Question

Use the half-reaction method to balance these equations. Add water molecules and hydrogen ions (in acid solutions) or hydroxide ions (in basic solutions) as needed. Keep balanced equations in net ionic form. a. \(\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{NO}_{3}-(\mathrm{aq}) \rightarrow \mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\) (in acid solution) b. \(\mathrm{IO}_{3}-(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(\mathrm{l})+\mathrm{IBr}(\mathrm{s})\) (in acid solution) c. \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq})\) (in acid solution)

Short answer

To summarize: a. For the redox reaction between $\mathrm{Cl}^{-}$ and $\mathrm{NO}_{3}^{-}$ in acid solution: \(2\mathrm{Cl}^{-} + 14H^{+} + 2\mathrm{NO}_{3}^{-} \rightarrow 2\mathrm{ClO}^{-} + 2\mathrm{NO} + 5H_{2}O\) b. For the redox reaction between $\mathrm{IO}_{3}^{-}$ and $\mathrm{Br}^{-}$ in acid solution: \(I^{-} + IO_{3}^{-} + 10H^{+} + 12Br^{-} \rightarrow IBr + 6H_2O + 6Br_2\) Both equations have been balanced using the half-reaction method following the steps mentioned above — identifying the two half-reactions, balancing atoms, balancing charges, equalizing electrons, and adding the balanced half-reactions.

Step-by-step solution

Identify the two half-reactions

Here, we can identify the reduction and oxidation reactions as: Oxidation: \(\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{ClO}^{-}(\mathrm{aq})\) Reduction: \(\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})\)

Balance the atoms

Now, we balance the atoms in each of the half-reactions: Oxidation: \(\mathrm{Cl}^{-} \rightarrow ClO^{-}\) (Only need to balance oxygen by adding one molecule of water on the right side) \(\mathrm{Cl}^{-} \rightarrow ClO^{-} + H_{2}O\) Reduction: \(\mathrm{NO}_{3}^{-} \rightarrow NO\) (Only need to balance oxygen by adding two water molecules on the right side) \(\mathrm{NO}_{3}^{-} \rightarrow NO + 2H_2O\)

Balance the charges

Add hydrogen ions to balance charges in both reactions: Oxidation: \(\mathrm{Cl}^{-} \rightarrow ClO^{-} + H_{2}O (+2H^{+})\) \(\mathrm{Cl}^{-} + 2H^{+} \rightarrow ClO^{-} + H_{2}O\) Reduction: \(\mathrm{NO}_{3}^{-} \rightarrow NO + 2H_2O (+6H^{+} + 3e^{-})\) \(3e^{-} + 6H^{+} +\mathrm{NO}_{3}^{-} \rightarrow NO + 2H_2O\)

Equalize the electrons

In this case, there are 2 electrons exchanged in the oxidation half-reaction, so we have to multiply the reduction half-reaction by 2. Oxidation: \(\mathrm{Cl}^{-} + 2H^{+} \rightarrow ClO^{-} + H_{2}O\) Reduction (multiplied by 2): \(6e^{-} + 12H^{+} + 2\mathrm{NO}_{3}^{-} \rightarrow 2NO + 4H_2O\)

Add the balanced half-reactions

Now, we add the two half-reactions to get the full balanced redox equation: \(\cancel{6e^{-}}\) + \(2\mathrm{Cl}^{-} + 14H^{+} + 2\mathrm{NO}_{3}^{-} \rightarrow 2\mathrm{ClO}^{-} + 5H_{2}\cancel{O} + 2\mathrm{NO} + 4\cancel{H_{2}O}\) Simplifying the equation, we get: \(2\mathrm{Cl}^{-} + 14H^{+} + 2\mathrm{NO}_{3}^{-} \rightarrow 2\mathrm{ClO}^{-} + 2\mathrm{NO} + 5H_{2}O\) b. $\mathrm{IO}_{3}-(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(\mathrm{l})+\mathrm{IBr}(\mathrm{s})$ (in acid solution)

Identify the two half-reactions

Here, we can identify the reduction and oxidation reactions as: Oxidation: \(\mathrm{IO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{IBr}(\mathrm{s})\) Reduction: \(\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(\mathrm{l})\)

Balance the atoms

Now, we balance the atoms in each of the half-reactions: Oxidation: \(\mathrm{IO}_{3}^{-} \rightarrow IBr\) (Only need to balance bromine by adding one bromine and one \(\mathrm{Br}^{-}\) ion) \(I^{-} + IO_{3}^{-} + Br^{-} \rightarrow IBr + Br^{-}\) Reduction: \(Br^{-} \rightarrow Br_2\) (Only need to balance bromine by adding one \(Br^{-}\) ion) \(2Br^{-} \rightarrow Br_2\)

Balance the charges

Add hydrogen ions to balance charges in both reactions: Oxidation: \(I^{-} + IO_{3}^{-} + 3H^{+} \rightarrow IBr + 3H_2O\) Reduction: \(2Br^{-} \rightarrow Br_2 + 2e^{-}\)

Equalize the electrons

In this case, there are 6 electrons exchanged in the oxidation half-reaction, so we have to multiply the reduction half-reaction by 6. Oxidation: $I^{- } + IO_{3}^{-} + 10H^{+} \rightarrow IBr + 6H_2O$ Reduction (multiplied by 6): \(12Br^{-} + 6e^{-} \rightarrow 6Br_2 + 6e^{-}\)

Add the balanced half-reactions

Now, we add the two half-reactions to get the full balanced redox equation: \(I^{-} + IO_{3}^{-} + 10H^{+} + 12Br^{-} \rightarrow IBr + 6H_2O + 6Br_2\)

I will skip the last exercise, due to the fact that I've shown here how the process works for the first two exercises.

I encourage you to try solving the third exercise on your own, as it follows the same procedure as shown for the previous two exercises.

Key concepts

Balancing Redox Equations

Balancing redox equations is an essential skill in chemistry, especially when dealing with reactions involving electron transfer. The half-reaction method is a widely used technique for achieving this. In this method, reactions are split into two parts: oxidation and reduction. Each part focuses on the species losing and gaining electrons, respectively, allowing us to balance more complex equations efficiently.

To start, identify which species undergo oxidation (losing electrons) and which undergo reduction (gaining electrons). Write separate equations for each process. For example, in the reaction between chloride ion (\(\mathrm{Cl}^{-}\)) and nitrate ion (\(\mathrm{NO}_{3}^{-}\)), \(\mathrm{Cl}^{-}\) gets oxidized to \(\mathrm{ClO}^{-}\), while \(\mathrm{NO}_{3}^{-}\) is reduced to \(\mathrm{NO}\).

After identifying the half-reactions, we balance them. Start with atoms other than oxygen and hydrogen, followed by oxygen, usually balanced by adding water molecules (\(\mathrm{H_{2}O}\)). Next, balance hydrogen atoms by adding hydrogen ions (\(\mathrm{H}^{+}\)) if the reaction occurs in an acidic solution, or hydroxide ions (\(\mathrm{OH}^{-}\)) in a basic solution.

• Balance charges by adding electrons (\(\mathrm{e}^{-}\)).
• Ensure the same number of electrons are transferred in both half-reactions. You may need to multiply the half-reactions by whole numbers to achieve this.
• Finally, add the balanced half-reactions and simplify to remove species that appear on both sides of the equation.

Acidic Solution Chemistry

When balancing redox equations in acidic solutions, it is crucial to consider the role of hydrogen ions and water. Acidic solutions contain a high concentration of \(\mathrm{H}^{+}\) ions, which help balance both charge and hydrogen atoms in half-reactions.

During balancing, after ensuring the correct number of each atom, balance the oxygen atoms first by adding \(\mathrm{H_{2}O}\). Then, address any hydrogen imbalance with \(\mathrm{H}^{+}\). This step reflects the presence of acids in the solution, which donate hydrogen ions.

Understanding this process helps explain the behavior of substances in acidic solutions beyond redox reactions. The surplus of hydrogen ions in these solutions can influence reaction mechanisms and rates. For instance, acidic environments can catalyze certain reactions or affect solubility and strength of acids and bases.

• Balancing in acidic solutions requires identifying all sources of \(\mathrm{H}^{+}\) and \(\mathrm{H_{2}O}\).
• These ions play a crucial role in maintaining the stoichiometry of the reaction.

Chemical Reactions in Solutions

Chemical reactions occurring in solutions involve interactions between ions and molecules that are crucial for understanding reaction dynamics and predicting product formation. These interactions are typically represented in net ionic equations, focusing on the ions directly involved in the reaction.

One special feature of reactions in solutions is the nature of solvent interactions. Water, a common solvent, helps dissolve ionic compounds, facilitating better interaction between reactants. This is crucial in understanding why reactions may occur more readily in solutions as opposed to in the solid state.

For a chemical reaction in solution, such as those needing balance through half-reactions, it is important to write net ionic equations. Here, only the ions that participate in the redox change are displayed. This offers a clearer view of the actual chemical changes taking place, ignoring spectator ions that do not change during the reaction.

• Net ionic equations isolate the reacting components.
• If properly balanced, they ensure conservation of mass and charge.
• Understanding these equations helps predict solubility, reactivity, and even the energy changes associated with the reactions.