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) \(\mathrm{Ni}^{+}(a q)-\rightarrow \rightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)\) (acidic solution) (d) \(\mathrm{Cl}_{2}(a q)-\rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q)\) (basic solution)

Short answer

The balanced disproportionation reactions are: a) \(2Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\) b) \(3MnO_{4}^{2-}(aq) + 4H^+ \rightarrow 2MnO_4^-(aq) + MnO_2(s) + 2H_2O(l)\) c) \(3H_{2}SO_{3}(aq) + 4H^+ \rightarrow 2HSO_4^-(aq) + S(s)\) d) \(2Cl_{2}(aq) + 2OH^-(aq) \rightarrow ClO^-(aq) + 2Cl^-(aq) + H_2O(l)\)

Step-by-step solution

a) Ni\(^{+}(aq)\rightarrow Ni^{2+}(aq)+Ni(s)\) (acidic solution)

1. Identify half-reactions: Oxidation half-reaction: Ni\(^{+}(aq)\rightarrow Ni^{2+}(aq)\) Reduction half-reaction: Ni\(^{+}(aq)\rightarrow Ni(s)\) 2. Analyze the oxidation states: Oxidation states: Ni\(^+\) has an oxidation state of +1, Ni\(^{2+}\) has an oxidation state of +2, and Ni(s) has an oxidation state of 0. 3. Balance the half-reactions: Oxidation: Ni\(^{+}(aq)\rightarrow Ni^{2+}(aq) + e^-\) Reduction: Ni\(^{+}(aq) + e^- \rightarrow Ni(s)\) Combine and cancel electrons: Ni\(^{+}(aq) \rightarrow Ni^{2+}(aq) + e^-\) Ni\(^{+}(aq) + e^- \rightarrow Ni(s)\) ------------------------------ Ni\(^{+}(aq) + Ni^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\) The balanced reaction is: 2Ni\(^{+}(aq) \rightarrow Ni^{2+}(aq) + Ni(s)\)

b) MnO\(_4^{2-}(aq) \longrightarrow MnO_4^-(aq)+MnO_2(s)\) (acidic solution)

1. Identify half-reactions: Oxidation half-reaction: MnO\(_4^{2-}(aq) \longrightarrow MnO_4^-(aq)\) Reduction half-reaction: MnO\(_4^{2-}(aq) \longrightarrow MnO_2(s)\) 2. Analyze the oxidation states: Oxidation states: Mn in MnO\(_4^{2-}\) has an oxidation state of +6, Mn in MnO\(_4^-\) has an oxidation state of +7, and Mn in MnO\(_2\) has an oxidation state of +4. 3. Balance the half-reactions: Oxidation: MnO\(_4^{2-}(aq) \rightarrow MnO_4^-(aq) + e^-\) Reduction: MnO\(_4^{2-}(aq) + 2e^- \rightarrow MnO_2(s) + 2OH^-(aq)\) (for acidic solution, balance O by adding H\(_2\)O and then H\(^+\)) MnO\(_4^{2-}(aq) + 2e^- + 4H^+ \rightarrow MnO_2(s) + 2H_2O(l)\) Combine and cancel electrons: 2MnO\(_4^{2-}(aq) \rightarrow 2MnO_4^-(aq) + 2e^-\) MnO\(_4^{2-}(aq) + 2e^- + 4H^+ \rightarrow MnO_2(s) + 2H_2O(l)\) ------------------------------ 3MnO\(_4^{2-}(aq) + 4H^+ \rightarrow 2MnO_4^-(aq) + MnO_2(s) + 2H_2O(l)\) The balanced reaction is: 3MnO\(_4^{2-}(aq) + 4H^+ \rightarrow 2MnO_4^-(aq) + MnO_2(s)+ 2H_2O(l)\)

c) H\(_2\)SO\(_3(aq) \longrightarrow S(s)+HSO\)_4^-(aq) (acidic solution)

1. Identify half-reactions: Oxidation half-reaction: H\(_2\)SO\(_3(aq) \longrightarrow S(s)\) Reduction half-reaction: H\(_2\)SO\(_3(aq) \longrightarrow HSO\)_4^-(aq) 2. Analyze the oxidation states: Oxidation states: S in H\(_2\)SO\(_3\) has an oxidation state of +4, S in S(s) has an oxidation state of 0, and S in HSO\(_4^-\) has an oxidation state of +6. 3. Balance the half-reactions: Oxidation: H\(_2\)SO\(_3(aq) \rightarrow S(s) + 4H^+(aq) + 4e^-\) Reduction: H\(_2\)SO\(_3(aq) + 2H^+ + 2e^- \rightarrow HSO\)_4^-(aq) Combine and cancel electrons: H\(_2\)SO\(_3(aq) \rightarrow S(s) + 4H^+(aq) + 4e^-\) 2H\(_2\)SO\(_3(aq) + 4H^+ + 4e^- \rightarrow 2HSO\)_4^-(aq) ------------------------------ 3H\(_2\)SO\(_3(aq) + 4H^+ \rightarrow 2HSO\)_4^-(aq) + S(s)$ The balanced reaction is: 3H\(_2\)SO\(_3(aq) + 4H^+ \rightarrow 2HSO\)_4^-(aq) + S(s)$

d) Cl\(_2(aq) \rightarrow Cl^-(aq)+ClO^-(aq)\) (basic solution)

1. Identify half-reactions: Oxidation half-reaction: Cl\(_2(aq) \rightarrow ClO^-(aq)\) Reduction half-reaction: Cl\(_2(aq) \rightarrow Cl^-(aq)\) 2. Analyze the oxidation states: Oxidation states: Cl in Cl\(_2\) has an oxidation state of 0, Cl in ClO\(^-\) has an oxidation state of +1, and Cl in Cl\(^-\) has an oxidation state of -1. 3. Balance the half-reactions: Oxidation: Cl\(_2(aq) \rightarrow ClO^-(aq) + 2e^-\) Reduction: Cl\(_2(aq) + 2e^- \rightarrow 2Cl^-(aq)\) Combine and cancel electrons and balance for basic solution: Cl\(_2(aq) \rightarrow ClO^-(aq) + 2e^-\) Cl\(_2(aq) + 2e^- \rightarrow 2Cl^-(aq)\) ------------------------------ 2Cl\(_2(aq) \rightarrow ClO^-(aq) + 2Cl^-(aq)\) 2Cl\(_2(aq) + 2OH^-(aq) \rightarrow ClO^-(aq) + 2Cl^-(aq) + H_2O(l)\) (Add 2 OH\(^-\) to balance O atoms and form H\(_2\)O) The balanced reaction is: 2Cl\(_2(aq) + 2OH^-(aq) \rightarrow ClO^-(aq) + 2Cl^-(aq) + H_2O(l)\)

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often known as redox reactions, involve the transfer of electrons between species, leading to changes in their oxidation states. In the context of disproportionation reactions, a single element undergoes both oxidation and reduction simultaneously. For example, when nickel(I) ion is involved in a disproportionation reaction, it's both getting oxidized to nickel(II) ion and reduced to nickel metal.

In a redox reaction, the substance that loses electrons is said to be oxidized, while the one that gains electrons is reduced. It's essential to remember: Oxidation Is Loss, and Reduction Is Gain of electrons, often remembered by the acronym 'OIL RIG'.

When working with redox reactions, we also need to ensure charge balance – the total charge on both sides of the equation should be equal – and mass balance – the same number of each type of atom on both sides of the equation. Balancing these reactions often requires a step-by-step approach, starting with the identification of half-reactions, analysis of oxidation states, balancing of the half-reactions, and lastly, combining and canceling electrons to balance the overall reaction.

Chemical Equilibrium

Chemical equilibrium refers to a state in which the forward and reverse reactions occur at the same rate, leading to no net change in the concentrations of the reactants and products over time. It is a dynamic state, meaning the reactions are still occurring, but since they are at an equal pace, the system appears static.

In a disproportionation reaction, equilibrium is not the focus because disproportionation is a special type of redox reaction where a substance is both oxidized and reduced, typically leading to complete consumption of the original species. However, understanding the concept of equilibrium is crucial when considering reversible reactions and the conditions under which they occur.

In redox reactions, like other types of chemical reactions, changing the conditions such as concentration, temperature, or pressure can shift the equilibrium, favoring either the reactants or the products. This is described by Le Chatelier’s Principle, which tells us that a system at equilibrium will adjust to counteract any changes imposed upon it.

Balancing Chemical Equations

The ability to balance chemical equations is foundational in chemistry. It requires that the number of atoms for each element, and the total charge, is the same on both sides of the equation. This represents the law of conservation of mass and charge. Disproportionation reactions present an interesting challenge, as the same element must be balanced in its varying oxidation states.

Here are some tips for balancing chemical equations:

• Write the correct formulas for all reactants and products.
• Identify the number of atoms of each element on both sides and start by balancing the element that appears in the greatest complexity.
• For redox reactions, balance each half-reaction for mass and charge separately; add electrons to balance charge, and use water and hydrogen ions (or hydroxide ions in basic solutions) to balance oxygen and hydrogen atoms, respectively.
• Combine the half-reactions, ensuring that electrons cancel, and then make sure the overall equation is balanced for both mass and charge.

Remember, practice is key to mastering balancing equations, and step-by-step approaches as illustrated in disproportionation problems are particularly useful in more complex scenarios.