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Predict whether, to any significant extent, (a) \(\mathrm{Fe}(\mathrm{s})\) will displace \(\mathrm{Zn}^{2+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution; (c) \(\mathrm{Ag}(\mathrm{s})\) will react with \(1 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (d) \(\mathrm{O}_{2}(\mathrm{g})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution.

Short Answer

Expert verified
(a) No (b) Yes (c) No (d) No

Step by step solution

01

Interpret the Given Problem

In this task, four scenarios are presented where it needs to be determined if a reaction will occur, based on the properties of the substances involved, particularly their position in the electrochemical series.
02

Predict Reaction for Scenario (a)\( \mathrm{Fe}(\mathrm{s}) \) and \( \mathrm{Zn}^{2+}(\mathrm{aq}) \)

Refer to the reduction potential table. Iron (Fe) has a standard reduction potential of -0.44 V and Zinc (Zn) has a reduction potential of -0.76 V. In a redox reaction, the substance with the higher reduction potential will be reduced. Since Zinc has a lower reduction potential, it will not be displaced by Iron. Therefore, the reaction will not occur to a significant extent.
03

Predict Reaction for Scenario (b)\( \mathrm{MnO}_{4}^{-}(\mathrm{aq}) \) and \( \mathrm{Cl}^{-}(\mathrm{aq}) \) to \( \mathrm{Cl}_{2}(\mathrm{g}) \) in acidic solution

In this scenario, we have \( \mathrm{MnO}_{4}^{-} \) and \( \mathrm{Cl}^{-} \). Under acidic conditions, \( \mathrm{MnO}_{4}^{-} \) is a strong oxidizing agent with a reduction potential of +1.51 V while the reduction of \( \mathrm{Cl}_{2} \) to \( \mathrm{Cl}^{-} \) has a potential of +1.36 V. Given these potentials, \( \mathrm{MnO}_{4}^{-} \) can oxidize \( \mathrm{Cl}^{-} \) to \( \mathrm{Cl}_{2} \). Therefore, the reaction will occur.
04

Predict Reaction for Scenario (c)\( \mathrm{Ag}(\mathrm{s}) \) and \( \mathrm{HCl}(\mathrm{aq}) \)

Silver (Ag) has a reduction potential of +0.80 V. Hydrogen (H) in \( \mathrm{HCl} \) has a reduction potential of 0 V. The substance with the higher reduction potential is reduced. Therefore, Ag will not react with \( \mathrm{HCl} \). Thus, the reaction will not proceed to a significant extent.
05

Predict Reaction for Scenario (d)\( \mathrm{O}_{2}(\mathrm{g}) \) and \( \mathrm{Cl}^{-}(\mathrm{aq}) \) to \( \mathrm{Cl}_{2} \)

In this scenario, oxygen \( \mathrm{O}_{2} \) is expected to oxidize \( \mathrm{Cl}^{-} \) to \( \mathrm{Cl}_{2} \). Oxygen in acidic solution has a reduction potential of +1.23 V and the reduction of \( \mathrm{Cl}_{2} \) to \( \mathrm{Cl}^{-} \) has a reduction potential of +1.36 V. The substance with the higher reduction potential will be reduced, hence \( \mathrm{O}_{2} \) cannot oxidize \( \mathrm{Cl}^{-} \) to \( \mathrm{Cl}_{2} \). Therefore, this reaction will not occur.

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Key Concepts

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

Electrochemical Series
The electrochemical series, also known as the activity series, is a list that ranks elements in order of their standard electrode potentials. This list is pivotal in determining which element will act as an oxidizing or reducing agent in a redox reaction.

Elements at the top of the series have higher tendencies to gain electrons and are thus good oxidizing agents. Conversely, elements at the bottom are more likely to lose electrons and serve as reducing agents. The series allows us to predict the spontaneity of redox reactions: a substance can oxidize any other substance below it in the series.
Standard Reduction Potential
Standard reduction potential is a measurement that indicates how likely a chemical species will gain electrons and be reduced. Expressed in volts, it compares the tendency of a substance to reduce, relative to the standard hydrogen electrode which is arbitrarily given a potential of 0 volts.

When comparing two substances, the one with the higher reduction potential will be reduced, and the one with the lower potential will be oxidized. For instance, in the provided scenarios, Zinc with a standard reduction potential of -0.76 V is less likely to be reduced than Iron which has a potential of -0.44 V, leading to the conclusion that Iron will not replace Zinc to a significant extent.
Oxidation and Reduction Processes
Oxidation and reduction processes are at the heart of redox reactions, which involve the transfer of electrons between substances. Oxidation is the process whereby an element loses electrons, while reduction is the gain of electrons. These processes always occur simultaneously - if one substance is oxidized, another must be reduced.

Identifying the oxidizing and reducing agents is key to understanding redox reactions. An oxidizing agent gains electrons and is reduced, whereas a reducing agent loses electrons and is oxidized. An element’s position in the electrochemical series and its standard reduction potential are used to determine its role in a redox reaction, as illustrated in the exercise solutions.
Predicting Chemical Reactions
Predicting whether a chemical reaction will occur involves a clear understanding of electrochemical series and standard reduction potentials. By looking at these values, one can determine if a given substance can act as a sufficient oxidizing or reducing agent in a reaction.

In practical scenarios such as the exercise provided, comparing reduction potentials helps predict the feasibility of reactions. For instance, since Silver (Ag) has a higher reduction potential than Hydrogen (H), Silver will not react with Hydrochloric acid (HCl) to any significant degree. Similarly, predicting whether Oxygen (O2) will oxidize Chloride (Cl-) ions depends on their respective potentials, leading to the conclusion that this reaction will not happen under acidic conditions.

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Most popular questions from this chapter

For the reaction \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq}), E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) (a) can a solution be prepared at \(298 \mathrm{K}\) that is \(0.500 \mathrm{M}\) in each of the four ions? (b) If not, in which direction will a reaction occur?

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

In each of the following examples, sketch a voltaic cell that uses the given reaction. Label the anode and cathode; indicate the direction of electron flow; write a balanced equation for the cell reaction; and calculate \(E_{\mathrm{cell}}^{\circ}\). (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Pb}^{2+}(\mathrm{aq})\) is displaced from solution by \(\mathrm{Al}(\mathrm{s})\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\) \(\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g})\)

Your task is to determine \(E^{\circ}\) for the reduction of \(\mathrm{CO}_{2}(\mathrm{g})\) to \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) in two different ways and to explain why each gives the same result. (a) Consider a fuel cell in which the cell reaction corresponds to the complete combustion of propane gas. Write the half-cell reactions and the overall reaction. Determine \(\Delta G^{\circ}\) and \(E_{\text {cell }}^{\circ}\) for the reaction, then obtain \(E_{\mathrm{CO}_{2} / \mathrm{C}_{3} \mathrm{H}_{8}^{*}}^{\circ}\) (b) Without considering the oxidation that occurs simultaneously, obtain \(E_{\mathrm{CO}_{2} / \mathrm{C}_{3} \mathrm{H}_{8}}^{\circ}\) directly from tabulated thermodynamic data for the reduction half-reaction.

Dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) in acidic solution is a good oxidizing agent. Which of the following oxidations can be accomplished with dichromate ion in acidic solution? Explain. (a) \(\operatorname{sn}^{2+}\left(\text { aq) to } \operatorname{Sn}^{4+}(\text { aq })\right.\) (b) \(\mathrm{I}_{2}(\mathrm{s})\) to \(\mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Mn}^{2+}(\mathrm{aq})\) to \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\)

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