Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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
Among the given species, sodium \((\mathrm{Na})\) is the best reducing agent, and fluorine \((\mathrm{F}_{2})\) is the best oxidizing agent. This is because \(\mathrm{Na}\) has a higher tendency to lose electrons and get oxidized, while \(\mathrm{F}_{2}\) has a greater ability to gain electrons and get reduced due to its electronegativity. From Table 18.1, the strongest reducing agent is lithium \((\mathrm{Li})\), and the strongest oxidizing agent remains to be fluorine \((\mathrm{F}_{2})\).

Step by step solution

01

Identify the given species

The given species are: \(\mathrm{F}_{2}, \mathrm{H}_{2}, \mathrm{Na}, \mathrm{Na}^{+}, \mathrm{F}^{-}\).
02

Determine the best reducing agent

To determine the best reducing agent, we need to consider which species has the greatest tendency to lose electrons and get oxidized. Since metallic elements, such as sodium \((\mathrm{Na})\), are more likely to lose electrons and form cations, \(\mathrm{Na}\) can be considered the best reducing agent among the given species.
03

Order as oxidizing agents

Now, we need to order the given species based on their oxidizing ability. The order will be as follows: 1. \(\mathrm{F}_{2}\) 2. \(\mathrm{H}_{2}\) The \(\mathrm{F}_{2}\) is the strongest oxidizing agent among the given species, as it is the most electronegative element and thus has the highest tendency to gain electrons. The \(\mathrm{H}_{2}\) is second in the list of oxidizing agents, as it can gain electrons to form \(\mathrm{H}^-\) ions. We cannot order the other species because \(\mathrm{Na}\) is not an oxidizing agent and will not gain electrons, and the remaining ions \(\mathrm{Na}^{+}\) and \(\mathrm{F}^-\) are already in their oxidized and reduced forms, respectively.
04

Choose the best oxidizing agent and reducing agent from Table 18.1

From Table 18.1, we can determine the strongest reducing and oxidizing agents: Best oxidizing agent: \(\mathrm{F}_2\) Best reducing agent: \(\mathrm{Li}\) (lithium is at the top of the reactivity series) In summary, among the given species, \(\mathrm{Na}\) is the best reducing agent, while \(\mathrm{F}_{2}\) is the best oxidizing agent. From Table 18.1, the strongest reducing agent is \(\mathrm{Li}\), and the strongest oxidizing agent is still \(\mathrm{F}_{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium metal. c. Calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accident at Three Mile Island, Pennsylvania, in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{~kg} \mathrm{Zr}\) reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at \(1.0 \mathrm{~atm}\) and \(1000 .{ }^{\circ} \mathrm{C}\) is produced? e. At Chernobyl, USSR, in 1986 , hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ A chemical explosion involving the hydrogen gas did occur at Chernobyl. In light of this fact, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

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 \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) \(\mathrm{mol} \mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \operatorname{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(+0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). 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 \(25^{\circ} \mathrm{C}\).

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. Ag j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) 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 \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{KF}\) b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free