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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 \(\mathrm{O}_{3}(g)\)

Short Answer

Expert verified
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 by step solution

01

(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.
02

(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.
03

(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.
04

(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)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider a redox reaction for which \(E^{\circ}\) is a negative number. (a) What is the sign of \(\Delta G^{\circ}\) for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than \(1 ?\) (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings?

Indicate whether each statement is true or false: (a) The anode is the electrode at which oxidation takes place. (b) A voltaic cell always has a positive emf. (c) A salt bridge or permeable barrier is necessary to allow a voltaic cell to operate.

Li-ion batteries used in automobiles typically use a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode in place of the \(\mathrm{LiCoO}_{2}\) cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a \(\mathrm{LiCoO}_{2}\) cathode, approximately \(50 \%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

(a) Aluminum metal is used as a sacrificial anode to protect offshore pipelines in salt water from corrosion. Why is the aluminum referred to as a "sacrificial anode"? (b) Looking in Appendix E, suggest what metal the pipelines could be made from in order for aluminum to be successful as a sacrificial anode.

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