Problem 42 - Chemistry: The Central Science

Question

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{IO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

Step-by-Step Solution

Answer

The stronger oxidizing agents from each pair are: (a) \(\mathrm{Cl}_{2}(g)\), (b) \(\mathrm{Cd}^{2+}(a q)\), (c) \(\mathrm{BrO}_{3}^{-}(a q)\), and (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\), based on their higher standard reduction potentials.

Step 1: Finding Standard Reduction Potentials

Find the respective standard reduction potentials for \(\mathrm{Cl}_{2}(g)\) and \(\mathrm{Br}_{2}(l)\) in Appendix E. Refer to their half-reactions: \(\mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(a q)\) with \(E^{\circ}= +1.36\,\mathrm{V}\) \(\mathrm{Br}_{2}(l) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(a q)\) with \(E^{\circ} = +1.087\,\mathrm{V}\)

Step 2: Compare Reduction Potentials

We have the respective standard reduction potentials for the given substances. Compare the values: \(E^{\circ}(\mathrm{Cl}_{2}) > E^{\circ}(\mathrm{Br}_{2})\) As \(\mathrm{Cl}_{2}(g)\) has a higher reduction potential, it is a stronger oxidizing agent among the two. (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\)

Step 1: Finding Standard Reduction Potentials

Find the respective standard reduction potentials for \(\mathrm{Zn}^{2+}(a q)\) and \(\mathrm{Cd}^{2+}(a q)\) in Appendix E. Refer to their half-reactions: \(\mathrm{Zn}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s)\) with \(E^{\circ}= -0.763\,\mathrm{V}\) \(\mathrm{Cd}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) with \(E^{\circ} = -0.402\,\mathrm{V}\)

Step 2: Compare Reduction Potentials

We have the respective standard reduction potentials for the given substances. Compare the values: \(E^{\circ}(\mathrm{Zn}^{2+}) < E^{\circ}(\mathrm{Cd}^{2+})\) As \(\mathrm{Cd}^{2+}(a q)\) has a higher reduction potential, it is a stronger oxidizing agent among the two. (c) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{IO}_{3}^{-}(a q)\)

Step 1: Finding Standard Reduction Potentials

Find the respective standard reduction potentials for \(\mathrm{BrO}_{3}^{-}(a q)\) and \(\mathrm{IO}_{3}^{-}(a q)\) in Appendix E. Refer to their half-reactions: \(\mathrm{BrO}_{3}^{-}(a q) + 6 \mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}(a q) + 3\mathrm{O}_{2}(g)\) with \(E^{\circ}= +1.52\,\mathrm{V}\) \(\mathrm{IO}_{3}^{-}(a q) + 6 \mathrm{e}^{-} \rightarrow \mathrm{I}^{-}(a q) + 3\mathrm{O}_{2}(g)\) with \(E^{\circ} = +1.2\,\mathrm{V}\)

Step 2: Compare Reduction Potentials

We have the respective standard reduction potentials for the given substances. Compare the values: \(E^{\circ}(\mathrm{BrO}_{3}^{-}) > E^{\circ}(\mathrm{IO}_{3}^{-})\) As \(\mathrm{BrO}_{3}^{-}(a q)\) has a higher reduction potential, it is a stronger oxidizing agent among the two. (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

Step 1: Finding Standard Reduction Potentials

Find the respective standard reduction potentials for \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) and \(\mathrm{O}_{3}(g)\) in Appendix E. Refer to their half-reactions: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2 \mathrm{e}^{-} \rightarrow 2\mathrm{OH}^{-}(a \thinspace q)\) with \(E^{\circ}= +0.878\,\mathrm{V}\) \(\mathrm{O}_{3}(g) + 2 \mathrm{e}^{-} \rightarrow 3\mathrm{O}_{2}(g)\) with \(E^{\circ} = +0.683\,\mathrm{V}\)

Step 2: Compare Reduction Potentials

We have the respective standard reduction potentials for the given substances. Compare the values: \(E^{\circ}(\mathrm{H}_{2}\mathrm{O}_{2}) > E^{\circ}(\mathrm{O}_{3})\) As \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) has a higher reduction potential, it is a stronger oxidizing agent among the two.

Key Concepts

Standard Reduction Potential

When we talk about standard reduction potential (denoted as E°), we are referring to the inherent tendency of a chemical species to acquire electrons and thereby be reduced. This is measured under standard conditions, which include a temperature of 298 K, a 1 M concentration for each ion participating in the reaction, and a partial pressure of 1 bar for any gases involved.

The standard reduction potential is critical in determining the strength of an oxidizing agent. In general, the higher the E° value, the greater the species' tendency to gain electrons and oxidize other substances. This is why it can be used to predict the outcome of redox reactions. For instance, if we compare chlorine gas (Cl₂) with E° = +1.36 V to bromine liquid (Br₂) with E° = +1.087 V, we can identify chlorine as the stronger oxidizing agent because it has a higher standard reduction potential.

Understanding standard reduction potentials also helps in ranking substances in terms of their oxidizing power. The table or Appendix E in a textbook often lists these values for various substances, providing a valuable reference for students and chemists alike.

Half-Reactions

At the heart of the analysis of redox reactions are half-reactions. These are the two parts that make up the overall redox process: one for the reduction and one for the oxidation. A half-reaction explicitly shows the electrons as either reactants (in reduction) or products (in oxidation).

When solving for the stronger oxidizing agent between two substances, it's vital to consider their respective half-reactions. For example, the half-reaction for the reduction of H₂O₂ (aqueous) is written as H₂O₂ (aq) + 2 e^- → 2 OH^- (aq) with an E° = +0.878 V. By writing out these half-reactions, we can more easily visualize and compare the changes occurring during the reaction, particularly the transfer of electrons, which is crucial for identifying the oxidizing and reducing agents in the process.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the interconversion of chemical energy and electrical energy. The principles of electrochemistry are directly applied when analyzing redox (oxidation-reduction) reactions, as such reactions involve electron transfer between substances.

In the context of our discussion on oxidizing agents, electrochemistry helps us understand how substances that act as oxidizing agents are reduced during electrochemical reactions, and in turn, how they can cause other substances to be oxidized. Electrochemical cells, such as galvanic or voltaic cells, are practical applications of these principles. The cell is constructed with two half-cells where oxidation and reduction half-reactions occur separately, leading to a flow of electrons through an external circuit, thus generating an electrical current.

Overall, a firm grasp of electrochemistry is essential for interpreting redox reactions and predicting the sequence of events when different substances are involved, especially in the selection of stronger oxidizing agents, which depends on understanding both standard reduction potential and half-reactions within an electrochemical context.