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Chapter 17: Problem 62

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \quad \mathrm{Br}_{2}, \quad \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s) ?\)

Short Answer

Expert verified
(a) The strongest oxidizing agent is \(\mathrm{Br}_{2}\). (b) The strongest reducing agent is \(\mathrm{Ca}\). (c) \(\mathrm{H}^{+}\) and \(\mathrm{Cd}^{2+}\) can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid. (d) No species given can be reduced by \(\mathrm{Zn}(s)\).

Step by step solution

01

Consult the Standard Reduction Potential Table

Find the standard reduction potentials for each species in the table. The values are given below: \[ \begin{aligned} \mathrm{Br}^{-} / \mathrm{Br}_{2} &: +1.087 \, V \\ \mathrm{H}^{+} / \mathrm{H}_{2} &: 0 \, V \\ \mathrm{La}^{3+} / \mathrm{La} &: -2.52 \, V \\ \mathrm{Ca}^{2+} / \mathrm{Ca} &: -2.87 \, V \\ \mathrm{Cd}^{2+} / \mathrm{Cd} &: -0.402 \, V \end{aligned} \] (Note that we only search for half-reactions concerning the species mentioned in the exercise)
02

Identify the Strongest Oxidizing Agent

An oxidizing agent is a species that gains electrons and is reduced during the reaction. The stronger the oxidizing agent, the greater its tendency to gain electrons. Species with higher (more positive) reduction potentials have a greater tendency to be reduced. The highest reduction potential in the list above is for the reaction \(\mathrm{Br}^{-} / \mathrm{Br}_{2}\), with a value of +1.087 V. Therefore, the strongest oxidizing agent is \(\mathrm{Br}_{2}\). Answer (a): The strongest oxidizing agent is \(\mathrm{Br}_{2}\).
03

Identify the Strongest Reducing Agent

A reducing agent is a species that loses electrons and is oxidized during the reaction. The stronger the reducing agent, the greater its tendency to give up electrons. Species with lower (more negative) reduction potentials have a greater tendency to be oxidized. The lowest reduction potential in the list above is for the reaction \(\mathrm{Ca}^{2+} / \mathrm{Ca}\), with a value of -2.87 V. Therefore, the strongest reducing agent is \(\mathrm{Ca}\). Answer (b): The strongest reducing agent is \(\mathrm{Ca}\).
04

Identify the Species Oxidized by MnO4- in Acid

To determine which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid, we must first find the standard reduction potential for the reaction involving \(\mathrm{MnO}_{4}^{-}\) as the oxidizing agent: \(\mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 e^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\) : \(+1.51 \, V\) Now we are looking for any species (in our list) that can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acidic solution. In other words, we need to find species with a reduction potential lower than +1.51 V (because the oxidizing agent must have a higher reduction potential than the species being oxidized). Comparing the reduction potentials from Step 1 with +1.51 V, we can see that \(\mathrm{H}^{+}\) (0 V) and \(\mathrm{Cd}^{2+}\) (-0.402 V) can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid. Answer (c): \(\mathrm{H}^{+}\) and \(\mathrm{Cd}^{2+}\) can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid.
05

Identify the Species Reduced by Zn(s)

To determine which species can be reduced by \(\mathrm{Zn}(s)\), we first find the standard reduction potential for the reaction involving \(\mathrm{Zn}^{2+}\) as the reducing agent: \(\mathrm{Zn}^{2+} + 2 e^{-} \rightarrow \mathrm{Zn}\) : \(-0.76 \, V\) Now we are looking for any species (in our list) that can be reduced by \(\mathrm{Zn}(s)\). In other words, we need to find species with a reduction potential higher than -0.76 V (because the reducing agent must have a lower reduction potential than the species being reduced). Comparing the reduction potentials from Step 1 with -0.76 V, we can see that no species can be reduced by \(\mathrm{Zn}(s)\). Answer (d): No species given can be reduced by \(\mathrm{Zn}(s)\).

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Key Concepts

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

Oxidizing Agent
An oxidizing agent is a vital player in redox reactions, known for its ability to take on electrons from other substances. It essentially drives the oxidation process by undergoing reduction itself. The strength of an oxidizing agent is determined by its standard reduction potential; the more positive the potential, the stronger the agent's electron-grabbing ability.
In electrochemistry, the standard reduction potential table helps identify which substances make stronger oxidizing agents. For instance, in our exercise, \(\mathrm{Br}_{2}\) stands out with the most positive reduction potential of +1.087 V, making it the most potent oxidizer among the listed species. This means it is very efficient at gaining electrons, thereby facilitating the oxidation of other species.
Understanding oxidizing agents is crucial for predicting the behavior of chemical reactions, especially in electrochemical cells. They effectively "force" other substances to oxidize, driving the reactions forward.
Reducing Agent
Reducing agents play a critical role in redox reactions by donating electrons to other substances. This initiation of electron loss leads to the agent's own oxidation. The strength of a reducing agent is inversely related to its standard reduction potential; agents with more negative values are stronger because they more readily lose electrons.
In the given exercise, calcium (\(\mathrm{Ca}\)) emerges as the strongest reducing agent, with a standard reduction potential of -2.87 V. This low potential means calcium is highly eager to lose electrons, making it effective in reducing other substances by providing electrons.
Knowledge of reducing agents is essential when studying electrochemical processes such as battery operation and corrosion. They support reactions by bestowing electrons, thus, facilitating reduction in other species.
Electrochemistry
Electrochemistry is the branch of chemistry focused on the relationship between electrical energy and chemical change. It encompasses the study of electrochemical cells, where chemical energy is converted to electrical energy and vice-versa.
The principles of electrochemistry are governed by redox reactions, where oxidation and reduction occur simultaneously, transferring electrons between chemical species. This transfer can produce an electric current, which is harnessed in technologies like batteries and electrolysis.
Electrochemistry relies heavily on the concept of standard reduction potentials to predict how reactions will proceed within cells. These values help determine the feasibility of a cell reaction and the direction in which electrons will flow. The application of these concepts is widespread, spanning energy storage, metal refining, and even biological systems.
Redox Reactions
Redox reactions are chemical processes that involve the transfer of electrons between substances. They are characterized by their dual nature: one substance undergoes oxidation (loses electrons) and the other reduction (gains electrons).
Balancing redox reactions involves ensuring that the number of electrons lost equals those gained, maintaining charge neutrality in the equation. These reactions are pivotal in energy conversion processes, such as in electrochemical cells and metabolism.
The exercise highlights the importance of redox reactions in identifying which species can act as oxidizing agents or reducing agents. By looking at the standard reduction potentials, one can predict which substances are likely to be reduced or oxidized. This deep understanding is fundamental for manipulating chemical reactions purposely in industrial applications and natural processes.

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Most popular questions from this chapter

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}\)

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called cytochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the +3 to the +2 oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): $$\begin{aligned} &\text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.385 \mathrm{V}\\\ &\text { cytochrome } \mathbf{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.030 \mathrm{V}\\\ &\text { cytochrome } c\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.254 \mathrm{V} \end{aligned}$$ In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\)

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)
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