Problem 72 - Chemistry Matter and Change

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. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})} \\ {\text { (in acid solution) }} \\ {\text { b. } \mathrm{IO}_{3}-(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Br}_{2}(1)+\mathrm{IBr}(\mathrm{s}) \\ {\text { (in acid solution) }}} \\ {\text { 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})} \\\ {(\text { in acid solution })} \end{array} \end{equation}

Step-by-Step Solution

Answer

a) \( \mathrm{Cl}^- + 2 \mathrm{H}^+ + \mathrm{NO}_3^- \rightarrow \mathrm{ClO}^- + \mathrm{NO} + \mathrm{H}_2\mathrm{O} \); b) \( \mathrm{IO}_3^- + 6\mathrm{H}^+ + 6\mathrm{Br}^- \rightarrow \mathrm{IBr} + 3\mathrm{Br}_2 + 3\mathrm{H}_2\mathrm{O} \); c) \( \mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\mathrm{I}^- + \mathrm{S}_2\mathrm{O}_4^{2-} \)."}

Step 1: Write Unbalanced Half-reactions (a)

For the given half-reactions for reaction (a): \( \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}^{-} \) and \( \mathrm{NO}_3^{-} \rightarrow \mathrm{NO} \). Identify oxidation states: Chlorine goes from -1 to +1, and Nitrogen goes from +5 to +2.

Step 2: Balance Atoms Other Than O and H (a)

For chlorine: \( \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}^{-} \) is already balanced. For nitrogen: \( \mathrm{NO}_3^{-} \rightarrow \mathrm{NO} \), NO and \( \mathrm{NO}_3^{-} \) both have 1 N atom.

Step 3: Balance O Atoms with Water (a)

Add water to balance the oxygens. For \( \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}^{-}+\mathrm{H}_2\mathrm{O} \). For \( \mathrm{NO}_3^{-} \rightarrow \mathrm{NO} \), add two water molecules to the right: \( \mathrm{NO}_3^{-} \rightarrow \mathrm{NO} + 2 \mathrm{H}_2\mathrm{O} \).

Step 4: Balance H Atoms with H+ (a)

For \( \mathrm{Cl}^- \rightarrow \mathrm{ClO}^- \), add 2 \( \mathrm{H}^+ \) to the left: \( \mathrm{Cl}^- + 2 \mathrm{H}^+ \rightarrow \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O} \). For \( \mathrm{NO}_3^- \rightarrow \mathrm{NO} \), add 4 \( \mathrm{H}^+ \) to the left: \( \mathrm{NO}_3^- + 4 \mathrm{H}^+ \rightarrow \mathrm{NO} + 2 \mathrm{H}_2\mathrm{O} \).

Step 5: Balance Charge with Electrons (a)

Add electrons to balance the charge. \( \mathrm{Cl}^- + 2 \mathrm{H}^+ + 2e^- \rightarrow \mathrm{ClO^-} + \mathrm{H}_2\mathrm{O} \) and \( \mathrm{NO}_3^- + 4 \mathrm{H}^+ + 3e^- \rightarrow \mathrm{NO} + 2 \mathrm{H}_2\mathrm{O} \). Multiply Cl half-reaction by 3 and NO reaction by 2.

Step 6: Combine and Simplify (a)

Combining gives: \( 2 \mathrm{Cl}^{-} + 6 \mathrm{H}^{+} + 6e^{-} + 3 \mathrm{NO}_3^{-} + 12 \mathrm{H}^{+} + 6e^{-}\rightarrow 2 \mathrm{ClO}^{-} + 2\mathrm{H}_2\mathrm{O} + 3 \mathrm{NO} + 6\mathrm{H}_2\mathrm{O} \). Simplifying, cancel electrons and combine H+'s and water. Final equation: \( \mathrm{Cl}^{-} + 2 \mathrm{H}^{+} + \mathrm{NO}_3^{-} \rightarrow \mathrm{ClO}^{-} + \mathrm{NO} + \mathrm{H}_2\mathrm{O} \).

Step 1: Write Unbalanced Half-reactions (b)

For the given half-reactions for reaction (b): \( \mathrm{IO}_3^{-} \rightarrow \mathrm{IBr} \) and \( \mathrm{Br}^{-} \rightarrow \mathrm{Br}_2 \). Iodine changes oxidation state from +5 to +1. Bromine changes from -1 to 0.

Step 2: Balance Atoms Other Than O and H (b)

For iodine: Increase iodine atoms to balance reactants with products: \( \mathrm{IO}_3^{-} \rightarrow \mathrm{IBr} \) has one iodine atom. For bromine: \( 2 \mathrm{Br}^- \rightarrow \mathrm{Br}_2 \) is balanced.

Step 3: Balance O Atoms with Water (b)

Add water molecules to balance oxygen. For \( \mathrm{IO}_3^{-} \rightarrow \mathrm{IBr} \), add 3 water molecules to the right: \( \mathrm{IO}_3^{-} \rightarrow \mathrm{IBr} + 3 \mathrm{H}_2\mathrm{O} \).

Step 4: Balance H Atoms with H+ (b)

For \( \mathrm{IO}_3^{-} \rightarrow \mathrm{IBr} \), add 6 \( \mathrm{H}^+ \) to the left: \( \mathrm{IO}_3^{-} + 6\mathrm{H}^+ \rightarrow \mathrm{IBr} + 3 \mathrm{H}_2\mathrm{O} \).

Step 5: Balance Charge with Electrons (b)

Add electrons to balance charge. For iodine: \( \mathrm{IO}_3^{-} + 6\mathrm{H}^+ + 6e^- \rightarrow \mathrm{IBr} + 3 \mathrm{H}_2\mathrm{O} \). For bromine: \( \mathrm{2Br}^- \rightarrow \mathrm{Br}_2 + 2e^- \). Multiply bromine reaction by 3.

Step 6: Combine and Simplify (b)

Combining gives: \( \mathrm{IO}_3^{-} + 6 \mathrm{H}^{+} + 6e^{-} + 6\mathrm{Br}^{-} \rightarrow \mathrm{IBr} + 3 \mathrm{H}_2\mathrm{O} + 3 \mathrm{Br}_2 + 6e^- \). Cancel electrons and simplify: \( \mathrm{IO}_3^{-} + 6\mathrm{H}^{+} + 6\mathrm{Br}^{-} \rightarrow \mathrm{IBr} + 3\mathrm{Br}_2 + 3\mathrm{H}_2\mathrm{O} \).

Step 1: Write Unbalanced Half-reactions (c)

For the given half-reactions for reaction (c): \( \mathrm{I}_2 \rightarrow \mathrm{I}^- \) and \( \mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} \). Iodine goes from 0 to -1 and sulfur goes from +2 to +3.

Step 2: Balance Atoms Other Than O and H (c)

Balance the Iodine by having 2 atoms on the right side: \( \mathrm{I}_2 \rightarrow 2\mathrm{I}^- \). For sulfur, it is already balanced: \( \mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} \).

Step 3: Balance O Atoms with Water (c)

In this case, both iodine and sulfur reactions are balanced for oxygen.

Step 4: Balance H Atoms with H+ (c)

Since there are no hydrogen atoms to balance initially in this reaction when using the half-reaction method, this step is skipped.

Step 5: Balance Charge with Electrons (c)

For iodine: \( \mathrm{I}_2 + 2e^- \rightarrow 2\mathrm{I}^- \). For sulfur: add 2 electrons: \( \mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{S}_2\mathrm{O}_4^{2-} + 2 e^- \).

Step 6: Combine and Simplify (c)

Combine the equations: \( \mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\mathrm{I}^- + \mathrm{S}_2\mathrm{O}_4^{2-} \). The electrons are balanced and there is no need for further simplification. Final equation: \( \mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\mathrm{I}^- + \mathrm{S}_2\mathrm{O}_4^{2-} \).

Key Concepts

Redox Reactions

In chemistry, redox reactions, or oxidation-reduction reactions, are processes in which electrons are transferred between chemical species. These reactions involve two main parts: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. Understanding these concepts is key to mastering redox reactions, as they are foundational to many chemical processes in both biological and industrial systems.
To identify redox reactions, observe changes in the oxidation state of the elements participating in the reaction. A change in oxidation state indicates that a redox reaction has occurred.

• Oxidizing agents gain electrons and are reduced in the process.
• Reducing agents lose electrons and are oxidized in the process.

Balancing redox reactions ensures the conservation of charge and mass, typically accomplished through methods such as the half-reaction method.

Acidic and Basic Solutions

Chemistry often involves reactions taking place in various pH conditions, especially within acidic and basic solutions. These conditions fundamentally affect the way reactions, particularly redox reactions, are balanced.
In acidic solutions, hydrogen ions (\(H^+\)) play a crucial role. The use of water molecules and hydrogen ions is critical when balancing half-reactions in such solutions. They help in balancing oxygen and hydrogen atoms, respectively.

• Add\(H_2O\) to balance oxygens.
• Use\(H^+\) to balance hydrogens.

Conversely, in basic solutions, hydroxide ions (\(OH^-\)) are utilized. Water and hydroxide ions balance the necessary oxygen and hydrogen atoms respectively. Understanding the specific requirements of acidic and basic environments helps ensure the correct balancing of chemical equations.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, crucial for representing conservation of mass and charge accurately in a reaction. In redox reactions, this process can be more complex due to the need to balance electron transfer between species.
The half-reaction method simplifies this task in redox equations by focusing separately on the oxidation and reduction processes:

• Separate the overall reaction into two half-reactions based on oxidation and reduction.
• Balance each half-reaction for atoms, starting with elements other than oxygen and hydrogen.
• Use water, hydrogen ions, or hydroxide ions to balance oxygen and hydrogen atoms.
• Balance the charges by adding electrons.
• Combine half-reactions, ensuring electrons cancel out for a complete balanced equation.

This method ensures all atoms and charges are balanced appropriately, aligning with the physical laws of conservation.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical changes, often involving redox reactions. It provides insights into how chemical energy is converted into electrical energy and vice versa, which is fundamental in batteries and fuel cells.
In electrochemistry, the principles of redox reactions are applied to create electrochemical cells:

• Oxidation occurs at the anode, whereas reduction occurs at the cathode.
• Electric current flows from the anode to the cathode.

Through the half-reaction method, we see these principles in action as electrons are transferred in a controlled manner, illustrating how energy can be harnessed. Understanding the balancing of redox reactions provides essential insights into designing efficient electrochemical systems like modern-day batteries and the process of electrolysis.