Question

The \(\mathrm{Mn}^{3+}\) ion is unstable in solution and undergoes disproportionation to give \(\mathrm{Mn}^{2+}, \mathrm{MnO}_{2}\), and \(\mathrm{H}^{+}\) ion. Write a balanced ionic equation for the reaction.

Short answer

The balanced equation is \(2\mathrm{Mn}^{3+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2} + 4\mathrm{H}^{+}\).

Step-by-step solution

Identify Reactants and Products

Recognize the substances involved in the reaction. The reactant is the unstable ion \(\mathrm{Mn}^{3+}\), which disproportionates into \(\mathrm{Mn}^{2+}\), \(\mathrm{MnO}_{2}\), and \(\mathrm{H}^{+}\).

Write the Unbalanced Reaction Equation

Start by writing the unbalanced ionic equation showing \(\mathrm{Mn}^{3+}\) converting into \(\mathrm{Mn}^{2+}\), \(\mathrm{MnO}_{2}\), and \(\mathrm{H}^{+}\). This can be expressed as:\[\mathrm{Mn}^{3+} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2} + \mathrm{H}^{+}\]

Count Atoms and Charges

Examine the unbalanced equation to count the number of each type of atom and the charges on each side. There is one Mn atom on the left, but there are two Mn atoms on the right. The charge on the left is 3+, while we have 2+, 0, and 1+ on the right, which add to a total of 3+. Oxygen and hydrogen atoms are not yet balanced.

Balance Manganese Atoms

To balance the manganese atoms, we need two \(\mathrm{Mn}^{3+}\) ions on the reactant side. This ensures one will reduce to \(\mathrm{Mn}^{2+}\) and the other will oxidize to \(\mathrm{MnO}_{2}\).\[2\mathrm{Mn}^{3+} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2}\]

Balance Oxygen Atoms with Water

The appearance of \(\mathrm{MnO}_{2}\) introduces oxygen, so add 2 water molecules on the left side to balance the oxygens:\[2\mathrm{Mn}^{3+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2}\]

Balance Hydrogen Atoms with Protons

Since we added 2 water molecules introducing 4 hydrogen atoms, balance the hydrogens by adding 4 protons \((\mathrm{H}^{+})\) on the right:\[2\mathrm{Mn}^{3+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2} + 4\mathrm{H}^{+}\]

Check Charges and Atom Balance

Both manganese and hydrogen atoms are balanced, and the charge on each side of the equation is also balanced. The left side has a total charge of 6+ and the right side balances because 2+, 0, and 4+ add to 6+. The final balanced equation is:\[2\mathrm{Mn}^{3+} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MnO}_{2} + 4\mathrm{H}^{+}\]

Key concepts

Balanced Ionic Equation

When solving a chemical equation, balancing it is crucial. It ensures that matter is neither created nor destroyed during the reaction. A balanced ionic equation specifically deals with ions. It keeps track of both the number of each type of ion and the overall charge balance between reactants and products.
A balanced equation demonstrates that both the mass and charge are preserved throughout the reaction. Begin by writing down the unbalanced version, listing all reactants and products. Then, adjust the coefficients to ensure that the same number of each type of atom appears on both sides of the equation. Additionally, the charges on each side must be equal.
In a disproportionation reaction, like the one involving \( \mathrm{Mn}^{3+} \), balancing not only involves checking atoms but also the electrons lost and gained. Typically, you deal with a single reactant that splits into two different products, each with different oxidation states. The final balanced equation will accurately reflect both conservation laws by showing equal numbers of each atom and the same total charge on both sides.

Oxidation-Reduction

Oxidation-reduction (redox) reactions are essential in chemistry. They involve the transfer of electrons between substances. One substance gets oxidized (loses electrons) while another gets reduced (gains electrons). In the given reaction of \( \mathrm{Mn}^{3+} \), both processes occur simultaneously.
The \( \mathrm{Mn}^{3+} \) ion undergoes disproportionation, meaning it gets both oxidized and reduced. A part of \( \mathrm{Mn}^{3+} \) reduces to \( \mathrm{Mn}^{2+} \), and the other part oxidizes to form \( \mathrm{MnO}_2 \). This results in a self-contained redox process, with electron transfer occurring internally within the ions.
To tackle a redox equation, identify the oxidation states of the substances involved and track changes. This will help you balance the equation accurately. Ensure that the number of electrons lost in oxidation equals those gained in reduction, as this keeps the charge balanced.

Chemical Equilibrium

Chemical equilibrium is a concept where the rates of forward and reverse reactions are equal. In a balanced state, the concentrations of reactants and products stay constant over time. In the context of a disproportionation reaction, equilibrium is essential.
When setting up the reaction equation for \( \mathrm{Mn}^{3+} \), achieving balance aligns with the concept of equilibrium. It signifies that all elements eventually achieve a steady state, even in a disproportionation context. The completion of balancing implies that no further net change happens in concentrations if the reaction reaches equilibrium.
Though in practice, equilibrium allows reversibility within reactions, it's particularly relevant in systems where conditions, like concentration or temperature, change. Remember that the optimal balanced state reflects the system's natural balance of energy, leading to a stable configuration.