Question

Potassium permanganate, \(\mathrm{KMnO}_{4}\), is one of the most widely used oxidizing agents. Balance each of the following oxidation-reduction reactions of the permanganate ion in acidic solution by using the "half-reaction" method. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)\) c. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)\)

Short answer

The balanced reactions are: a) \(4\mathrm{MnO}_{4}^{-}+ 10\mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 32\mathrm{H}^{+} \rightarrow 4\mathrm{Mn}^{2+} + 20\mathrm{CO}_{2} + 16\mathrm{H}_{2}\mathrm{O}\) b) \(\mathrm{MnO}_{4}^{-}+ 5\mathrm{Fe}^{2+} + 8\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 5\mathrm{Fe}^{3+} + 4\mathrm{H}_{2}\mathrm{O}\) c) \(2\mathrm{MnO}_{4}^{-}+ 10\mathrm{Cl}^{-} + 16\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{Cl}_{2} + 8\mathrm{H}_{2}\mathrm{O}\)

Step-by-step solution

Identify the Half-Reactions

The half-reactions are as follows: Oxidation: \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}\) Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)

Balance Atoms and Charges in half-reactions

Balancing the atoms and charges, we get: Oxidation: \(2 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 4 \mathrm{CO}_{2} + 4e^{-}\) Reduction: \(5e^{-} + 8\mathrm{H}^{+} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\)

Combine half-reactions

Multiplying the half-reactions to make electron gain and loss equal: Oxidation: \(10 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 20 \mathrm{CO}_{2} + 20e^{-}\) Reduction: \(20e^{-} + 32\mathrm{H}^{+} + 4\mathrm{MnO}_{4}^{-} \rightarrow 4 \mathrm{Mn}^{2+} + 16 \mathrm{H}_{2}\mathrm{O}\) Adding both half-reactions and simplifying: \(4\mathrm{MnO}_{4}^{-}+ 10\mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 32\mathrm{H}^{+} \rightarrow 4\mathrm{Mn}^{2+} + 20\mathrm{CO}_{2} + 16\mathrm{H}_{2}\mathrm{O}\) b. $\mathrm{MnO}_{4}^{-}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}$

Identify the Half-Reactions

The half-reactions are as follows: Oxidation: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}\) Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) (the same as in part a)

Balance Atoms and Charges in half-reactions

Balancing the atoms and charges, we get: Oxidation: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^{-}\) Reduction: (already balanced in part a): \(5e^{-} + 8\mathrm{H}^{+} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\)

Combine half-reactions

Multiplying the half-reactions to make electron gain and loss equal: Oxidation: \(5\mathrm{Fe}^{2+} \rightarrow 5\mathrm{Fe}^{3+} + 5e^{-}\) Reduction: (unchanged): \(5e^{-} + 8\mathrm{H}^{+} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\) Adding both half-reactions and simplifying: \(\mathrm{MnO}_{4}^{-}+ 5\mathrm{Fe}^{2+} + 8\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 5\mathrm{Fe}^{3+} + 4\mathrm{H}_{2}\mathrm{O}\) c. $\mathrm{MnO}_{4}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Cl}_{2}$

Identify the Half-Reactions

The half-reactions are as follows: Oxidation: \(2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}\) Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) (the same as in parts a and b)

Balance Atoms and Charges in half-reactions

Balancing the atoms and charges, we get: Oxidation: \(2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2e^{-}\) Reduction: (already balanced in previous parts): \(5e^{-} + 8\mathrm{H}^{+} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2}\mathrm{O}\)

Combine half-reactions

Multiplying the half-reactions to make electron gain and loss equal: Oxidation: \(5(2\mathrm{Cl}^{-}) \rightarrow 5\mathrm{Cl}_{2} + 10e^{-}\) Reduction: \(2(5e^{-} + 8\mathrm{H}^{+} + \mathrm{MnO}_{4}^{-}) \rightarrow 2\mathrm{Mn}^{2+} + 8 \mathrm{H}_{2}\mathrm{O}\) Adding both half-reactions and simplifying: \(2\mathrm{MnO}_{4}^{-}+ 10\mathrm{Cl}^{-} + 16\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{Cl}_{2} + 8\mathrm{H}_{2}\mathrm{O}\)

Key concepts

Half-Reaction Method

In the world of chemistry, the half-reaction method is an essential technique for balancing oxidation-reduction (redox) reactions. It involves breaking down the reaction into two parts: the oxidation half and the reduction half. This separation allows for a more straightforward balancing of both atoms and charges.

First, identify the substances being oxidized and reduced. For instance, in the reaction of potassium permanganate with oxalate ions,

• The oxidation half-reaction is: \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}\)
• The reduction half-reaction is: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)

Next, focus on balancing each half-reaction independently. This usually involves adding electrons, water, and hydrogen ions to balance the charges and substances involved.

This method not only ensures that the overall equation is balanced but also emphasizes the electron transfer, which is crucial in oxidation-reduction reactions. By balancing the electrons lost in oxidation with those gained in reduction, the integrity of the reaction is maintained.

Balancing Equations

Balancing redox reactions is fundamental in chemistry, as it ensures that the law of conservation of mass and charge is followed. The half-reaction method makes it easier by splitting the oxidation and reduction processes, allowing distinct balancing before recombining them.

For the oxidation half-reaction involving oxalate ions, balance carbons first to form carbon dioxide. Then, add electrons to account for the charge difference. Likewise, for the reduction process, adding water molecules balances oxygen, while hydrogen ions balance the hydrogen in an acidic solution.

To combine the half-reactions, equalize the electrons transferred by multiplying each half-reaction by appropriate coefficients. For example:

• Oxidation: Multiply if needed to equalize the electron exchange.
• Reduction: Do the same, ensuring electrons gained equals electrons lost in oxidation.

Finally, add both half-reactions together. This step involves cancellation of electrons, leading to a fully balanced redox equation. Such care in balancing caters to both quantitative and qualitative aspects of the chemical reaction.

Acidic Solution

Balancing redox equations often takes place in different media, and the most common is the acidic solution medium. Here, hydrogen ions (\(\mathrm{H}^{+}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)) play crucial roles in ensuring complete balance of the reactants and products.

In an acidic solution, when balancing the reduction half-reaction, add hydrogen ions to balance hydrogen atoms and add water molecules to balance oxygen atoms. For instance, when permanganate ion reduces to manganese ion (\(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)), adding \(8\mathrm{H}^{+}\) balances the hydrogens, while \(4\mathrm{H}_{2}\mathrm{O}\) balances oxygens.

• Hydrogen ions help in adjusting the hydrogen count without disrupting the overall charge balance.
• Water is added to counter any excess oxygen atoms, again maintaining balanced charges.

These adjustments are vital, especially in reactions in acidic environments, as the availability of hydrogen ions can greatly affect the feasibility and progression of a reaction.