Problem 43
Question
From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Fe}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Ca}(s)\) or \(\mathrm{Al}(s)\) (c) \(\mathrm{H}_{2}(g,\) acidic solution \()\) or \(\mathrm{H}_{2} \mathrm{~S}(g)\) (d) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{IO}_{3}^{-}(a q)\)
Step-by-Step Solution
Verified Answer
The stronger reducing agents for each pair are: (a) \(\mathrm{Mg}(s)\), (b) \(\mathrm{Ca}(s)\), (c) \(\mathrm{H}_{2}\mathrm{S}(g)\), and (d) \(\mathrm{IO}_{3}^{-}(a q)\).
1Step 1: Identify reduction potentials
Using Appendix E, find the reduction potentials of each substance in the given pairs. Make sure to pick the correct half-reactions for the species involved.
(a) \(\mathrm{Fe}(s)\): \(\mathrm{Fe^{2+}}(aq) + 2\mathrm{e^{-}} \rightarrow \mathrm{Fe}(s)\); \(E^{\circ}=-0.44 \,\mathrm{V}\)
\(\mathrm{Mg}(s)\): \(\mathrm{Mg^{2+}}(aq) + 2\mathrm{e^{-}} \rightarrow \mathrm{Mg}(s)\); \(E^{\circ}=-2.37 \,\mathrm{V}\)
(b) \(\mathrm{Ca}(s)\): \(\mathrm{Ca^{2+}}(aq) + 2\mathrm{e^{-}} \rightarrow \mathrm{Ca}(s)\); \(E^{\circ}=-2.87 \,\mathrm{V}\)
\(\mathrm{Al}(s)\): \(\mathrm{Al^{3+}}(aq) + 3\mathrm{e^{-}} \rightarrow \mathrm{Al}(s)\); \(E^{\circ}=-1.66 \,\mathrm{V}\)
(c) \(\mathrm{H}_{2}(g,\) acidic solution \()\): \(2\mathrm{H^{+}}(aq) + 2\mathrm{e^{-}} \rightarrow \mathrm{H}_{2}(g)\); \(E^{\circ}=0.00 \,\mathrm{V}\)
\(\mathrm{H}_{2} \mathrm{S}(g)\): \(2\mathrm{e^{-}} + 2\mathrm{H^{+}}(aq) + \mathrm{S}^{2-}_\mathrm{(aq)} \rightarrow \mathrm{H}_{2}\mathrm{S}(g)\); \(E^{\circ}=-0.14 \,\mathrm{V}\)
(d) \(\mathrm{BrO}_{3}^{-}(a q)\): \(2\mathrm{BrO}_{3}^{-}(aq) + 12\mathrm{H^{+}}(aq) + 10\mathrm{e^{-}} \rightarrow \mathrm{Br^{-}}(aq) + 6\mathrm{H}_{2}\mathrm{O (l)}\); \(E^{\circ}=1.52 \,\mathrm{V}\)
\(\mathrm{IO}_{3}^{-}(a q)\): \(2\mathrm{IO}_{3}^{-}(aq) + 12\mathrm{H^{+}}(aq) + 10\mathrm{e^{-}} \rightarrow \mathrm{I^{-}}(aq) + 6\mathrm{H}_{2}\mathrm{O (l)}\); \(E^{\circ}=0.54\,\mathrm{V}\)
2Step 2: Compare reduction potentials
Now that you have the reduction potentials, compare them for each pair and identify the substance with the more negative reduction potential as the stronger reducing agent.
(a) \(\mathrm{Fe}(s)\): \(E^{\circ}=-0.44\,\mathrm{V}\) vs. \(\mathrm{Mg}(s)\): \(E^{\circ}=-2.37\,\mathrm{V}\)
(b) \(\mathrm{Ca}(s)\): \(E^{\circ}=-2.87\,\mathrm{V}\) vs. \(\mathrm{Al}(s)\): \(E^{\circ}=-1.66\,\mathrm{V}\)
(c) \(\mathrm{H}_{2}(g,\) acidic solution \()\): \(E^{\circ}=0.00\,\mathrm{V}\) vs. \(\mathrm{H}_{2}\mathrm{S}(g)\): \(E^{\circ}=-0.14\,\mathrm{V}\)
(d) \(\mathrm{BrO}_{3}^{-}(a q)\): \(E^{\circ}=1.52\,\mathrm{V}\) vs. \(\mathrm{IO}_{3}^{-}(a q)\): \(E^{\circ}=0.54\,\mathrm{V}\)
3Step 3: Identify the stronger reducing agents
Based on the comparisons, the following substances are the stronger reducing agents in each pair:
(a) \(\mathrm{Mg}(s)\)
(b) \(\mathrm{Ca}(s)\)
(c) \(\mathrm{H}_{2}\mathrm{S}(g)\)
(d) \(\mathrm{IO}_{3}^{-}(a q)\)
Key Concepts
Reduction PotentialsElectrochemistryHalf-reactions
Reduction Potentials
Reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. This is an essential concept in electrochemistry, as it helps determine the oxidizing or reducing strength of different substances. Every half-reaction has a standard reduction potential, denoted as \(E^{\circ}\), which is measured in volts (V).
Reduction potentials can be found in tables, often called electrochemical or reduction tables. These tables list half-reactions along with their \(E^{\circ}\) values. The more positive a reduction potential, the greater the affinity of the substance for electrons and the stronger it acts as an oxidizing agent. Conversely, a more negative reduction potential implies a substance is a powerful reducing agent because it more readily donates electrons.
To find which substance in a pair is the stronger reducing agent, always select the one with the more negative \(E^{\circ}\). In the original problem, substances such as \(\text{Mg}(s)\) and \(\text{Ca}(s)\) have more negative reduction potentials than their counterparts, making them stronger reducers.
Understanding this concept allows us to predict the direction of electron flow in redox reactions, which is fundamental for designing batteries and understanding reactions in chemistry.
Reduction potentials can be found in tables, often called electrochemical or reduction tables. These tables list half-reactions along with their \(E^{\circ}\) values. The more positive a reduction potential, the greater the affinity of the substance for electrons and the stronger it acts as an oxidizing agent. Conversely, a more negative reduction potential implies a substance is a powerful reducing agent because it more readily donates electrons.
To find which substance in a pair is the stronger reducing agent, always select the one with the more negative \(E^{\circ}\). In the original problem, substances such as \(\text{Mg}(s)\) and \(\text{Ca}(s)\) have more negative reduction potentials than their counterparts, making them stronger reducers.
Understanding this concept allows us to predict the direction of electron flow in redox reactions, which is fundamental for designing batteries and understanding reactions in chemistry.
Electrochemistry
Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is what drives the production of electricity and the processes of oxidation and reduction occurring. These are often referred to as redox reactions.
In electrochemistry, the entire flow of reactions is governed by the transfer of electrons between chemical species. This electron flow can be harnessed and used to perform work, which is the basic principle behind devices like batteries and fuel cells.
An electrochemical cell comprises two half-cells connected by a conductive solution. Each half-cell contains an electrode and an electrolyte where either oxidation or reduction occurs. Electrons are transferred externally via a circuit between the two electrodes to perform work. This movement translates into electrical energy that can be used to power devices.
Understanding electrochemical series and half-cell reactions provides insight into which substances can serve as effective reducing or oxidizing agents. This understanding is crucial not only for educational purposes, as seen in the textbook exercise, but also for real-world applications in energy conversion and storage.
In electrochemistry, the entire flow of reactions is governed by the transfer of electrons between chemical species. This electron flow can be harnessed and used to perform work, which is the basic principle behind devices like batteries and fuel cells.
An electrochemical cell comprises two half-cells connected by a conductive solution. Each half-cell contains an electrode and an electrolyte where either oxidation or reduction occurs. Electrons are transferred externally via a circuit between the two electrodes to perform work. This movement translates into electrical energy that can be used to power devices.
Understanding electrochemical series and half-cell reactions provides insight into which substances can serve as effective reducing or oxidizing agents. This understanding is crucial not only for educational purposes, as seen in the textbook exercise, but also for real-world applications in energy conversion and storage.
Half-reactions
In the context of electrochemistry, chemical reactions are often divided into two parts known as half-reactions. These are the oxidation half-reaction, where a substance loses electrons, and the reduction half-reaction, where a substance gains electrons.
Half-reactions simplify the evaluation of reactions by isolating the “half” of the total reaction that involves electron gain or loss. Each half-reaction provides the specific details on how electrons are transferred during the redox process.
Writing half-reactions involves:
For example, in the reduction half-reaction \(\text{Fe}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Fe}(s)\), iron ions gain electrons to form solid iron metal. Conversely, the oxidation counterpart would involve losing electrons.
In redox reactions, the sum of the oxidation and reduction half-reactions gives a full picture of the redox system. These half-equations help predict reaction possibilities and are crucial when calculating electrode potentials, providing a clearer understanding of electrochemical processes.
Half-reactions simplify the evaluation of reactions by isolating the “half” of the total reaction that involves electron gain or loss. Each half-reaction provides the specific details on how electrons are transferred during the redox process.
Writing half-reactions involves:
- Identifying the species being oxidized or reduced.
- Balancing the half-reaction for mass (atoms) and charge (electrons).
For example, in the reduction half-reaction \(\text{Fe}^{2+}(aq) + 2\text{e}^{-} \rightarrow \text{Fe}(s)\), iron ions gain electrons to form solid iron metal. Conversely, the oxidation counterpart would involve losing electrons.
In redox reactions, the sum of the oxidation and reduction half-reactions gives a full picture of the redox system. These half-equations help predict reaction possibilities and are crucial when calculating electrode potentials, providing a clearer understanding of electrochemical processes.
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