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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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Key Concepts

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

Standard Reduction Potential
Standard reduction potential is a crucial concept in understanding redox reactions in electrochemistry. It represents the tendency of a chemical species to be reduced, measured in volts. The standard reduction potential is determined under standard conditions, which include 25掳C, 1 atm pressure, and 1 M concentration of all reacting species. This value can help predict the direction of a redox reaction, showing which species will be reduced and which will be oxidized.
A larger positive value indicates a higher tendency to gain electrons (reduction), making that species a strong oxidizing agent. Conversely, a more negative value signifies a weaker inclination to gain electrons, meaning that species will likely act as a reducing agent. In the exercise, for example,
  • Fluorine ( \( \text{F}_{2} \) ) has a very positive standard reduction potential (+2.87 V), making it a strong oxidizing agent.
  • Conversely, sodium ions ( \( \text{Na}^{+} \) ) have a lower reduction potential (-2.71 V), indicating their role as a reducing agent.
Understanding these values helps in predicting the spontaneity and direction of chemical reactions.
Oxidizing Agents
Oxidizing agents are substances that accept electrons during a chemical reaction. By gaining electrons, they reduce themselves while oxidizing another species. Their strength as oxidizing agents is determined by their standard reduction potentials; the higher the potential, the stronger the oxidizing agent.
In our exercise,
  • Fluorine ( \( \text{F}_{2} \) ) with its standard reduction potential of +2.87 V is the most powerful oxidizing agent among the given species. It easily accepts electrons to form fluoride ions ( \( \text{F}^{-} \) ).
  • Sodium ions ( \( \text{Na}^{+} \) ), although primarily a reducing agent, also serves as an oxidizing agent but is less effective than fluorine.
Some species cannot be ranked due to absent reduction potentials, showcasing the limits of available data.
Half-reactions in Electrochemistry
Half-reactions are a fundamental part of understanding redox reactions in electrochemistry. They split the overall reaction into two parts: oxidation and reduction. Each half-reaction shows the transfer of electrons and can be used to balance redox equations.
When analyzing reducing and oxidizing agents, these half-reactions help determine how electrons are gained or lost. In the exercise, half-reactions illustrate:
  • Fluorine being reduced by gaining electrons ( \( \text{F}_{2} + 2\text{e}^{-} \rightarrow 2\text{F}^{-} \) ). This half-reaction emphasizes fluorine's role as an oxidizing agent due to its acceptance of electrons.
  • Sodium ions being reduced to sodium ( \( \text{Na}^{+} + \text{e}^{-} \rightarrow \text{Na} \) ). This shows sodium's tendency to donate electrons, performing as a reducing agent.
These half-reactions are balanced using electrons and are essential for predicting the feasibility and direction of electrochemical processes.

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Most popular questions from this chapter

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ 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 \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22,1987 ) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q)\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of \(\mathrm{HCl}\) ? How can you harness this reaction to do useful work?
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