/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Can the displacement of \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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Can the displacement of \(\mathrm{Pb}(\mathrm{s})\) from \(1.0 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) be carried to completion by tin metal? Explain.

Short Answer

Expert verified
No, the displacement of Lead from Lead(II) nitrate cannot be carried to completion by tin metal because the standard reduction potential for Lead is less than for Tin.

Step by step solution

01

Determining the standard reduction potentials

Consult a table of standard reduction potentials. We find that \(Pb^{2+}(aq) + 2e^- \rightarrow Pb(s)\) has a standard reduction potential (\(E^0\)) of -0.13 V, while \(Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)\) has a standard reduction potential of -0.14 V.
02

Comparing the reduction potentials

Compare the standard reduction potentials. The reaction will proceed to the right (displacement will occur) if the oxidizing agent (in this case Lead) has a higher standard reduction potential than the reducing agent (Tin). However, the standard reduction potential for Lead is smaller than that for Tin.
03

Concluding

Since the standard reduction potential for Lead is less than for Tin, the reaction will not go to completion, meaning that the displacement of Lead by Tin will not be completed.

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Key Concepts

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

Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. It encompasses the study of chemical changes caused by the movement of electrons from one molecule or ion to another—an essential process in batteries, electrolysis, and various types of sensors and electrodes. At its core, electrochemistry looks at how electric current can drive a chemical reaction which might not occur spontaneously as well as how chemical reactions can generate electric current.
Redox Reactions
Redox reactions, short for 'reduction-oxidation reactions', are a family of chemical reactions involving the transfer of electrons. One substance gets reduced by gaining electrons, while another gets oxidized by losing electrons. This electron dance is crucial in electrochemical processes, including the operation of batteries and corrosion. The understanding of redox reactions is crucial when we analyze displacement reactions and their feasibilities, such as in the given exercise involving lead and tin.
Chemical Displacement
Chemical displacement, or single replacement, occurs when an element displaces another in a compound, typically as a result of differences in reactivity. Metals can replace less reactive metals from their compounds. For instance, in the exercise question, tin (Sn) metal is considered for the displacement of lead (Pb) from lead nitrate solution. Whether such a displacement reaction will successfully occur can be predicted by looking at the reactivity of metals, which is generally informed by standard reduction potential values.
Standard Electrode Potentials
Standard Electrode Potentials, represented by the symbol \(E^0\), measure the tendency of a chemical species to be reduced, and by extension, the strength of its oxidizing ability. It's tabulated under standard conditions: a 1 M concentration, 25°C temperature, and a pressure of 1 atm. When comparing two standard reduction potentials, the substance with the higher (more positive) potential will act as the oxidizing agent and tends to gain electrons, being reduced in the process. During a displacement reaction, a metal with a less negative \(E^0\) can be displaced by a metal with a more negative \(E^0\). However, in the exercise, since Pb has a less negative standard reduction potential than Sn, the displacement will not be completed as one might initially think.

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Most popular questions from this chapter

If a chemical reaction is carried out in a fuel cell, the maximum amount of useful work that can be obtained is (a) \(\Delta G ;\) (b) \(\Delta H ;\) (c) \(\Delta G / \Delta H ;\) (d) \(T \Delta S\).

In your own words, define the following symbols or terms: (a) \(E^{\circ} ;\) (b) \(F ;\) (c) anode; (d) cathode.

Consider the reaction \(\operatorname{Co}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{Ni}(\mathrm{s}), \quad\) with \(\quad E_{\mathrm{cell}}^{\circ}=0.02 \mathrm{V} . \quad\) If \(\quad \mathrm{Co}(\mathrm{s}) \quad\) is added to a solution with \(\left[\mathrm{Ni}^{2+}\right]=1 \mathrm{M},\) should the reaction go to completion? Explain.

Derive a balanced equation for the reaction occurring in the cell: $$\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(\mathrm{aq}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$$ (a) If \(E_{\text {cell }}^{\circ}=1.21 \mathrm{V},\) calculate \(\Delta G^{\circ}\) and the equilibrium constant for the reaction. (b) Use the Nernst equation to determine the potential for the cell: $$\begin{array}{r} \mathrm{Fe}(\mathrm{s}) | \mathrm{Fe}^{2+}\left(\mathrm{aq}, 1.0 \times 10^{-3} \mathrm{M}\right) \| \mathrm{Fe}^{3+}\left(\mathrm{aq}, 1.0 \times 10^{-3} \mathrm{M}\right) \\ \mathrm{Fe}^{2+}(\mathrm{aq}, 0.10 \mathrm{M}) | \mathrm{Pt}(\mathrm{s}) \end{array}$$ (c) In light of (a) and (b), what is the likelihood of being able to observe the disproportionation of \(\mathrm{Fe}^{2+}\) into \(\mathrm{Fe}^{3+}\) and Fe under standard conditions?

The theoretical \(E_{\text {cell }}^{\circ}\) for the methane-oxygen fuel cell is \(1.06 \mathrm{V} .\) What is \(E^{\circ}\) for the reduction half-reaction \(\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) ?\)

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