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.

Reducing agents are species that donate electrons in a redox reaction, causing the other species to be reduced (gain electrons). Conversely, oxidizing agents are species that accept electrons in a redox reaction, causing the other species to be oxidized (lose electrons). A good reducing agent has a strong tendency to lose electrons, while a good oxidizing agent has a strong tendency to gain electrons.

To determine the strength of the given species as reducing agents, we need to refer to Table 18.1 or other sources of standard reduction potentials. The more negative the standard reduction potential, the stronger the reducing agent. The more positive the standard reduction potential, the stronger the oxidizing agent.

From Table 18.1, we find the standard reduction potentials for the given species: F₂ + 2e⁻ → 2F⁻ Eº = +2.87 V H₂ + 2e⁻ → 2H⁻ Eº = 0 V (by definition) Na⁺ + e⁻ → Na Eº = -2.71 V For Na, we could use the reaction: Na → Na⁺ + e⁻. In this case, Eº = +2.71 V. However, to make it comparable to the other reactions, we need to reverse the equation to have the species as a reactant and not as a product: Na⁺ + e⁻ → Na with Eº = -2.71 V. Now, we can compare the standard reduction potentials.

The most negative standard reduction potential indicates the strongest reducing agent. Comparing the values for \(\mathrm{F}_{2}\), \(\mathrm{H}_{2}\) and \(\mathrm{Na}^{+}\), we find that Na+ has the most negative reduction potential, at -2.71 V. Therefore, \(\mathrm{Na}^{+}\) is the best reducing agent among the given species.

To order the species as oxidizing agents, we need to look at the reverse reactions where they act as reactants: 2F⁻ → F₂ + 2e⁻ Eº = -2.87 V 2H⁻ → H₂ + 2e⁻ Eº = 0 V Na → Na⁺ + e⁻ Eº = +2.71 V An oxidizing agent has a strong tendency to gain electrons, so we should look for the species with the most positive standard reduction potential. However, the species F⁻ and Na⁺, as anions and cations respectively, do not feature in the list of effective oxidizing agents since they do not typically accept electrons to undergo a reaction. Therefore, we can only order \(\mathrm{F}_{2}\) and \(\mathrm{H}_{2}\). Comparing their standard reduction potentials: \(\mathrm{F}_{2}\) is a better oxidizing agent than \(\mathrm{H}_{2}\) since its standard reduction potential (+2.87 V) is more positive than that of \(\mathrm{H}_{2}\) (0 V). So the order of oxidizing agents from best to worst would be \(\mathrm{F}_{2} > \mathrm{H}_{2}\).