Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

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) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

Step-by-Step Solution

Answer

In summary, using the reduction potentials found in Appendix E, the stronger oxidizing agents are: (a) \(Cl_{2}(g)\) (b) \(Cd^{2+}(aq)\) (c) \(ClO_{3}^{-}(aq)\) (d) \(O_{3}(g)\)

Step 1: (a) Cl2(g) or Br2(l)

\ First, we need to find the reduction potential for both substances in the pair. In Appendix E, we find the following half-reactions and their respective standard reduction potentials (E°): 1. \(\mathrm{Cl_{2}(g)} + 2 \mathrm{e^{-}} \to 2 \mathrm{Cl^{-}(aq)}\), \(E_{1}^\circ = +1.36V\) 2. \(\mathrm{Br_{2}(l)} + 2 \mathrm{e^{-}} \to 2 \mathrm{Br^{-}(aq)}\), \(E_{2}^\circ = +1.07V\) Now compare the reduction potentials (E°). Since \(E_{1}^\circ > E_{2}^\circ\), Cl2(g) is the stronger oxidizing agent.

Step 2: (b) Zn2+(aq) or Cd2+(aq)

\ Similarly, find the reduction potential for both substances in the pair: 1. \(\mathrm{Zn^{2+}(aq)} + 2 \mathrm{e^{-}} \to\mathrm{Zn(s)}\), \(E_{3}^\circ = -0.76V\) 2. \(\mathrm{Cd^{2+}(aq)} + 2 \mathrm{e^{-}} \to\mathrm{Cd(s)}\), \(E_{4}^\circ = -0.40V\) Comparing the reduction potentials, we find that \(E_{4}^\circ > E_{3}^\circ\), so Cd2+(aq) is the stronger oxidizing agent.

Step 3: (c) Cl-(aq) or ClO3-(aq)

\ Find the reduction potential for both substances in the pair: 1. \(2 \mathrm{Cl}^{-}(a q) \to \mathrm{Cl}_{2}(g) + 2 \mathrm{e^{-}}\), \(E_{5}^\circ = -1.36V\) 2. \(6 \mathrm{H}^{+}(a q)+2 \mathrm{ClO}_{3}^{-}(a q)+6 e^{-} \to 3\mathrm{H}_{2}\mathrm{O}(l) + 2 \mathrm{Cl}^{-}(a q)\), \(E_{6}^\circ = +1.47V\) Comparing the reduction potentials, we find that \(E_{6}^\circ > E_{5}^\circ\), so ClO3⁻(aq) is the stronger oxidizing agent.

Step 4: (d) H2O2(aq) or O3(g)

\ Finally, find the reduction potential for both substances in the pair: 1. \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2\mathrm{H}^{+}(a q) +2 \mathrm{e^{-}} \to 2 \mathrm{H}_{2}\mathrm{O}(l)\), \(E_{7}^\circ = +1.77V\) 2. \(\mathrm{O}_{3}(g) + 2 \mathrm{H}^{+}(a q)+2 \mathrm{e^{-}} \to \mathrm{O}_{2}(g)+ \mathrm{H}_{2}\mathrm{O}(l)\), \(E_{8}^\circ = +2.07V\) Comparing the reduction potentials, we find that \(E_{8}^\circ > E_{7}^\circ\), so O3(g) is the stronger oxidizing agent. In summary, the stronger oxidizing agents are: (a) Cl2(g) (b) Cd2+(aq) (c) ClO3⁻(aq) (d) O3(g)

Key Concepts

Reduction Potential

The concept of reduction potential is a cornerstone in understanding how substances behave in redox reactions. Reduction potential, often symbolized as E°, is the measure of the tendency of a chemical species to acquire electrons and be reduced. It is expressed in volts (V). A positive reduction potential signifies a strong tendency to gain electrons, making the species a strong oxidizing agent. Conversely, a negative value means the species is more likely to lose electrons, showing weaker oxidizing power.
To determine which substance in a pair is the stronger oxidizing agent, one must compare their standard reduction potentials. In essence:

• A higher (more positive) reduction potential indicates a stronger oxidizing agent.
• This means the substance more easily receives electrons during a reaction.

Reduction potentials can be found in tables that list standard electrode potentials. These values are measured under standard conditions (25°C, 1 atm pressure, 1 M concentrations). By using these tables, chemists can predict reaction directions and calculate cell potentials in electrochemical cells. Remember, comparing reduction potentials allows us to determine how readily one substance can oxidize another.

Redox Reactions

Redox reactions are integral to the study of chemistry, especially in understanding how substances interact electronically. The term "redox" is derived from the combination of 'reduction' and 'oxidation' – two processes that occur simultaneously during chemical reactions. In these reactions, one substance loses electrons (oxidation) and another gains electrons (reduction).
A simple way to understand redox reactions is through the mnemonic "OIL RIG":

• Oxidation Is Losing electrons.
• Reduction Is Gaining electrons.

For example, when zinc metal reacts with copper(II) sulfate, the zinc loses electrons to form zinc ions (oxidation), and the copper ions gain electrons to form copper metal (reduction). This interplay of electron transfer is what facilitates energy transformations seen in various biological and industrial processes.
In electrochemical terms, redox reactions form the basis for batteries and electrochemical cells, where they convert chemical energy into electrical energy. Understanding and balancing redox reactions are crucial for analyzing how substances can act as oxidizing or reducing agents.

Electrochemistry

Electrochemistry is the branch of chemistry that explores the relationship between electricity and chemical reactions. It focuses on how electron transfer occurs between different substances and how this can be harnessed to produce electric current in electrochemical cells. Electrochemical processes underpin technologies such as batteries, electrolysis, and corrosion prevention.
In an electrochemical cell, redox reactions are split into two half-reactions: oxidation occurs at the anode, and reduction occurs at the cathode. This separation allows electrons to flow through an external circuit from the anode to the cathode, generating electricity.
Electrochemistry allows the study and manipulation of these reactions, giving insights into:

• How chemical energy is converted to electrical energy.
• The mechanisms of reactions within batteries and fuel cells.
• Methods for plating metals through electrolysis.

It influences numerous aspects of daily life, from powering devices with portable batteries to refining metals like aluminum and copper. Understanding electrochemistry is vital for innovation in energy storage and conversion technologies.