Question

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

Short answer

The balanced disproportionation reactions are: (a) \( 2Fe^{2+}(aq) \longrightarrow Fe(s) + 2Fe^{3+}(aq) \). (b) \(3Br_{2}(l) + 12H^{+}(aq) \longrightarrow 6Br^{-}(aq) + 2BrO_3^{-}(aq) + 6H_{2}O(l)\). (c) \( 3Cr^{3+}(aq) + 14H^{+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + 2Cr(s) + 7H_{2}O(l)\). (d) \(6NO(g) + 6H_{2}O(l) \longrightarrow 2N_{2}(g) + 6NO_{3}^{-}(aq) + 12H^{+}(aq)\).

Step-by-step solution

Work on reaction (a)

\(Fe^{2+}(aq) \longrightarrow Fe(s) + Fe^{3+}(aq)\) First, identify the changes in oxidation numbers for the iron atoms: \( Fe^{2+}(aq)\) is reduced to \( Fe(s)\): Oxidation number decreases from +2 to 0. \( Fe^{2+}(aq)\) is oxidized to \( Fe^{3+}(aq)\): Oxidation number increases from +2 to +3. To balance the reaction, we have to make sure that the number of electrons transferred is the same during oxidation and reduction. The difference in electrons here is 1: Reduction: \( Fe^{2+}(aq) \longrightarrow Fe(s) + 1e^{-}\). Oxidation: \( Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + 1e^{-}\). Now we can write the balanced disproportionation reaction: \( 2Fe^{2+}(aq) \longrightarrow Fe(s) + 2Fe^{3+}(aq) \).

Work on reaction (b)

\(Br_{2}(l) \longrightarrow Br^{-}(aq) + BrO_3^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the bromine atoms: \( Br_{2}(l)\) is reduced to \( Br^{-}(aq)\): Oxidation number decreases from 0 to -1. \( Br_{2}(l)\) is oxidized to \( BrO_3^{-}(aq)\): Oxidation number increases from 0 to +5. To balance the reaction: Reduction: \( Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-} \). Oxidation: \( Br_{2}(l) + 10e^{-} \rightarrow 2BrO{_3}^{-}(aq) + 12H^{+}(aq) \) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 5: \(5 (Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-})\). Now, we can write the balanced disproportionation reaction in acidic solution: \(3Br_{2}(l) + 12H^{+}(aq) \longrightarrow 6Br^{-}(aq) + 2BrO_3^{-}(aq) + 6H_{2}O(l)\).

Work on reaction (c)

\(Cr^{3+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + Cr(s)\) (acidic solution) First, identify the changes in oxidation numbers for the chromium atoms: \( Cr^{3+}(aq)\) is reduced to \( Cr(s)\): Oxidation number decreases from +3 to 0. \( Cr^{3+}(aq)\) is oxidized to \( Cr_{2}O_{7}^{2-}(aq)\): Oxidation number increases from +3 to +6. To balance the reaction: Reduction: \(2Cr^{3+}(aq) + 6e^{-} \rightarrow 2Cr(s)\). Oxidation: \(2Cr^{3+}(aq) \rightarrow Cr_{2}O_{7}^{2-}(aq) + 6e^{-} + 14H^{+}(aq)\) (Considering the acidic solution). Now we can write the balanced disproportionation reaction in acidic solution: \( 3Cr^{3+}(aq) + 14H^{+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + 2Cr(s) + 7H_{2}O(l)\).

Work on reaction (d)

\(NO(g) \longrightarrow N_{2}(g) + NO_{3}^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the nitrogen atoms: \( NO(g)\) is reduced to \( N_{2}(g)\): Oxidation number decreases from +2 to 0. \( NO(g)\) is oxidized to \( NO_{3}^{-}(aq)\): Oxidation number increases from +2 to +5. To balance the reaction: Reduction: \(2NO(g) + 2e^{-} \rightarrow N_{2}(g)\). Oxidation: \(4NO(g) + 6H_2O(l) \rightarrow 4NO_{3}^{-}(aq) + 12H^{+}(aq) + 6e^{-}\) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 3: \(3 (2NO(g) + 2e^{-} \rightarrow N_{2}(g))\). Now, we can write the balanced disproportionation reaction in acidic solution: \(6NO(g) + 6H_{2}O(l) \longrightarrow 2N_{2}(g) + 6NO_{3}^{-}(aq) + 12H^{+}(aq)\).

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often referred to as redox reactions, are a fundamental type of chemical reaction where there is a transfer of electrons between two species. These reactions are characterized by changes in oxidation states. The substance that loses electrons is oxidized, while the substance that gains electrons is reduced.
In a disproportionation reaction, which is a special type of redox reaction, a single substance undergoes both oxidation and reduction simultaneously. For example, in reaction (a):

• The iron ion, Fe^{2+}$, is both oxidized to \( ext{Fe}^{3+} \) and reduced to \( ext{Fe}(s)\).

Understanding these changes requires recognizing how to track oxidation numbers and balance the resulting equation to reflect the conservation of mass and charge.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the reaction. The process respects the Law of Conservation of Mass, where mass is neither created nor destroyed during a chemical reaction.
When balancing a disproportionation reaction, one must balance both mass and charge. In reaction (b), \( ext{Br}_2 \) disproportionates into \( ext{Br}^- \) and \( ext{BrO}_3^- \), requiring careful adjustment of coefficients to match electron transfer between reactants and products:

• Balance the electrons transferred: Ensure the electrons lost during oxidation match those gained during reduction.
• Add necessary ions (such as \( ext{H}^+ \)) and water molecules to balance hydrogen and oxygen in the context of an acidic medium.

This comprehensive balancing makes sure no excess charge is left unaccounted for.

Oxidation Numbers

Oxidation numbers act as guidelines to understanding electron transfer in redox reactions. Each atom in a chemical compound is assigned an oxidation number that reflects its electron density relative to a pure element state. The key rules for determining oxidation numbers include:

• The oxidation number of a free element (not combined with other elements) is always 0.
• For ions, it is equal to the charge of the ion.
• In compounds, the sum of oxidation numbers for all atoms equals the compound’s overall charge.

For instance, in reaction (c), the chromium ion \( ext{Cr}^{3+}\) has an oxidation number of +3, which decreases to 0 when reduced to \( ext{Cr}(s)\) and increases to +6 when it forms \( ext{Cr}_2 ext{O}_7^{2-}\). These variations signify electron gain or loss and are essential for reaction balancing.

Half-Reactions in Acidic Solutions

Redox reactions in acidic solutions often involve breaking the overall reaction into half-reactions to simplify the balancing process. A half-reaction reveals either the oxidation or reduction part of the full reaction. Typically, one starts by identifying the species undergoing oxidation and reduction and then explicitly writing out the half-reactions:

• For the reduction half: Electrons are gained by the molecule or ion.
• For the oxidation half: Electrons are lost by the molecule or ion.

In reaction (d), the nitric oxide \( ext{NO}(g)\) reduces to \( ext{N}_2(g)\) and oxidizes to \( ext{NO}_3^-\), essentially performing both functions.
After writing each half-reaction, balance the atoms and charges by adding \( ext{H}^+ \) ions (to balance hydrogen) and \( ext{H}_2 ext{O} \) (to balance oxygen). Finally, adjust for equal electron flow between oxidation and reduction to maintain electroneutrality.