Problem 46 - Chemistry The Central Science

Question

An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q),(\mathbf{b})\) acidic aqueous \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to aqueous \(\mathrm{Cr}^{3+},(\mathrm{c})\) aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) to mercury metal. Write balanced equations for these reactions.

Step-by-Step Solution

Answer

The balanced equations for the given redox reactions are as follows: Reaction 1: \(5SO_{2} + 2KMnO_{4} + 2H_{2}O \to K_{2}Mn_{2}(SO_{4})_{5} + 6H_{2}O\) Reaction 2: \(6SO_{2} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} \to 6H_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 4H_{2}O\) Reaction 3: \(SO_{2} + Hg_{2}(NO_{3})_{2} + 2H_{2}O \to H_{2}SO_{4} + 2Hg + 2HNO_{3}\)

Step 1: Balancing Reaction 1: Aqueous SO2 reducing aqueous KMnO4 to MnSO4

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (MnO4^- to Mn^2+): \(MnO_{4}^{-} \to Mn^{2+}\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(5e^{-} + 8H^{+} + MnO_{4}^{-} \to Mn^{2+} + 4H_{2}O\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 5 and reduction half-reaction by 2, and then add. \(5(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(2(5e^{-} + 8H^{+} + MnO_{4}^{-} \to Mn^{2+} + 4H_{2}O)\) Reaction 1 Balanced Equation: \(5SO_{2} + 2KMnO_{4} + 2H_{2}O \to 5SO_{4}^{2-} + 2Mn^{2+} + 12H^{+} + 2K^{+}\) \(5SO_{2} + 2KMnO_{4} + 2H_{2}O \to K_{2}Mn_{2}(SO_{4})_{5} + 6H_{2}O\)

Step 2: Balancing Reaction 2: Aqueous SO2 reducing acidic aqueous K2Cr2O7 to Cr3+

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (Cr2O7^2- to Cr^3+): \(Cr_{2}O_{7}^{2-} \to 2Cr^{3+}\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \to 2Cr^{3+} + 7H_{2}O\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 6 and reduction half-reaction by 1, and then add. \(6(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \to 2Cr^{3+} + 7H_{2}O\) Reaction 2 Balanced Equation: \(6SO_{2} + K_{2}Cr_{2}O_{7} + 14H^{+} \to 6SO_{4}^{2-} + 4H_{2}O + 2Cr^{3+} + 2K^{+}\) \(6SO_{2} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} \to 6H_{2}SO_{4} + 4H_{2}O + 2Cr^{3+} + 2K^{+}\) \(6SO_{2} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} \to 6H_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 4H_{2}O\)

Step 3: Balancing Reaction 3: Aqueous SO2 reducing aqueous Hg2(NO3)2 to mercury metal

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (Hg^2+ to Hg): \(Hg^{2+} \to Hg\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(2e^{-} + Hg^{2+} \to Hg\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 1 and reduction half-reaction by 4, and then add. \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(4(2e^{-} + Hg^{2+} \to Hg)\) Reaction 3 Balanced Equation: \(SO_{2} + 4Hg^{2+} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+} + 4Hg\) \(SO_{2} + Hg_{2}(NO_{3})_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+} + 2Hg + 2NO_{3}^{-}\) \(SO_{2} + Hg_{2}(NO_{3})_{2} + 2H_{2}O \to H_{2}SO_{4} + 2Hg + 2HNO_{3}\)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is essential for accurately representing a chemical reaction. It ensures that the same number of each type of atom appears on both sides of the equation. This adherence to the Law of Conservation of Mass is crucial, as atoms cannot be created or destroyed in a chemical reaction.
The process involves several steps which simplify balancing complex reactions, such as oxidation-reduction reactions. These steps are:

• Identifying the reactants and products.
• Writing the unbalanced chemical equation.
• Using coefficients to balance the number of atoms for each element on both sides of the equation.
• Ensuring that charges are balanced if dealing with ionic compounds.

By following these steps, you can systematically approach and balance any chemical equation. Remember, practice is essential for mastering the concept!

Half-Reaction Method

The half-reaction method is a systematic approach used to balance redox (oxidation-reduction) reactions. It individually balances the oxidation and reduction parts of a redox process, underlining changes in oxidation states.
This method provides a clear framework and consists of these steps:

• Divide the reaction into two half-reactions - one for oxidation and one for reduction.
• Balance the atoms in each half-reaction, usually starting with elements other than oxygen and hydrogen.
• Balance the oxygen atoms by adding water molecules if necessary, then balance hydrogen atoms by adding hydrogen ions ( H⁺).
• Balance the charges by adding electrons.
• Multiply each half-reaction by appropriate factors so that the number of electrons is equal in both reactions.
• Combine the balanced half-reactions to form the overall balanced equation.

This method not only equilibrates atoms but also ensures charge neutrality, making it an analytic way for balancing redox equations.

Oxidation-Reduction

Oxidation-reduction, often referred to as redox, reactions are processes where electrons are transferred between chemical species. This results in changes in the oxidation states of atoms involved in the reaction. Understanding redox reactions involves recognizing two key processes:

• Oxidation: This process involves the loss of electrons by a molecule, atom, or ion, which increases its oxidation state. An easy way to remember this is: "LEO" (Loss of Electrons is Oxidation).
• Reduction: In this process, there is a gain of electrons, decreasing the oxidation state of the substance. A memory aid for this is: "GER" (Gain of Electrons is Reduction).

Together, these terms frame redox reactions as an electron exchange. One component donates electrons (gets oxidized) while another accepts electrons (gets reduced). Recognizing and balancing redox reactions is vital in chemistry, especially in fields like electrochemistry and metabolic processes.