Problem 59 - Chemistry Matter and Change

Question

Use the oxidation-number method to balance these redox equations. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{HOCl}} \\ {\text { b. } \mathrm{HBrO}_{3} \rightarrow \mathrm{Br}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. \( \mathrm{Cl}_{2} + 2\mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{HOCl} + \mathrm{NaOH} \); b. \( 2\mathrm{HBrO}_{3} \rightarrow \mathrm{Br}_{2} + 3\mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \).

Step 1: Assign Oxidation Numbers (Equation a)

For equation (a), identify the oxidation states for each element. In \( \mathrm{Cl}_{2} \), chlorine is in the elemental state so its oxidation number is 0. In \( \mathrm{NaOH} \), \( \mathrm{Na} \) is +1, \( \mathrm{O} \) is -2, and \( \mathrm{H} \) is +1. In \( \mathrm{NaCl} \), \( \mathrm{Cl} \) is -1. In \( \mathrm{HOCl} \), \( \mathrm{Cl} \) is +1.

Step 2: Determine Changes in Oxidation Numbers (Equation a)

Chlorine changes from 0 in \( \mathrm{Cl}_{2} \) to -1 in \( \mathrm{NaCl} \) and +1 in \( \mathrm{HOCl} \). There is a reduction from 0 to -1 and oxidation from 0 to +1.

Step 3: Balance Oxidation and Reduction Changes (Equation a)

Chlorine is both oxidized and reduced, generating disproportionation. To balance chlorine atoms, ensure that the total increase in oxidation number (+2 for two \( \mathrm{Cl} \ ightarrow \mathrm{HOCl} \)) matches the total decrease (-2 for two \( \mathrm{Cl} \ ightarrow \mathrm{NaCl} \)). The balanced equation is \( \mathrm{Cl}_{2} + 2\mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{HOCl} + \mathrm{NaOH} \).

Step 4: Assign Oxidation Numbers (Equation b)

For equation (b), assign oxidation states as follows: In \( \mathrm{HBrO}_{3} \), \( \mathrm{Br} \) is +5. In \( \mathrm{Br}_{2} \), \( \mathrm{Br} \) is 0. In \( \mathrm{H}_{2}\mathrm{O} \), \( \mathrm{H} \) is +1 and \( \mathrm{O} \) is -2. In \( \mathrm{O}_{2} \), \( \mathrm{O} \) is 0.

Step 5: Determine Changes in Oxidation Numbers (Equation b)

Bromine in \( \mathrm{HBrO}_{3} \) is reduced from +5 to 0 in \( \mathrm{Br}_{2} \). Oxygen in \( \mathrm{HBrO}_{3} \) is oxidized from -2 to 0 in \( \mathrm{O}_{2} \).

Step 6: Balance Oxidation and Reduction Changes (Equation b)

For every 2 Bromines reduced, 5 Oxygens need to be oxidized to conserve mass. Multiply equation to account for electrons transferred. The balanced equation is \( 2\mathrm{HBrO}_{3} \rightarrow \mathrm{Br}_{2} + 3\mathrm{H}_{2}\mathrm{O} + \mathrm{O}_{2} \).

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is essential for balancing redox reactions. Oxidation numbers indicate the degree of oxidation or reduction of an element within a compound. This concept helps to track electron transfer during chemical reactions.
To assign oxidation numbers, follow these simple rules:

• For pure elements, the oxidation number is always 0.
• In simple ions, the oxidation number equals the charge on the ion.
• Oxygen typically has an oxidation number of -2, while hydrogen is usually +1.
• For neutral compounds, the sum of oxidation numbers must be zero.
• For ions, the sum of oxidation numbers equals the overall charge.

Using these rules helps identify which elements are oxidized and which are reduced during a reaction.

Balancing Equations

Balancing chemical equations ensures that the same number of atoms of each element appear on both sides of the equation. It's a crucial step in confirming that mass and charge are conserved.
When balancing equations, follow these guidelines:

• First, write down the unbalanced chemical equation.
• Next, balance the number of atoms for each element, focusing on different elements sequentially.
• Check that charges are balanced, particularly important in ionic equations.
• Finally, review the balanced equation to ensure the principles of conservation of mass and charge are satisfied.

Balancing becomes especially important in redox reactions, where electrons are transferred, affecting oxidation numbers.

Oxidation-Reduction

Oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. In these reactions, one species gets oxidized (loses electrons) while another gets reduced (gains electrons).
Key aspects of redox reactions include:

• Oxidation: An increase in oxidation number, indicating loss of electrons.
• Reduction: A decrease in oxidation number, indicating gain of electrons.
• Redox Couples: The oxidizing and reducing agents involved in electron transfer.

A proper understanding of redox principles helps in identifying how elements change their oxidation states and ensures proper balancing of chemical equations.

Chemical Equations

Chemical equations represent the substances involved in a chemical reaction, showing reactants and products with their respective quantities. Writing equations accurately is vital for understanding the reaction dynamics.
Some tips for working with chemical equations include:

• Identify and list all reactants and products before writing the equations.
• Make sure to include correct subscripts and superscripts for compounds and ions.
• Ensure all chemical symbols and formulas are correct to avoid errors during balancing.
• Write the physical states of compounds where necessary, e.g., (s) for solids, (l) for liquids, (g) for gases, and (aq) for aqueous solutions.

By following these guidelines, you can translate a verbal description of a chemical reaction into a precise chemical equation that illustrates the process taking place.