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Why is \(\mathrm{O}_{2}\) a stronger oxidizing agent in acid than in base? Use standard reduction potentials from Appendix 6 to support your answer.

Short Answer

Expert verified
Answer: O鈧 is a stronger oxidizing agent in acidic solutions compared to basic solutions due to the higher standard reduction potential in acidic conditions (1.23 V) compared to basic conditions (0.40 V). This is because the acidic environment reaction with O鈧 is more favorable, with a higher attraction of O鈧 for the H鈦 ions, making it a stronger oxidizing agent in acid than in base.

Step by step solution

01

Identify the half-reactions in acidic and basic environments

To proceed with the problem, first, we need to find the half-reactions of oxygen in both acidic and basic environments. These half-reactions can be found in the standard reduction potentials table (Appendix 6). For acidic conditions: O鈧 + 4H鈦 + 4e鈦 鈫 2H鈧侽 E掳 = 1.23 V For basic conditions: O鈧 + 4e鈦 +2H鈧侽 鈫 4OH鈦 E掳 = 0.40 V
02

Compare the standard reduction potentials

Now that we have identified the two half-reactions for acidic and basic environments, let's compare their standard reduction potentials. The standard reduction potential for O鈧 in the acidic environment is 1.23 V, and in the basic environment, it's 0.40 V.
03

Explain the differences in oxidizing strength

The higher the standard reduction potential, the stronger the oxidizing agent. Since the standard reduction potential of O鈧 in acidic conditions (1.23 V) is higher than that in basic conditions (0.40 V), it indicates that O鈧 is a stronger oxidizing agent in acidic conditions compared to basic conditions. This difference in the oxidizing strength can also be explained by looking at the chemical equations: Acidic conditions: O鈧 + 4H鈦 + 4e鈦 鈫 2H鈧侽 Basic conditions: O鈧 + 4e鈦 +2H鈧侽 鈫 4OH鈦 In acidic conditions, O鈧 gets reduced by gaining 4 electrons and reacting with 4 H鈦 ions to form water, while in basic conditions, O鈧 gets reduced by gaining 4 electrons and reacting with only 2 water molecules to produce 4 OH鈦 ions. The reaction occurring in the acidic environment is more favorable due to the higher attraction of O鈧 for the H鈦 ions. This makes O鈧 a stronger oxidizing agent in acid than in base.

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Key Concepts

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

Standard Reduction Potentials
Standard reduction potentials are a way of measuring the tendency for a specific substance to gain electrons in a redox reaction. This is a crucial piece of electrochemical data that helps predict the direction of electron flow and the relative strengths of oxidizing agents.
To get a clearer understanding, imagine a chart listing substances with their corresponding reduction potentials. Each potential is measured in volts (V) relative to the standard hydrogen electrode, which is set at 0 V.

In the context of oxygen in acidic and basic environments, standard reduction potentials can explain how effectively oxygen acts as an oxidizing agent in each setting. Higher potentials indicate a greater ability to attract and gain electrons.
  • Acidic environment: O鈧 has a standard reduction potential of 1.23 V.
  • Basic environment: The potential drops to 0.40 V.
These values reveal that oxygen is far more effective as an oxidizing agent in acidic conditions.
Acidic and Basic Environments
In chemistry, the environment in which a reaction takes place can have substantial effects on the outcome. Acidic and basic environments refer to the pH conditions of the solution in which reactions occur.

Acidic conditions are characterized by an abundance of hydrogen ions (H鈦), while basic conditions have a surplus of hydroxide ions (OH鈦). These differing environments can alter the behavior of reactants and products significantly.

For example, the greater concentration of H鈦 ions in an acidic environment increases the likelihood of these ions participating in redox reactions. This participation often enhances the oxidizing capacity of substances like \( ext{O}_2\), which can readily accept electrons in the presence of additional H鈦 ions to form water.
  • Acidic: Higher concentration of H鈦 ions
  • Basic: Higher concentration of OH鈦 ions
Understanding the environment enables us to predict and manipulate chemical reactions for desired outcomes.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. It鈥檚 like the study of how chemical reactions can produce electricity and how electricity can cause chemical reactions.

This field relies heavily on concepts such as standard reduction potentials and is key to understanding the behavior of oxidizing agents like oxygen. When a substance undergoes a redox reaction, electrons are transferred, leading to changes in chemical and electrical energy.

For example, in electrochemical cells, electron flow from one substance to another generates an electric current. This principle is harnessed in batteries, electroplating, and even metabolic processes within the human body.
  • In electrochemical cells, reduction occurs at the cathode while oxidation happens at the anode.
  • Applications: Batteries, corrosion, and electrolysis.
  • Key to balancing energy efficiency and reaction effectiveness.
Through electrochemistry, we gain insights into the efficiency and dynamics of real-world processes.
Half-Reactions
In a redox reaction, two processes occur simultaneously: oxidation and reduction. These can be broken down into half-reactions, which describe the donor and acceptor of electrons separately.

A half-reaction highlights either the release or acceptance of electrons, helping to understand how oxidation and reduction contribute to the overall chemical reaction. For example,

The acidic half-reaction for oxygen looks like:
  • O鈧 + 4H鈦 + 4e鈦 鈫 2H鈧侽
While the basic half-reaction is:
  • O鈧 + 4e鈦 + 2H鈧侽 鈫 4OH鈦
By examining these, we see how hydrogen ions (H鈦) in an acidic environment influence the reduction of oxygen, leading to a more favorable scenario in terms of gaining electrons.

Breaking down reactions into half-reactions helps in correctly balancing chemical equations and understanding the mechanics of electron transfer. It illustrates:
  • Electron gain (reduction) and electron loss (oxidation).
  • The role of surrounding ions in the redox process.
Comprehending half-reactions clarifies complex chemical interactions, simplifying them into manageable parts.

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Most popular questions from this chapter

Thomas Edison, the inventor of the incandescent lightbulb, developed a voltaic cell that delivers \(1.4 \mathrm{V}\) of cell potential in an alkaline electrolyte based on the following cell reaction: \(3 \mathrm{Fe}(s)+8 \mathrm{NiO}(\mathrm{OH})(s)+4 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 8 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Fe}_{3} \mathrm{O}_{4}(s) $$ a. Assign oxidation numbers to each element in each of the nickel and iron compounds. b. How many electrons are transferred in the overall reaction? c. What is the value of \(\Delta G_{\text {cell }} ?\)

Quantitative Analysis Electrolysis can be used to determine the concentration of \(\mathrm{Cu}^{2+}\) in a given volume of solution by electrolyzing the solution in a cell equipped with a platinum cathode. If all of the \(\mathrm{Cu}^{2+}\) is reduced to \(\mathrm{Cu}\) metal at the cathode, the increase in mass of the electrode provides a measure of the concentration of \(\mathrm{Cu}^{2+}\) in the original solution. To ensure the complete (99.9996) removal of the \(\mathrm{Cu}^{2+}\) from a solution in which \(\left[\mathrm{Cu}^{2+}\right]\) is initially about \(1.0 M,\) will the potential of the cathode (versus SHE) have to be more negative or less negative than \(0.34 \mathrm{V}\) (the standard potential for \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) )?

Voltaic cells based on the following pairs of half-reactions are constructed. For each pair, write a balanced equation for the cell reaction, and identify which half-reaction takes place at each anode and cathode. a. \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) \(\mathrm{Ag}^{+}(a q)+\mathrm{c}^{-} \rightarrow \mathrm{Ag}(s)\) b. \(\mathrm{AgBr}(s)+\mathrm{c}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q)\) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\operatorname{Pt} C l_{4}^{2-}(a q)+2 e^{-} \rightarrow \operatorname{Pt}(s)+4 C l^{-}(a q)\) \(\mathrm{AgCl}(s)+\mathrm{c}^{-} \rightarrow \mathrm{Ag (s)+\mathrm{Cl}^{-}(a q)\)

Starting with the appropriate standard free energies of formation from Appendix \(4,\) calculate the valucs of \(\Delta G^{\circ}\) and \(E_{\text {cell }}\) of the following reactions: a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q)\)

Nickel-Sodium Batteries Researchers in England are developing a battery for electric vehicles based on the reaction between \(\mathrm{NiCl}_{2}(s)\) and \(\mathrm{Na}(\mathrm{s}):\) $$ 2 \mathrm{Na}(s)+\mathrm{NiCl}_{2}(s) \rightarrow \mathrm{Ni}(s)+2 \mathrm{NaCl}(s) $$ The cells in the battery produce \(2.58 \mathrm{V}\) a. Assign oxidation numbers to each element in the nickel and sodium compounds. b. How many electrons are transferred in the overall reaction? c. What is the value of \(\Delta G_{\text {cen }} ?\)

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