Question

From each of the following pairs of substances, use data in Appendix 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 \(\operatorname{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{2}(\mathrm{~g})\)

Short answer

The stronger oxidizing agents for each pair are: (a) \(\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{Cd}^{2+}(aq)\) (c) \(\mathrm{ClO}_{3}^{-}(aq)\) (d) \(\mathrm{O}_{2}(g)\)

Step-by-step solution

(a) - Analyzing \(\mathrm{Cl}_{2}(g)\) and \(\mathrm{Br}_{2}(l)\)

The reduction half-reactions for the substances are: 1. \(\mathrm{Cl}_{2}(g) + 2e^{-} \rightarrow 2\mathrm{Cl}^{-}(aq)\) 2. \(\mathrm{Br}_{2}(l) + 2e^{-} \rightarrow 2\mathrm{Br}^{-}(aq)\) Refer to Appendix E to find the standard reduction potentials: 1. \(E^{\circ}_{\mathrm{Cl}_{2}} = +1.36 \ \mathrm{V}\) 2. \(E^{\circ}_{\mathrm{Br}_{2}} = +1.07 \ \mathrm{V}\)

(a) - Choosing the stronger oxidizing agent

Comparing the standard reduction potentials, \(\mathrm{Cl}_{2}(g)\) has a higher potential than \(\mathrm{Br}_{2}(l)\). Therefore, \(\mathrm{Cl}_{2}(g)\) is the stronger oxidizing agent.

(b) - Analyzing \(\mathrm{Zn}^{2+}(a q)\) and \(\operatorname{Cd}^{2+}(a q)\)

The reduction half-reactions for the substances are: 1. \(\mathrm{Zn}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Zn}(s)\) 2. \(\mathrm{Cd}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Cd}(s)\) Refer to Appendix E to find the standard reduction potentials: 1. \(E^{\circ}_{\mathrm{Zn}^{2+}} = -0.76 \ \mathrm{V}\) 2. \(E^{\circ}_{\mathrm{Cd}^{2+}} = -0.40 \ \mathrm{V}\)

(b) - Choosing the stronger oxidizing agent

Comparing the standard reduction potentials, \(\mathrm{Cd}^{2+}(aq)\) has a higher potential than \(\mathrm{Zn}^{2+}(aq)\). Therefore, \(\mathrm{Cd}^{2+}(aq)\) is the stronger oxidizing agent.

(c) - Analyzing \(\mathrm{Cl}^{-}(a q)\) and \(\mathrm{ClO}_{3}^{-}(a q)\)

The reduction half-reactions for the substances are: 1. \(2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{Cl}_{2}(g) + 2e^{-}\) 2. \(\mathrm{ClO}_{3}^{-}(aq) + 6\mathrm{H}_{2}\mathrm{O}(l) + 6e^{-} \rightarrow \mathrm{Cl}^{-}(aq) + 12\mathrm{OH}^{-}(aq)\) Refer to Appendix E to find the standard reduction potentials: 1. \(E^{\circ}_{\mathrm{Cl}^{-}} = -1.36 \ \mathrm{V}\) 2. \(E^{\circ}_{\mathrm{ClO}_{3}^{-}} = +0.89 \ \mathrm{V}\)

(c) - Choosing the stronger oxidizing agent

Comparing the standard reduction potentials, \(\mathrm{ClO}_{3}^{-}(aq)\) has a higher potential than \(\mathrm{Cl}^{-}(aq)\). Therefore, \(\mathrm{ClO}_{3}^{-}(aq)\) is the stronger oxidizing agent.

(d) - Analyzing \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) and \(\mathrm{O}_{2}(\mathrm{~g})\)

The reduction half-reactions for the substances are: 1. \(2\mathrm{H}_{2} \mathrm{O}_{2}(aq) + 2e^{-} \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g)\) 2. \(\mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) + 4e^{-} \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\) Refer to Appendix E to find the standard reduction potentials: 1. \(E^{\circ}_{\mathrm{H}_{2} \mathrm{O}_{2}} = +0.68 \ \mathrm{V}\) 2. \(E^{\circ}_{\mathrm{O}_{2}} = +1.23 \ \mathrm{V}\)

(d) - Choosing the stronger oxidizing agent

Comparing the standard reduction potentials, \(\mathrm{O}_{2}(g)\) has a higher potential than \(\mathrm{H}_{2} \mathrm{O}_{2}(aq)\). Therefore, \(\mathrm{O}_{2}(g)\) is the stronger oxidizing agent.

Key concepts

Reduction Potentials

Reduction potentials are a key concept in electrochemistry. They are used to understand how readily a chemical species gains electrons. When we talk about reduction potentials, we're referring to the voltage associated with a reduction reaction, where a substance gains electrons.

The reduction potential is essentially an indicator of the ability of a compound to act as an oxidizing agent, which means it can oxidize other substances by taking electrons from them.

Think of it like a battery: the higher the potential, the more energy it has to provide. In chemistry, it works similarly, where a higher reduction potential means the species is more likely to "pull" electrons toward itself. This concept is crucial when determining which substances will dominate in redox reactions.

Standard Reduction Potential

The standard reduction potential is another critical concept in electrochemistry. It differs from reduction potential as it is measured under specific, standard conditions.

• Temperature: 298 Kelvin (or 25°C)
• Pressure: 1 atmosphere
• Concentration: 1 Molar solutions

These standard conditions are crucial for the comparison of different half-reactions. The values are usually given with symbols like \(E^{\circ}\) to denote it’s under standard conditions.

These potentials help us understand which reactions are more likely to occur naturally. A higher \(E^{\circ}\) means the substance is a stronger oxidizing agent compared to one with a lower \(E^{\circ}\) under the same conditions. This helps chemists predict the flow of electrons in electrochemical cells, and ultimately, the feasibility of chemical reactions.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical changes. It plays a vital role in a wide range of processes among various fields.

In electrochemical reactions, oxidation and reduction occur. \( \text{Oxidation} \) is when a species loses electrons, while \( \text{reduction} \) is when a species gains electrons. Electrochemical cells, like voltaic or galvanic cells, involve redox reactions that generate an electrical current.

Understanding electrochemistry is fundamental for applications such as:

• Battery technology
• Corrosion prevention
• Industrial electrolysis - used in processes like extracting metals

It helps in designing systems where chemical energy is converted into electrical energy and vice versa. By mastering electrochemistry, one can innovate in creating efficient energy solutions, among many other technological advances.