Problem 57 - Chemistry and Chemical Reactivity

Question

The halogen oxides and oxoanions are good oxidizing agents. For example, the reduction of bromate ion has an \(E^{\circ}\) value of \(1.44 \mathrm{V}\) in acid solution: $$2 \mathrm{BrO}_{3}^{-}(\mathrm{aq})+12 \mathrm{H}^{+}(\mathrm{aq})+10 e^{-} \rightarrow{\mathrm{Br}_{2}(\mathrm{aq})}+6 \mathrm{H}_{2} \mathrm{O}(\ell)$$ Is it possible to oxidize aqueous \(1.0 \mathrm{M} \mathrm{Mn}^{2+}\) to aqueous \(\mathrm{MnO}_{4}^{-}\) with \(1.0 \mathrm{M}\) bromate ion?

Step-by-Step Solution

Answer

No, bromate ions cannot oxidize Mn^{2+} to MnO_{4}^{-} under standard conditions.

Step 1: Determine the Half-Reactions

First, identify the oxidation and reduction half-reactions involved. The given reduction reaction for bromate ion is:\[2 \mathrm{BrO}_{3}^{-} + 12 \mathrm{H}^{+} + 10 e^{-} \rightarrow \mathrm{Br}_{2} + 6 \mathrm{H}_{2} \mathrm{O}\]The oxidation of \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\) can be represented as:\[\mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} + 5 e^{-}\]

Step 2: Find the Standard Electrode Potentials

Using a standard table for electrode potentials, look for the values:- The standard reduction potential \(E^{\circ}\) for the reduction of \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\) is given as \(1.51\, \mathrm{V}\).- The given potential for the reduction of bromate ion is \(1.44\, \mathrm{V}\).

Step 3: Compare the Potentials

Determine if the oxidation of \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\) is feasible using bromate ions by comparing the standard electrode potentials:If the \(E^{\circ}\) for \(\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}\) which is \(1.51\, \mathrm{V}\) is higher than \(E^{\circ}\) for \(\mathrm{BrO}_{3}^{-}/\mathrm{Br}_{2}\), which is \(1.44\, \mathrm{V}\), then it is not feasible since the conversion requires a more positive reduction potential than the available bromate ion potential.

Step 4: Conclusion

Since the \(E^{\circ}\) for the \(\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}\) pair is higher than the \(\mathrm{BrO}_{3}^{-}/\mathrm{Br}_{2}\) pair, bromate ions cannot oxidize \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\). The reaction will not occur under standard conditions.

Key Concepts

Standard Electrode Potentials

Standard electrode potentials are crucial for understanding how oxidation-reduction reactions work. They are represented by the symbol \(E^{\circ}\) and measured in volts (V). Electrode potentials reflect a substance's tendency to gain or lose electrons compared to a standard hydrogen electrode, which is set at 0 V. A more positive \(E^{\circ}\) value indicates a stronger tendency to undergo reduction, whereas a more negative value suggests a stronger tendency for oxidation.

To compare whether one substance can oxidize another, we use these potentials. In the exercise, we have the bromate ion with a standard potential of 1.44 V and the permanganate ion with 1.51 V. To determine if an oxidation reaction can proceed, the standard electrode potential of the oxidizing agent (bromate ion) must be higher than the potential of the species it is attempting to oxidize (\(\mathrm{Mn}^{2+}\)). Since 1.44 V is less than 1.51 V, bromate cannot oxidize \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\), as it lacks the necessary energy.

Understanding standard electrode potentials helps in predicting the feasibility of chemical reactions and is key in applications like electrochemistry and battery technology.

Oxidizing Agents

Oxidizing agents are substances that gain electrons during a chemical reaction, causing another substance to be oxidized. They are reduced in the process. In the context of the original exercise, the bromate ion \(\mathrm{BrO}_{3}^{-}\) is considered the oxidizing agent because it gains electrons and is reduced to bromine \(\mathrm{Br}_{2}\).

Oxidizing agents are pivotal in many chemical reactions, including industrial processes like bleach production and wastewater treatment. They are also key components in biological systems, playing crucial roles in cellular respiration and photosynthesis.

Analyzing an oxidizing agent's effectiveness involves comparing standard electrode potentials. A strong oxidizing agent typically has a higher (more positive) standard electrode potential, signaling a robust ability to capture electrons from other reactants. This is why standard potentials are instrumental in predicting reaction direction and feasibility.

Understanding oxidizing agents and balancing their half-reactions enable chemists and scientists to harness these reactions in controlled settings for various applications.

Half-Reactions

Half-reactions allow you to break down a redox reaction into two parts: oxidation half and reduction half. Each half-reaction represents the transfer of electrons. This breakdown is essential for balancing redox reactions and understanding electron flow.

In the original exercise, the reduction half-reaction involves the bromate ion \(\mathrm{BrO}_{3}^{-}\), which gains electrons and is converted into bromine \(\mathrm{Br}_{2}\). The oxidation half-reaction involves \(\mathrm{Mn}^{2+}\) being converted into \(\mathrm{MnO}_{4}^{-}\) by losing electrons.

Balancing these half-reactions involves making sure the number of atoms and charges are equal on both sides. Traditionally, you balance the atoms, then the charges using electrons, water, and hydrogen ions in acidic solutions.

Understanding half-reactions is important because it helps illuminate the underlying process of electron transfer, making complex redox reactions more manageable. Whether in academia or industry, this knowledge aids in the design of electrochemical cells, corrosion prevention, and even energy storage systems like batteries.