Question

Balance the following oxidation–reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}-(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

Short answer

The balanced redox reactions in basic solutions are: a. \(\mathrm{Al}(s)+ \mathrm{MnO}_{4}^{-}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) + 2\mathrm{OH}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) c. \(4\mathrm{Al}(s) +6\mathrm{NO}_{2}^{-}(a q) +18\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 6\mathrm{NH}_{3}(g)+ 4\mathrm{AlO}_{2}^{-}(a q) +12\mathrm{OH}^{-}(a q)\)

Step-by-step solution

Step 1-3: Identify half-reactions and balance atoms other than Hydrogen and Oxygen

Oxidation half-reaction: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\) Notice that the Aluminum and Manganese atoms are already balanced.

Balance Oxygen and Hydrogen

Oxidation half-reaction (Balanced): \(\mathrm{Al}(s) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) Reduction half-reaction (Balanced): \(\mathrm{MnO}_{4}^{-}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + 4\mathrm{OH}^{-}(a q)\)

Balance charges by adding electrons

Oxidation half-reaction (Balanced): \(\mathrm{Al}(s) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) + 3\mathrm{e}^{-}\) Reduction half-reaction (Balanced): \(3\mathrm{e}^{-} + \mathrm{MnO}_{4}^{-}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + 4\mathrm{OH}^{-}(a q)\)

Combine half-reactions

Overall equation (Balanced): \(\mathrm{Al}(s)+ \mathrm{MnO}_{4}^{-}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) Following the same steps for this reaction: Oxidation half-reaction (Balanced): \(\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{OCl}^{-}(a q)+2\mathrm{e}^{-}\) Reduction half-reaction (Balanced): \(\mathrm{Cl}_{2}(g) + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-}(a q)\) Overall equation (Balanced): \(\mathrm{Cl}_{2}(g) + 2\mathrm{OH}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\) Following the same steps for this reaction: Oxidation half-reaction (Balanced): \(\mathrm{Al}(s) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+3\mathrm{e}^{-}\) Reduction half-reaction (Balanced): \(6\mathrm{NO}_{2}^{-}(a q) + 18\mathrm{H}_{2}\mathrm{O}(l) + 12\mathrm{e}^{-} \rightarrow 6\mathrm{NH}_{3}(g)+12\mathrm{OH}^{-}(a q)\) Note that we need to multiply the oxidation half-reaction by 4 to equalize the number of electrons in both half-reactions: Overall equation (Balanced): \(4\mathrm{Al}(s) +6\mathrm{NO}_{2}^{-}(a q) +18\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 6\mathrm{NH}_{3}(g)+ 4\mathrm{AlO}_{2}^{-}(a q) +12\mathrm{OH}^{-}(a q)\)

Key concepts

Half-Reactions

In balancing oxidation-reduction (redox) reactions, identifying half-reactions is a crucial step. A half-reaction represents either the oxidation or reduction component of a full redox reaction. These are split into two parts:

• The oxidation half-reaction, where a substance loses electrons.
• The reduction half-reaction, where a substance gains electrons.

For example, in the reaction of Aluminum with Permanganate, the oxidation half-reaction can be written as: \[ \mathrm{Al}(s) \rightarrow \mathrm{Al(OH)}_{4}^{-}(aq) + 3\,\mathrm{e}^{-} \] This indicates that Aluminum is oxidized by losing three electrons. The reduction half-reaction can be expressed as:\[ 3\,\mathrm{e}^{-} + \mathrm{MnO}_{4}^{-}(aq) + 2\,\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{MnO}_{2}(s) + 4\,\mathrm{OH}^{-}(aq) \] This shows that Manganese is reduced as it gains electrons. Proper balancing of these half-reactions involves tracking changes in both atoms and charges.

Basic Solution

Balancing redox reactions occurring in a basic solution introduces a unique aspect due to the presence of hydroxide ions (OH⁻). Unlike in acidic solutions, where hydrogen ions (H⁺) are abundant, basic solutions require adding OH⁻ ions to balance proton excess.

• Add OH⁻ to both sides of the equation to neutralize excess H⁺.
• Combine H⁺ and OH⁻ to form water where possible.

For instance, in the Chlorine reaction, where \( \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(aq) + \mathrm{OCl}^{-}(aq) \), by adding OH⁻ ions appropriately, the balanced equation becomes:\[ \mathrm{Cl}_{2}(g) + 2\,\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Cl}^{-}(aq) + \mathrm{OCl}^{-}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \] This incorporation of OH⁻ ensures that all protons are accounted for, forming stable water molecules. Appreciating the role of the basic environment is crucial to achieving a proper balance of redox equations.

Electron Transfer

Electron transfer is at the core of redox reactions, driving the changes that occur during these processes. In every redox reaction, electrons are transferred from one species to another:

• Oxidation entails losing electrons.
• Reduction involves gaining electrons.

A balanced reaction ensures that the total number of electrons lost is equal to the number of electrons gained. Take the example of the Aluminum and Nitrite reaction, where:Oxidation: \[ \mathrm{Al}(s) \rightarrow \mathrm{AlO}_{2}^{-}(aq) + 3\,\mathrm{e}^{-} \] Reduction:\[ 6\,\mathrm{NO}_{2}^{-}(aq) + 18\,\mathrm{H}_{2}\mathrm{O}(l) + 12\,\mathrm{e}^{-} \rightarrow 6\,\mathrm{NH}_{3}(g) + 12\,\mathrm{OH}^{-}(aq) \] For the reaction to be fully balanced, the half-reactions are multiplied to equalize electron transfer, thereby maintaining stoichiometric balance. Mastering electron transfer principles greatly facilitates the accurate balancing of complex redox reactions.