Question

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Short answer

a. \(3I^{-}(aq) + ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq)\) b. \(2As_{2}O_{3}(s) + 12NO_{3}^{-}(aq) + 12H^{+}(aq) \rightarrow 4H_{3}AsO_{4}(aq) + 6NO(g) + 6H_{2}O(l)\) c. \(10Br^{-}(aq) + 2MnO_{4}^{-}(aq) + 16H^{+}(aq) \rightarrow 5Br_{2}(l) + 2Mn^{2+}(aq) + 8H_{2}O(l)\) d. \(3CH_{3}OH(aq) + Cr_{2}O_{7}^{2-}(aq) + 8H^{+}(aq) \rightarrow 3CH_{2}O(aq) + 2Cr^{3+}(aq) + 4H_{2}O(l)\)

Step-by-step solution

Assign oxidation numbers

First, assign oxidation numbers to all elements in the unbalanced equation: \(I^{-}: -1 \quad ClO^{-}: +1 \quad I_{3}^{-}: 0 \quad Cl^{-}: -1\)

Identify half-reactions

Next, we can see the oxidation half-reaction is: \(I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) And the reduction half-reaction is: \(ClO^{-}(aq) \rightarrow Cl^{-}(aq)\)

Balance atoms

Balance the atoms in both half-reactions: \(I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) (balance I atoms by multiplying the \(I_{3}^{-}\) by 3) \(3I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) The chlorine and oxygen atoms are already balanced in the reduction half-reaction.

Balance charges

Balance the charges in both half-reactions: For the oxidation half-reaction, add 2 electrons to the right side: \(3I^{-}(aq) \rightarrow I_{3}^{-}(aq) + 2e^{-}\) For the reduction half-reaction, add an electron to the left side: \(e^{-}+ClO^{-}(aq) \rightarrow Cl^{-}(aq)\)

Align electron counts

For this reaction, the electron counts are already the same (2) in both half-reactions.

Add half-reactions

Add the balanced half-reactions and cancel the electrons: \(3I^{-}(aq) + e^{-}+ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq) + 2e^{-}\) Cancel the electrons and obtain the final balanced equation: \(3I^{-}(aq) + ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq)\) To balance the remaining reactions, simply follow the same steps. b. \(As_{2}O_{3}(s)+NO_{3}^{-}(aq) \rightarrow H_{3}AsO_{4}(aq)+NO(g)\) c. \(Br^{-}(aq)+MnO_{4}^{-}(aq) \rightarrow Br_{2}(l)+Mn^{2+}(aq)\) d. \(CH_{3}OH(aq)+Cr_{2}O_{7}^{2-}(aq) \rightarrow CH_{2}O(aq)+Cr^{3+}(aq)\)

Key concepts

Half-Reaction Method

Breaking down complex reactions into simpler parts can be like solving a puzzle. The half-reaction method is one such exciting technique used for balancing oxidation-reduction reactions. This method involves splitting the entire chemical equation into two separate smaller reactions: the oxidation half and the reduction half. Each smaller reaction deals with either the loss or gain of electrons, allowing us to balance them separately.

To start, you identify the two pieces, or half-reactions, by observing which substances are gaining electrons and which are losing them. This is where understanding oxidation and reduction becomes key. Once you have these, you can address them individually by first balancing each half for atoms and then for charges.

By keeping the focus on the simple parts, the method allows us to work through balancing intricate chemical equations methodically and with clarity.

Oxidation Numbers

Oxidation numbers are a helpful tool to keep track of where electrons are headed during chemical reactions. Think of oxidation numbers as a bookkeeping system for electrons. By assigning these numbers to elements, we can more easily spot which ones are gaining or losing electrons.

Each element in a reaction has an assigned oxidation number, based on certain rules. For instance, the charge of a simple ion is its oxidation number.

• Neutral elements have an oxidation number of zero.
• Oxygen usually has an oxidation number of -2 (with exceptions like in peroxides).
• Hydrogen is often +1 except when bonded with metals, where it becomes -1.

Using these rules, we can determine changes in oxidation states, helping us identify the half-reactions needed for balancing.

Overall, understanding and assigning oxidation numbers effectively can reveal the hidden dance of electrons in a chemical equation.

Acidic Solution

In many oxidation-reduction problems, the solution's acid-base environment plays a crucial role. Here, we focus on reactions taking place in acidic solutions. The acidity affects how we balance equations, especially with oxygen and hydrogen atoms.

When working in acidic environments, water and hydrogen ions ( H^+ ) often come into play.

• We use water molecules to balance oxygen atoms if they're not already balanced in the half-reactions.
• Once oxygen is balanced, hydrogen ions ( H^+ ) are added to balance the hydrogen atoms.

This special consideration for the simplest ions and water stands as a reminder to always account for the context in which reactions occur. It helps ensure all parts of the equation are in harmony.

Electron Balancing

Keeping electron counts balanced is crucial when handling oxidation-reduction reactions. The goal is to reach a state where every electron lost in an oxidation event matches an electron gained in a reduction event.

To achieve this, we find the common factor between the electrons transferred in the half-reactions.

• Often, extra electrons are added to one side of a half-reaction to equal the count above or below in the other half-reaction.
• The two half-reactions are then simultaneously solved for both mass and charge, ensuring the entire equation is balanced.

Finally, these balanced half-reactions are recombined, dropping any electrons on each side, to reveal the full, balanced chemical equation.

Understanding and applying the principle of electron balance brings clarity and order to what otherwise might feel like another chemistry conundrum.