Question

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

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

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-)

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

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: \(\boxed{\mathrm{Na}}\).

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.

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.

Key concepts

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°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.