Problem 15
Question
Balance the following half-reactions by adding the appropriate number of electrons. Identify the oxidation half-reactions and the reduction half- reactions. a. \(\quad \mathrm{Br}_{2}(\ell) \rightarrow 2 \mathrm{Br}^{-}(a q)\) b. \(\mathrm{Pb}(s)+2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)\) c. \(\mathrm{O}_{3}(g)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) d. \(\mathrm{H}_{2} \mathrm{S}(g) \rightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)\)
Step-by-Step Solution
Verified Answer
Question: Based on the balanced half-reactions, identify the oxidation and reduction half-reactions and the number of electrons involved in each.
Answer: The oxidation and reduction half-reactions and number of electrons involved are as follows:
a. Reduction: Br₂(ℓ) + 2 e⁻ → 2 Br⁻(aq), 2 electrons are gained.
b. Oxidation: Pb(s) → PbCl₂(s) + 2 e⁻, 2 electrons are lost.
c. Reduction: O₃(g) + 2H⁺(aq) + 2 e⁻ → O₂(g) + H₂O(ℓ), 2 electrons are gained.
d. Oxidation: H₂S(g) → S(s) + 2 H⁺(aq) + 2 e⁻, 2 electrons are lost.
1Step 1: Identify Oxidation and Reduction Half-Reactions
Determine the change in oxidation state for each involved element and identify which half-reactions are oxidation (loss of electrons) and which are reduction (gain of electrons).
a. Oxidation state of Br in \(\mathrm{Br}_2\) is 0 and in \(\mathrm{Br}^-\) is -1. Since charge decreases, it is a reduction half-reaction.
b. Oxidation state of Pb in \(\mathrm{Pb}\) is 0 and in \(\mathrm{PbCl}_2\) is +2. Since charge increases, it is an oxidation half-reaction.
c. Oxidation state of O in \(\mathrm{O}_3\) is -2/3 and in \(\mathrm{O}_2\) is -2 and in \(\mathrm{H}_2\mathrm{O}\) is -2. Change from -2/3 to -2 is reduction.
d. Oxidation state of S in \(\mathrm{H}_2\mathrm{S}\) is -2 and in \(\mathrm{S}\) is 0. Since charge increases, it is an oxidation half-reaction.
2Step 2: Balance Half-Reactions by Adding Electrons
Add the number of electrons necessary for each half-reaction to ensure charged species are equal on both sides of the reaction:
a. Reduction half-reaction: \(\mathrm{Br}_2(\ell) + 2 e^- \rightarrow 2 \mathrm{Br}^-(aq)\)
b. Oxidation half-reaction: \(\mathrm{Pb}(s) \rightarrow \mathrm{PbCl}_{2}(s) + 2 e^-\)
c. Reduction half-reaction: \(\mathrm{O}_{3}(g) + 2\mathrm{H}^+(aq) + 2 e^- \rightarrow \mathrm{O}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(\ell)\)
d. Oxidation half-reaction: \(\mathrm{H}_{2}\mathrm{S}(g) \rightarrow \mathrm{S}(s) + 2 \mathrm{H}^{+}(aq) + 2 e^-\)
Key Concepts
Half-Reaction MethodOxidation StatesBalancing ReactionsElectrochemistry
Half-Reaction Method
The half-reaction method is an approach used to balance redox reactions by separately addressing the oxidation and reduction processes. It helps to identify the changes in the oxidation state of elements in a reaction, simplify the balancing process, and ensure the total charge is conserved. To balance a redox reaction using the half-reaction method, you need to follow these general steps:
- First, separate the reaction into two half-reactions: one for oxidation (loss of electrons) and one for reduction (gain of electrons).
- Next, balance the atoms in each half-reaction, starting with elements other than oxygen and hydrogen.
- Then, balance the oxygen atoms by adding water molecules, and balance hydrogen atoms by adding protons (H+).
- Finally, balance the charges by adding the appropriate number of electrons to one side of each half-reaction.
Oxidation States
Oxidation states, or oxidation numbers, are theoretical charges assigned to atoms within a molecule or ion. They provide a way to keep track of the electrons in redox reactions, helping us identify which atoms are oxidized and which are reduced.
- An element's oxidation state in its standard state is always zero, such as \(\mathrm{Br}_2\) or \(\mathrm{Pb}\) in their elemental forms.
- The oxidation state increases when an atom loses electrons, indicating oxidation.
- Conversely, the oxidation state decreases when an atom gains electrons, indicating reduction.
Balancing Reactions
Balancing reactions is a crucial part of ensuring that the law of conservation of mass and charge is satisfied. In redox reactions, each element must have the same number of atoms on both sides of the equation. Electrons transferred must also balance out to reflect the true nature of the chemical change occurring.
After identifying oxidation and reduction, here is how you balance:
- Balance all elements except oxygen and hydrogen first.
- Balance oxygen atoms by adding H2O.
- Balance hydrogen atoms by adding H+ (in acidic solutions) or OH- (in basic solutions).
- Balance the charges by adding electrons to the side that is more positive.
Electrochemistry
Electrochemistry is the branch of chemistry that studies chemical processes which involve the movement of electrons. This field is closely linked to redox reactions as these reactions either produce or require electricity. Understanding electrochemistry allows you to explore devices like batteries, fuel cells, and electrolysis units.
- In electrochemical cells, a redox reaction generates electrical energy, serving as the foundation for batteries.
- In electrolytic cells, electrical energy is used to drive a non-spontaneous chemical reaction, as seen in electrolysis.
- Key variables studied in electrochemistry include electrode potential, which is measured in volts, and the current, measured in amperes.
-
This understanding extends to industrial applications, environmental monitoring, and even biological systems like nerve conduction. The balance and flow of electrons in electrochemical systems are fundamental to numerous technologies and scientific studies.
Other exercises in this chapter
Problem 13
Why can't a wire perform the same function as a porous separator in an electrochemical cell?
View solution Problem 14
In a voltaic cell, why is the cathode labeled the positive terminal and the anode the negative terminal?
View solution Problem 16
Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction half- reactions?
View solution Problem 17
Write a half-reaction for the oxidation of magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)\) to hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\)
View solution