half-reaction (Balance Chlorine atoms, add water to balance Oxygen atom, and add electron to balance charges): \( 2 ClO^-(aq) + 2e^- \rightarrow 2 Cl^-(aq) + H_2O(l)\) Charge before: -2 Charge after: -2 + 0 = -2

Step 3: Balance the electrons in both half-reactions

In order to combine the balanced half-reactions, we must balance the number of electrons in both half-reactions: Oxidation: \(2 N_2O(g) \rightarrow 4 NO_2^-(aq) + 2 H_2O(l) + 4 OH^-(aq)\) (4 electrons) Reduction: \( 2 ClO^-(aq) + 2e^- \rightarrow 2 Cl^-(aq) + H_2O(l)\) We can see that the number of electrons is already balanced in both half-reactions.

Step 4: Combine the balanced half-reactions

Now, we can combine the balanced half-reactions to get the balanced redox equation: Oxidation: \(2 N_2O(g) \rightarrow 4 NO_2^-(aq) + 2 H_2O(l) + 4 OH^-(aq)\) Reduction: \( 2 ClO^-(aq) + 2e^- \rightarrow 2 Cl^-(aq) + H_2O(l)\) Summing up the half-reactions, we obtain: \(2 N_2O(g) + 2 ClO^-(aq) \rightarrow 4 NO_2^-(aq) + 2 Cl^-(aq) + 3 H_2O(l) + 4 OH^-(aq)\) The balanced redox equation in ionic form in basic solution is: \(2 N_2O(g) + 2 ClO^-(aq) \rightarrow 4 NO_2^-(aq) + 2 Cl^-(aq) + 3 H_2O(l) + 4 OH^-(aq)\)

Key Concepts

Half-Reaction Method

The half-reaction method is a powerful tool for balancing redox reactions, especially when chemical reactions involve electron transfer processes. Here's how it works:

• Identify the Redox Species: Separate the reaction into two half-reactions: the oxidation half and the reduction half.
• Assign Oxidation Numbers: This helps you identify which elements are being oxidized and which are being reduced.
• Balance Atoms and Charges: Each half-reaction must have balanced atoms and charges, usually involving adding water (H₂O), hydroxide ions (OH^-), or electrons (e^-).
• Combine Half-Reactions: Make sure the number of electrons lost in the oxidation equals the number of electrons gained in the reduction.

By focusing on one half-reaction at a time, you simplify the complex balancing process, ensuring both mass and charge are balanced in the resulting equation.

Oxidation Numbers

Oxidation numbers are a handy way of keeping track of electrons in atoms as they participate in chemical reactions. Understanding how to assign these numbers is crucial:

• Each atom in a pure element has an oxidation number of 0.
• For monoatomic ions, the oxidation number is equal to the ion's charge.
• Oxygen usually has an oxidation number of -2, and hydrogen is usually +1.
• In compounds, the more electronegative element receives a negative oxidation number.

By determining oxidation numbers, you can identify whether an atom is undergoing oxidation (increasing in oxidation number) or reduction (decreasing in oxidation number). In our exercise, nitrogen in N₂O goes from +1 to +3, indicating oxidation, and chlorine in ClO^- drops from +1 to -1, suggesting reduction.

Basic Solution

Balancing redox reactions in a basic solution involves special steps due to the nature of basic chemistry. Here's a guide:

• Add OH^-: When balancing oxygen, add H₂O to the side that requires more oxygen. If dealing with acids or bases, additional OH^- ions are used to maintain basic conditions.
• Balance Hydrogen with Water: Add H₂O, as needed, to balance the hydrogen atoms after adding OH^-.
• Ensure Charge Balance: The ions in solution must ultimately balance with electrons.

Basic solutions require careful steps to ensure the correct environment for the reaction, which may involve balancing extra steps not required in neutral or acidic solutions.

Electron Balance

Electron balance is central to the half-reaction method, ensuring that electron transfer adheres to conservation laws. Here's how to maintain it:

• For each half-reaction, count the total electrons gained by the reduction and lost by the oxidation.
• Multiply the half-reactions by appropriate whole numbers so the electrons canceled out when summing them.
• Ensure that no unbalanced electrons remain in the final equation.

A balanced redox reaction must have the same number of electrons involved in both the oxidation and the reduction processes. In the exercise provided, both half-reactions required no further adjustments for electron balance beyond their initial balancing.