Question

Which species in each pair is the stronger oxidizing agent? (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\) (c) \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\) (d) \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution

Short answer

a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\) b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\) c) \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\) d) \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution Answer: a) \(\mathrm{NO}_{3}^{-}\) is the stronger oxidizing agent. b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) is the stronger oxidizing agent. c) \(\mathrm{MnO}_{2}\) is the stronger oxidizing agent. d) \(\mathrm{ClO}_{3}^{-}\) has the same oxidizing strength in both acidic and basic solution.

Step-by-step solution

Pair (a) - \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{I}_{2}\)

First, let's write the half-reactions for each species: \(\mathrm{NO}_{3}^{-} + 2\mathrm{H}^{+} + 1\mathrm{e}^{-} \rightarrow \mathrm{NO}_{2} + \mathrm{H}_{2}\mathrm{O}\) (Reduction of \(\mathrm{NO}_{3}^{-}\)) \(\mathrm{I}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{I}^{-}\) (Reduction of \(\mathrm{I}_{2}\)) Now we will look up the standard reduction potentials for each reaction: \(E^{\circ}_{\mathrm{NO}_{3}^{-}} = 0.78\) V \(E^{\circ}_{\mathrm{I}_{2}} = 0.54\) V Since the reduction potential of \(\mathrm{NO}_{3}^{-}\) is greater than that of \(\mathrm{I}_{2}\), \(\mathrm{NO}_{3}^{-}\) is the stronger oxidizing agent.

Pair (b) - \(\mathrm{Fe}(\mathrm{OH})_{3}\) or \(\mathrm{S}\)

First, let's write the half-reactions for each species: \(\mathrm{Fe}(\mathrm{OH})_{3} + 3\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{3+} + 3\mathrm{H}_{2}\mathrm{O}\) (Reduction of \(\mathrm{Fe(OH)}_{3}\)) \(\mathrm{S} + 2\mathrm{e}^{-} \rightarrow \mathrm{S}^{2-}\) (Reduction of \(\mathrm{S}\)) Now we will look up the standard reduction potentials for each reaction: \(E^{\circ}_{\mathrm{Fe}(\mathrm{OH})_{3}} = 0.88\) V \(E^{\circ}_{\mathrm{S}} = -0.14\) V As the reduction potential of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is greater than that of \(\mathrm{S}\), \(\mathrm{Fe}(\mathrm{OH})_{3}\) is the stronger oxidizing agent.

Pair (c) - \(\mathrm{Mn}^{2+}\) or \(\mathrm{MnO}_{2}\)

First, let's write the half-reactions for each species: \(\mathrm{Mn}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Mn}\) (Reduction of \(\mathrm{Mn}^{2+}\)) \(\mathrm{MnO}_{2} + 4\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{H}_{2}\mathrm{O}\) (Reduction of \(\mathrm{MnO}_{2}\)) Now we will look up the standard reduction potentials for each reaction: \(E^{\circ}_{\mathrm{Mn}^{2+}} = -1.18\) V \(E^{\circ}_{\mathrm{MnO}_{2}} = 1.23\) V Since the reduction potential of \(\mathrm{MnO}_{2}\) is greater than that of \(\mathrm{Mn}^{2+}\), \(\mathrm{MnO}_{2}\) is the stronger oxidizing agent.

Pair (d) - \(\mathrm{ClO}_{3}^{-}\) in acidic solution or \(\mathrm{ClO}_{3}^{-}\) in basic solution

First, let's write the half-reactions for each species under acidic and basic conditions: Acidic solution: \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_{2}\mathrm{O}\) (Reduction of \(\mathrm{ClO}_{3}^{-}\) in acidic solution) \(E^{\circ}_{\mathrm{ClO}_{3}^{-}\ \text{acidic}} = 1.47\) V Basic solution: The reduction of \(\mathrm{ClO}_{3}^{-}\) in basic solution follows the same process as in acidic solution. Therefore, \(E^{\circ}_{\mathrm{ClO}_{3}^{-}\ \text{basic}} = 1.47\) V Since there is no difference in the reduction potential of \(\mathrm{ClO}_{3}^{-}\) in acidic or basic solution, both share the same oxidizing strength. In summary, the stronger oxidizing agents in each pair are: (a) \(\mathrm{NO}_{3}^{-}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{ClO}_{3}^{-}\) in both acidic and basic solution

Key concepts

Standard Reduction Potential

Understanding the standard reduction potential is crucial in determining the tendency of a chemical species to gain electrons and be reduced. It is measured in volts (V) and represented by the symbol E°. Each half-reaction has an associated standard reduction potential, which can be found in a table of standard electrode potentials. These values are determined under standard conditions, which typically means a concentration of 1 M for each ion, a pressure of 1 atmosphere for any gases involved, and a temperature of 25°C (298 K).

Higher standard reduction potential indicates a greater tendency to gain electrons and thus, a stronger ability to act as an oxidizing agent. In the context of the exercise, comparing the E° values reveals which species is the stronger oxidizer. For instance, since \(E^{\textbackslash{}\textbackslash{}circ}_{\textbackslash{}\textbackslash{}mathrm{NO}_{3}^{-}} = 0.78\textbackslash{}\textbackslash{}text{V}\) is higher than \(E^{\textbackslash{}\textbackslash{}circ}_{\textbackslash{}\textbackslash{}mathrm{I}_{2}} = 0.54\textbackslash{}\textbackslash{}text{V}\), nitrate (\(\textbackslash{}\textbackslash{}mathrm{NO}_{3}^{-}\)) is the stronger oxidizing agent compared to iodine (\(\textbackslash{}\textbackslash{}mathrm{I}_{2}\)).

Redox Reactions

Redox reactions are a family of chemical reactions where reduction and oxidation simultaneously occur. The term 'redox' is a portmanteau for reduction-oxidation. Reduction describes the gain of electrons by a molecule, atom, or ion, while oxidation refers to the loss of electrons. One species takes electrons from the other, which means there is a transfer of electrons between the two reacting species.

In the exercise, identifying the stronger oxidizing agent involves recognizing which species is more likely to gain electrons, thereby causing the other to lose electrons. The concept reinforces that in a redox reaction, one reactant is always oxidized (loses electrons) and the other is reduced (gains electrons). The species with higher standard reduction potential will be reduced and is thus the better oxidizing agent.

Half-Reactions

Half-reactions are equations that show either the oxidation or reduction component of a redox reaction. They are essential for understanding the electrons' pathway during the reaction, and they simplify balancing complex redox equations. Each half-reaction has its own standard reduction potential, and when paired, one half-reaction will occur in the forward direction (as a reduction), and the other will occur in the reverse direction (as an oxidation).

In solving the given exercise, writing half-reactions helps in isolating the reduction process for each species and facilitates the determination of their standard reduction potentials. For example, the half-reaction \( \textbackslash{}\textbackslash{}mathrm{NO}_{3}^{-} + 2\textbackslash{}\textbackslash{}mathrm{H}^{+} + 1\textbackslash{}\textbackslash{}mathrm{e}^{-} \textbackslash{}\textbackslash{}rightarrow \textbackslash{}\textbackslash{}mathrm{NO}_{2} + \textbackslash{}\textbackslash{}mathrm{H}_{2}\textbackslash{}\textbackslash{}mathrm{O} \) allows us to pinpoint the reduction of nitrate, making it clear why \( \textbackslash{}\textbackslash{}mathrm{NO}_{3}^{-} \) is the stronger oxidizing agent when compared to iodine.

Electrochemical Series

The electrochemical series, also known as the activity or reactivity series, lists elements in order of their standard reduction potentials. It is a practical tool in predicting the outcome of redox reactions and identifying strong oxidizing and reducing agents. Elements at the top of the series possess higher reduction potentials and are more likely to be reduced; consequently, they are strong oxidizing agents. Conversely, elements at the bottom have lower reduction potentials, indicating a propensity to lose electrons and thus act as reducing agents.

Using the electrochemical series, students can quickly determine which species in the exercise will act as the stronger oxidizing agent. For instance, comparing \( \textbackslash{}\textbackslash{}mathrm{Mn}^{2+} \) and \( \textbackslash{}\textbackslash{}mathrm{MnO}_{2} \) reveals that \( \textbackslash{}\textbackslash{}mathrm{MnO}_{2} \) has a much higher reduction potential and as such, stands higher in the electrochemical series, making it the stronger oxidizing agent.