Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state. (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)-\cdots\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \rightarrow-\rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(a q)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

Short answer

Balanced Equations and Oxidizing/Reducing Agents: (a) \(3 \mathrm{NO}_{2}^{-}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+12 \mathrm{H}^{+}+6 \mathrm{e}^{-}\rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{NO}_{3}^{-}+6 \mathrm{H}_{2}\mathrm{O}\) Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{NO}_{2}^{-}\) (b) \(3\mathrm{S}(s)+4\mathrm{HNO}_{3}(a q) \rightarrow 3\mathrm{H}_{2} \mathrm{SO}_{3}(a q)+4\mathrm{N}_{2} \mathrm{O}(g)\) Oxidizing agent: \(\mathrm{HNO}_{3}\) Reducing agent: \(\mathrm{S}\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3\mathrm{CH}_{3} \mathrm{OH}(a q)+8\mathrm{H}^{+}(a q) \rightarrow 2\mathrm{HCO}_{2} \mathrm{H}(a q)+2\mathrm{Cr}^{3+}(a q)+7\mathrm{H}_{2}\mathrm{O}(l)\) Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{CH}_{3} \mathrm{OH}\) (d) \(2\mathrm{MnO}_{4}^{-}(a q)+10\mathrm{Cl}^{-}(a q)+16\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Mn}^{2+}(a q)+5\mathrm{Cl}_{2}(a q)+8\mathrm{H}_{2}\mathrm{O}(l)\) Oxidizing agent: \(\mathrm{MnO}_{4}^{-}\) Reducing agent: \(\mathrm{Cl}^{-}\) (e) \(3\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) +6\mathrm{OH}^-(a q) \longrightarrow 3\mathrm{NH}_{4}{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\) Oxidizing agent: \(\mathrm{NO}_{2}^{-}\) Reducing agent: \(\mathrm{Al}\) (f) \(5\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2\mathrm{ClO}_{2}(a q) \longrightarrow 4\mathrm{H}_{2}\mathrm{O}(l) +2\mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) Oxidizing agent: \(\mathrm{ClO}_{2}\) Reducing agent: \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Step-by-step solution

Assign oxidation numbers

\(\mathrm{NO}_{2}^{-}\): Oxidation state of \(\mathrm{N}\): +3 \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): Oxidation state of \(\mathrm{Cr}\): +6 \(\mathrm{Cr}^{3+}\): Oxidation state of \(\mathrm{Cr}\): +3 \(\mathrm{NO}_{3}^{-}\): Oxidation state of \(\mathrm{N}\): +5

Balance elements other than H and O

Balance the \(\mathrm{Cr}\) and \(\mathrm{N}\) elements: \(3 \times \mathrm{NO}_{2}^{-}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\rightarrow 2 \times \mathrm{Cr}^{3+}+3 \times \mathrm{NO}_{3}^{-}\)

Balance O using H2O

Balance the \(\mathrm{O}\) atoms by using \(\mathrm{H}_{2}\mathrm{O}\). Add 6 water molecules to the product side to balance the \(\mathrm{O}\) atoms: \(3 \times \mathrm{NO}_{2}^{-}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\rightarrow 2 \times \mathrm{Cr}^{3+}+3 \times \mathrm{NO}_{3}^{-}+6 \times \mathrm{H}_{2}\mathrm{O}\)

Balance H using H+ ions

Balance the \(\mathrm{H}\) atoms using \(\mathrm{H}^{+}\) ions (since it's in an acidic solution). Add 12 \(\mathrm{H}^{+}\) ions to the reactant side to balance the \(\mathrm{H}\) atoms: \(3 \times \mathrm{NO}_{2}^{-}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+12 \times \mathrm{H}^{+}\rightarrow 2 \times \mathrm{Cr}^{3+}+3 \times \mathrm{NO}_{3}^{-}+6 \times \mathrm{H}_{2}\mathrm{O}\)

Balance charges using electrons

Balance the charges using electrons. By comparing both sides, we need 6 \(\mathrm{e}^{-}\) on the reactant side: \(3 \times \mathrm{NO}_{2}^{-}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+12 \times \mathrm{H}^{+}+6 \times \mathrm{e}^{-}\rightarrow 2 \times \mathrm{Cr}^{3+}+3 \times \mathrm{NO}_{3}^{-}+6 \times \mathrm{H}_{2}\mathrm{O}\)

Identifying Oxidizing and Reducing Agents

Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{NO}_{2}^{-}\) For the other reactions (due to similarity with reaction (a)): (b): Final balanced equation: \(3\mathrm{S}(s)+4\mathrm{HNO}_{3}(a q) \rightarrow 3\mathrm{H}_{2} \mathrm{SO}_{3}(a q)+4\mathrm{N}_{2} \mathrm{O}(g)\) Oxidizing agent: \(\mathrm{HNO}_{3}\) Reducing agent: \(\mathrm{S}\) (c): Final balanced equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3\mathrm{CH}_{3} \mathrm{OH}(a q)+8\mathrm{H}^{+}(a q) \rightarrow 2\mathrm{HCO}_{2} \mathrm{H}(a q)+2\mathrm{Cr}^{3+}(a q)+7\mathrm{H}_{2}\mathrm{O}(l)\) Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{CH}_{3} \mathrm{OH}\) (d): Final balanced equation: \(2\mathrm{MnO}_{4}^{-}(a q)+10\mathrm{Cl}^{-}(a q)+16\mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Mn}^{2+}(a q)+5\mathrm{Cl}_{2}(a q)+8\mathrm{H}_{2}\mathrm{O}(l)\) Oxidizing agent: \(\mathrm{MnO}_{4}^{-}\) Reducing agent: \(\mathrm{Cl}^{-}\) (e): Final balanced equation: \(3\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) +6\mathrm{OH}^-(a q) \longrightarrow 3\mathrm{NH}_{4}{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\) Oxidizing agent: \(\mathrm{NO}_{2}^{-}\) Reducing agent: \(\mathrm{Al}\) (f): Final balanced equation: \(5\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2\mathrm{ClO}_{2}(a q) \longrightarrow 4\mathrm{H}_{2}\mathrm{O}(l) +2\mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) Oxidizing agent: \(\mathrm{ClO}_{2}\) Reducing agent: \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Key concepts

Oxidation Numbers

Understanding oxidation numbers is crucial for grasping redox reactions. An oxidation number is a positive or negative number that indicates how many electrons an atom has gained, lost, or shared to become stable. In a formula, the sum of the oxidation numbers must equal the overall charge.

For instance, in a molecule of water (H2O), the oxidation number of hydrogen is +1 and of oxygen is -2. When assigning oxidation numbers, one must follow certain rules. The oxidation number of a pure element is always zero, the number for a monoatomic ion is equal to its charge, and specific elements like oxygen and hydrogen usually have oxidation numbers of -2 and +1, respectively, with exceptions in compounds like peroxides.

To balance redox reactions like the given examples, identifying the changes in oxidation numbers reveals which species are oxidized or reduced—key for later determining the oxidizing and reducing agents.

Oxidizing and Reducing Agents

In redox processes, oxidizing and reducing agents are the driving forces behind the transfer of electrons. An oxidizing agent gains electrons and is reduced, while a reducing agent loses electrons and is oxidized.

These agents are determined based on the changes in oxidation numbers observed across the reactants and products. For example, in the reaction between NO2− and Cr2O72−, where the oxidation state of N changes from +3 to +5 and that of Cr from +6 to +3, Cr2O72− is the oxidizing agent because it is reduced, and NO2− is the reducing agent as it is oxidized. Recognizing these agents is a fundamental step in understanding and balancing redox reactions, as it helps us identify the direction of electron flow.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox equations. It breaks down the overall reaction into two separate half-reactions—one for oxidation and one for the reduction. Each half-reaction is balanced for mass and charge independently before being combined to give the balanced overall reaction.

To employ this method, follow these steps:

• Write the unbalanced half-reactions for oxidation and reduction.
• Balance all atoms except hydrogen and oxygen.
• Balance oxygen atoms by adding water molecules.
• Balance hydrogen atoms by adding hydrogen ions (H+ in acidic solutions or OH− in basic solutions).
• Balance the charges by adding electrons (e−).
• Equalize the number of electrons in both half-reactions and combine them to cancel out the electrons, forming the complete reaction.

Applying the half-reaction method ensures that the conservation of mass and charge is maintained, providing a correctly balanced redox equation.