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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 lead storage battery is charged at too high a voltage, gases are produced at each electrode. (It is possible to recharge a lead-storage battery only because of the high overpotential for gas formation on the electrodes.) (a) What are these gases? (b) Write a cell reaction to describe their formation.

When deciding whether a particular reaction corresponds to a cell with a positive standard cell potential, which of the following thermodynamic properties would you use to get your answer without performing any calculations? Which would you not use? Explain. (a) \(\Delta G^{\circ} ;\) (b) \(\Delta S^{\circ} ;\) (c) \(\Delta H^{\circ} ;\) (d) \(\Delta U^{\circ} ;\) (e) \(K\).

\(\mathrm{Ni}^{2+}\) has a more positive reduction potential than \(\mathrm{Cd}^{2+}\) (a) Which ion is more easily reduced to the metal? (b) Which metal, Ni or Cd, is more easily oxidized?

You must estimate \(E^{\circ}\) for the half-reaction \(\operatorname{In}^{3+}(\mathrm{aq})+\) \(3 \mathrm{e}^{-} \longrightarrow \operatorname{In}(\mathrm{s}) .\) You have no electrical equipment, but you do have all of the metals listed in Table 20.1 and aqueous solutions of their ions, as well as \(\operatorname{In}(\mathrm{s})\) and \(\operatorname{In}^{3+}(\text { aq })\). Describe the experiments you would perform and the accuracy you would expect in your result.

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.

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