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: $$ \begin{array}{l}{\text { (a) } \mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)(\text { acidic solution })} \\ {\text { (b) } \operatorname{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s) \text { (acidic }} \\ \quad {\text { solution) }} \\\ {\text { (c) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)(\text { acidic solution })} \\ {\text { (d) } \mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q) \text { (basic solution) }}\end{array} $$

Short answer

The balanced disproportionation reactions are: (a) \(2Ni^+ \rightarrow Ni^{2+} + Ni\) (b) \(4MnO_4^{2-} + 4H^+ \rightarrow 3MnO_4^- + MnO_2 + 2H_2O\) (c) \(3H_2SO_3 + 2H^+ \rightarrow HSO_4^- + S + 4H_2O\) (d) \(2Cl_2 + 4OH^- \rightarrow 2Cl^- + ClO^- + 2H_2O\)

Step-by-step solution

(a) Balancing Ni+ disproportionation

: First, we'll split the reaction into two half-reactions: 1. Oxidation half-reaction: \(Ni^+ \rightarrow Ni^{2+}\) 2. Reduction half-reaction: \(Ni^+ \rightarrow Ni\) Next, balance the atoms and charges: 1. For the oxidation half-reaction: - Add one electron to the right side: \(Ni^+ \rightarrow Ni^{2+} + e^-\) 2. For the reduction half-reaction: - No balancing is required as both sides already have one Ni atom and same charge. Now, combine both half-reactions to yield the balanced overall reaction: \(Ni^+ + e^- + Ni^+ \rightarrow Ni^{2+} + Ni\)

(b) Balancing \(MnO_4^{2-}\) disproportionation

: First, we'll split the reaction into two half-reactions: 1. Oxidation half-reaction: \(MnO_4^{2-} \rightarrow MnO_4^-\) 2. Reduction half-reaction: \(MnO_4^{2-} \rightarrow MnO_2\) Next, balance the atoms and charges: 1. For the oxidation half-reaction: - Add one electron to the left side: \(MnO_4^{2-} + e^- \rightarrow MnO_4^-\) 2. For the reduction half-reaction: - Add 2 water molecules to the right side: \(MnO_4^{2-} \rightarrow MnO_2 + 2H_2O\) - Add 4 hydrogen ions to the left side: \(MnO_4^{2-} + 4H^+ \rightarrow MnO_2 + 2H_2O\) - Add 3 electrons to the left side: \(MnO_4^{2-} + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O\) Now, combine and balance both half-reactions: \(3(MnO_4^{2-} + e^-) + MnO_4^{2-} + 4H^+ + 3e^- \rightarrow 3MnO_4^- + MnO_2 + 2H_2O\) So, the balanced reaction is: \(4MnO_4^{2-} + 4H^+ \rightarrow 3MnO_4^- + MnO_2 + 2H_2O\)

(c) Balancing \(H_2SO_3\) disproportionation

: First, we'll split the reaction into two half-reactions: 1. Oxidation half-reaction: \(H_2SO_3 \rightarrow S\) 2. Reduction half-reaction: \(H_2SO_3 \rightarrow HSO_4^-\) Next, balance the atoms and charges: 1. For the oxidation half-reaction: - Add 4 hydrogen ions to the left side: \(4H^+ + H_2SO_3 \rightarrow S\) - Add 4 water molecules to the right side: \(4H^+ + H_2SO_3 \rightarrow S + 4H_2O\) - Add 6 electrons to the right side: \(4H^+ + H_2SO_3 + 6e^- \rightarrow S + 4H_2O\) 2. For the reduction half-reaction: - Add a water molecule to the left side: \(H_2SO_3 + H_2O \rightarrow HSO_4^-\) - Add 2 hydrogen ions to the right side: \(H_2SO_3 + H_2O \rightarrow HSO_4^- + 2H^+\) - Add 2 electrons to the right side: \(H_2SO_3 + H_2O + 2e^- \rightarrow HSO_4^- + 2H^+\) Now, combine and balance both half-reactions: \(3(H_2SO_3 + H_2O + 2e^-) + 4H^+ + H_2SO_3 + 6e^- \rightarrow HSO_4^- + 2H^+ + S + 4H_2O\) So, the balanced reaction is: \(3H_2SO_3 + 4H^+ \rightarrow HSO_4^- + 2H^+ + S + 4H_2O\) which simplifies to... \(3H_2SO_3 + 2H^+ \rightarrow HSO_4^- + S + 4H_2O\)

(d) Balancing \(Cl_2\) disproportionation

: First, we'll split the reaction into two half-reactions: 1. Oxidation half-reaction: \(Cl_2 \rightarrow Cl^-\) 2. Reduction half-reaction: \(Cl_2 \rightarrow ClO^-\) Next, balance the atoms and charges: 1. For the oxidation half-reaction: - Add an electron to the right side: \(Cl_2 + 2e^- \rightarrow 2Cl^-\) 2. For the reduction half-reaction: - Add 2 water molecules to the right side: \(Cl_2 \rightarrow ClO^- + 2H_2O\) - Add 4 hydroxide ions to the left side: \(Cl_2 + 4OH^- \rightarrow ClO^- + 2H_2O\) - Add 2 electrons to the left side: \(Cl_2 + 4OH^- + 2e^- \rightarrow ClO^- + 2H_2O\) Now, we can combine and balance both half-reactions: \(Cl_2 + 2e^- + Cl_2 + 4OH^- + 2e^- \rightarrow 2Cl^- + ClO^- + 2H_2O\) So, the balanced reaction is: \(2Cl_2 + 4OH^- \rightarrow 2Cl^- + ClO^- + 2H_2O\)

Key concepts

Understanding Oxidation-Reduction Reactions

An oxidation-reduction reaction, commonly referred to as a redox reaction, is a type of chemical reaction involving the transfer of electrons between two species. It is composed of two half-reactions: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. The relevance of redox reactions spans from biological systems, where they are vital to life processes such as cellular respiration, to industrial applications like battery operation.

In analyzing redox reactions, it's essential to identify the oxidation states of the elements involved. The oxidation state is a conceptual charge that an atom would have if the compound was composed of ions. Oxidation involves an increase in oxidation state, while reduction involves a decrease. Disproportionation reactions are a unique type of redox reaction where a single element undergoes both oxidation and reduction simultaneously, transforming into at least two different products with different oxidation states.

Chemical Equation Balancing

Why Balance Chemical Equations

The law of conservation of mass dictates that matter cannot be created or destroyed in an isolated system. This law applies to chemical reactions, so the atoms of the reactants should equal the atoms of the products. This is where balancing chemical equations comes into play. When a chemical equation is balanced, it accurately represents the conservation of atoms.

To balance a chemical equation, one must ensure that the number of atoms of each element is the same on both the reactant and product sides. In disproportionation reactions, the balancing act is slightly more complex due to the same element undergoing both oxidation and reduction. For instance, when balancing the disproportionation reaction of nickel (II) ion to nickel metal and nickel (II) ion, careful attention to both the elements and the charges is required for each half-reaction before combining them into a balanced overall equation.

The Half-Reaction Method

Applying the Half-Reaction Method

The half-reaction method is an invaluable tool for balancing complex redox reactions, particularly those involving disproportionation. It breaks down a redox reaction into its constituent half-reactions, one for oxidation and one for reduction. The steps involve balancing the atoms except for oxygen and hydrogen first, followed by adding water molecules to balance oxygen atoms and hydrogen ions (H+) or hydroxide ions (OH-) to balance hydrogen atoms, depending on whether the reaction is in an acidic or basic solution.

After atoms are balanced, the next step is to balance the charge by adding electrons. The number of electrons lost in the oxidation half-reaction should equal the number gained in the reduction half-reaction. By separately balancing and then combining the half-reactions, the overall chemical equation can be balanced. This method illuminates the electron transfer process, making it easier to understand the redox behavior of the disproportionating element, as exemplified in the textbook exercises provided.