Question

Use the oxidation-number method to balance these redox equations. a. \(\mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{HOCl}\) b. \(\mathrm{HBrO}_{3} \rightarrow \mathrm{Br}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\)

Short answer

a. The balanced redox equation for \(\mathrm{Cl}_2 + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{HOCl}\) is: \(\mathrm{Cl}_{2} + 2\mathrm{NaOH} \rightarrow 2\mathrm{NaCl} + 2\mathrm{HOCl}\) b. The balanced redox equation for \(\mathrm{HBrO}_3 \rightarrow \mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{O}_2\) is: \(6\mathrm{HBrO}_3 \rightarrow 6\mathrm{Br}_2 + 12\mathrm{H}_2\mathrm{O} + 5\mathrm{O}_2\)

Step-by-step solution

Assign oxidation numbers

Assign oxidation numbers to each element in the reaction as follows: For reactants: Cl in \(\mathrm{Cl}_2\) = 0, Na in \(\mathrm{NaOH}\) = +1, O in \(\mathrm{NaOH}\) = -2, H in \(\mathrm{NaOH}\) = +1 For products: Na in \(\mathrm{NaCl}\) = +1, Cl in \(\mathrm{NaCl}\) = -1, H in \(\mathrm{HOCl}\) = +1, O in \(\mathrm{HOCl}\) = -2, Cl in \(\mathrm{HOCl}\) = +1

Calculate the change in oxidation numbers

Change in oxidation numbers: Cl (0 to -1) and (0 to +1)

Balance half-reactions for oxidized and reduced species

Oxidation half-reaction: \(\mathrm{Cl}_{2} \rightarrow 2\mathrm{Cl}^{-}\) Reduction half-reaction: \(2\mathrm{Cl_2} + 4\mathrm{OH^-} \rightarrow 4\mathrm{Cl}^+ + 2\mathrm{H_2O}\)

Balance electrons

Ensure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. Oxidation half-reaction: 2 electrons lost Reduction half-reaction: 4 electrons gained (divide by 2) To balance electrons, multiply the oxidation half-reaction by 2: \(2\mathrm{Cl}_{2} \rightarrow 4\mathrm{Cl}^{-}\)

Add the half-reactions and obtain the balanced redox equation

Combined balanced redox equation: \(\mathrm{Cl}_{2} + 2\mathrm{NaOH} \rightarrow 2\mathrm{NaCl} + 2\mathrm{HOCl}\) b. Now let's analyze the redox reaction \(\mathrm{HBrO}_3 \rightarrow \mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{O}_2\)

Assign oxidation numbers

Assign oxidation numbers to each element in the reaction as follows: For reactants: H in \(\mathrm{HBrO_3}\) = +1, Br in \(\mathrm{HBrO_3}\) = +5, O in \(\mathrm{HBrO_3}\) = -2 For products: Br in \(\mathrm{Br_2}\) = 0, H in \(\mathrm{H_2O}\) = +1, O in \(\mathrm{H_2O}\) = -2, O in \(\mathrm{O_2}\) = 0

Calculate the change in oxidation numbers

Change in oxidation numbers: Br (+5 to 0) and O (-2 to 0)

Balance half-reactions for oxidized and reduced species

Oxidation half-reaction: \(4\mathrm{BrO_3}^- \rightarrow 6\mathrm{Br}_2 + 12\mathrm{H_2O}\) Reduction half-reaction: \(6\mathrm{BrO_3}^- + 12\mathrm{H_2O} \rightarrow 3\mathrm{Br}_2 + 36\mathrm{H^+} + 30\mathrm{O}^{2-}\) (Notice that 30 = 6x5-6x2)

Balance electrons

Ensure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. Oxidation half-reaction: 20 electrons lost Reduction half-reaction: 20 electrons gained (no need to divide, as both are equal)

Add the half-reactions and obtain the balanced redox equation

Combined balanced redox equation (Notice that the charges are already balanced, since both are equal): \(6\mathrm{HBrO}_3 \rightarrow 6\mathrm{Br}_2 + 12\mathrm{H}_2\mathrm{O} + 5\mathrm{O}_2\)

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes where electrons are transferred between two substances. These reactions involve two main parts: oxidation and reduction. During oxidation, a substance loses electrons, whereas reduction involves the gain of electrons. These electron exchanges are fundamental in many chemical reactions that power various systems, from cellular respiration in living organisms to the rusting of a nail.

• Oxidation: Refers to the loss of electrons in a substance, leading it to become more positively charged.
• Reduction: Involves the gain of electrons, resulting in a substance becoming more negatively charged.

Understanding redox reactions is crucial in chemistry because they explain how different substances interact at the molecular level. Recognizing the movement of electrons helps us predict the outcome of chemical reactions and their applications in real-world scenarios like battery operations and fuel combustion.

Balancing Equations

Balancing chemical equations is an essential skill in chemistry, ensuring that the amount of each element is conserved in a reaction. A balanced equation reflects the law of conservation of mass, meaning the same number of atoms of each element must be present on both sides of the equation.
In redox reactions, balancing equations becomes slightly more complex due to the electron exchange involved. Here, not only are the atoms balanced, but the charge is balanced as well.

• Start by writing the unbalanced equation.
• Assign oxidation numbers to identify which species are oxidized and reduced.
• Balance the atoms aside from oxygen and hydrogen.
• Use water, hydrogen ions, or hydroxide ions to balance oxygen and hydrogen atoms in acidic or basic solutions.
• Finally, adjust coefficients so the number of electrons lost equals the number gained.

Following these steps ensures that both the atoms and the charges are conserved in the reaction, providing an accurate representation of the chemical change happening.

Oxidation Numbers

Oxidation numbers provide a way to track how many electrons an atom is gaining or losing in a reaction. Assigning oxidation numbers to each element in a chemical equation helps identify which atoms are oxidized and which are reduced.

Here are key points about oxidation numbers:

• An element in its pure state (like \(\)Cl\(_2\)) has an oxidation number of 0.
• In compounds, oxidation numbers are generally based on a set of rules, where elements like oxygen typically have -2 and hydrogen +1.
• The sum of oxidation numbers in a molecule or ion must equal the overall charge. For example, the sum in a neutral compound is zero.

By calculating the change in oxidation numbers before and after the reaction, we can determine which elements are undergoing oxidation and reduction. This is the underlying principle of the oxidation-number method, crucial for correctly balancing redox reactions.

Half-Reactions

Half-reactions separate the oxidation and reduction processes into two different equations, making it easier to see how electrons are transferred in redox reactions. It is a fundamental step when using the oxidation-number method to balance redox equations.

Understanding half-reactions involves the following steps:

• Write separate equations for oxidation and reduction, listing the reactants and products for each.
• Balance all elements except for oxygen and hydrogen, then balance oxygen using water molecules and hydrogen using hydrogen ions or hydroxide ions if it’s in a basic solution.
• Ensure that the electrons lost in the oxidation half-reaction match the electrons gained in the reduction half-reaction.
• Finally, add the half-reactions back together, ensuring that atoms and charges are balanced.

Half-reactions are useful because they simplify complex reactions, breaking them into parts that can be easily understood and balanced. This approach provides clarity when studying the electron dynamics in redox reactions, crucial for understanding chemical processes at a detailed level.