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Chapter 18: Problem 6

Which of the following is the best reducing agent: \(\mathrm{F}_{2}, \mathrm{H}_{2}, \mathrm{Na}\), \(\mathrm{Na}^{+}, \mathrm{F}^{-}\) ? Explain. Order as many of these species as possible from the best to the worst oxidizing agent. Why can't you order all of them? From Table \(18.1\) choose the species that is the best oxidizing agent. Choose the best reducing agent. Explain.

Short Answer

Expert verified
The best reducing agent among the given species is Na, as it has the lowest reduction potential (-2.71 V). The best oxidizing agent is F2, with the highest reduction potential (+2.87 V). We can only order F2 and Na+ as oxidizing agents due to the limitations of the standard reduction potential table, which does not provide values for H2, Na, and F-.

Step by step solution

01

Identify the relevant half-reactions

Given species are: F2, H2, Na, Na+, and F-. For each species, we will write the half-reaction involving one electron: 1. \(\mathrm{F}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{F}^{-}\) (the reduction of F2 to F-) 2. \(\mathrm{H}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{H}^{-}\) (the reduction of H2 to H-) 3. \(\mathrm{Na} + \mathrm{e}^{-} \rightarrow \mathrm{Na}^{-}\) (the reduction of Na to Na-) 4. \(\mathrm{Na}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Na}\) (the reduction of Na+ to Na) 5. \(\mathrm{F}^{-} + \mathrm{e}^{-} \rightarrow \mathrm{F}^{2-}\) (the reduction of F- to F2-)
02

Determine the standard reduction potentials

Using a standard reduction potential table, we can find the standard reduction potentials for the above half-reactions: 1. \(\mathrm{F}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{F}^{-}\): +2.87 V 2. \(\mathrm{H}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{H}^{-}\): not found in standard tables 3. \(\mathrm{Na} + \mathrm{e}^{-} \rightarrow \mathrm{Na}^{-}\): not found in standard tables 4. \(\mathrm{Na}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Na}\): -2.71 V 5. \(\mathrm{F}^{-} + \mathrm{e}^{-} \rightarrow \mathrm{F}^{2-}\): not found in standard tables
03

Choose the best reducing agent

To identify the best reducing agent, we need to look for the lowest reduction potential. In this case, we have: - \(\mathrm{F}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{F}^{-}\): +2.87 V - \(\mathrm{Na}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Na}\): -2.71 V The lowest reduction potential is that of \(\mathrm{Na}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Na}\) (-2.71 V). The best reducing agent is the one that causes the reduction to occur most readily: \(\boxed{\mathrm{Na}}\).
04

Order the species as oxidizing agents

To order the species as oxidizing agents, we need to look for their reduction potential values. 1. \(\mathrm{F}_{2}\) has the highest reduction potential and is the best oxidizing agent. 2. \(\mathrm{Na}^{+}\) has a lower reduction potential than F2 and is ranked second as an oxidizing agent. However, for H2, Na, and F-, we were unable to find the reduction potentials from the available standard tables. Therefore, it isn't possible to order them among these given species.
05

Explain the limitations in ordering

We were unable to order all of the species as oxidizing agents because the standard reduction potential table does not provide values for all the half-reactions. H2, Na, and F- were not found in the table, which prevents us from ordering them among the given species. In conclusion, the best reducing agent among the given species is \(\mathrm{Na}\), and the best oxidizing agent is \(\mathrm{F}_{2}\). However, we can't order all of them due to the limitations of the standard reduction potential table.

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Most popular questions from this chapter

An unknown metal \(\mathrm{M}\) is electrolyzed. It took \(74.1 \mathrm{~s}\) for a current of \(2.00 \mathrm{~A}\) to plate out \(0.107 \mathrm{~g}\) of the metal from a solution containing \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{3} .\) Identify the metal.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M\). Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)
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