Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\mathrm{Cr}^{3+}\) (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 three reactions are: 1. \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) 2. \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) 3. \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\)

Step 1: In all three reactions, SO2 acts as a reducing agent, meaning it loses electrons. Thus, the SO2 species will be involved in the oxidation half-reaction. In each case, we need to find the oxidation state of the species involving KMnO4, K2Cr2O7, and Hg2(NO3)2. #Step 2: Write the oxidation half-reaction for SO2#

SO2 gets oxidized to SO4^2-. In this process, each S atom changes its oxidation state from +4 to +6, and loses 2 electrons: \(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^-\) #Step 3: Write the reduction half-reaction for KMnO4 reaction#

Step 2: KMnO4 gets reduced to MnSO4. In this process, each Mn atom changes its oxidation state from +7 to +2, and gains 5 electrons: \(MnO_{4}^{-} + 5 e^{-} \rightarrow Mn^{2+}\) #Step 4: Combine the oxidation and reduction half-reactions for the KMnO4 reaction#

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(5(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-})\) \(2(MnO_{4}^{-} + 5 e^{-} \rightarrow Mn^{2+})\) The final balanced equation is: \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) #Step 5: Write the reduction half-reaction for the K2Cr2O7 reaction#

Step 3: K2Cr2O7 gets reduced to Cr^3+. In this process, each Cr atom changes its oxidation state from +6 to +3, and gains 3 electrons: \(Cr_{2}O_{7}^{2-} + 6 e^{-} \rightarrow 2 Cr^{3+}\) #Step 6: Combine the oxidation and reduction half-reactions for the K2Cr2O7 reaction#

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(3(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-})\) \(2(Cr_{2}O_{7}^{2-} + 6 e^{-} \rightarrow 2 Cr^{3+})\) The final balanced equation is: \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) #Step 7: Write the reduction half-reaction for the Hg2(NO3)2 reaction#

Step 4: Hg2(NO3)2 gets reduced to Hg (mercury metal). In this process, each Hg atom changes its oxidation state from +2 to 0, and gains 2 electrons: \(Hg^{2+} + 2 e^{-} \rightarrow Hg\) #Step 8: Combine the oxidation and reduction half-reactions for the Hg2(NO3)2 reaction#

Multiply the oxidation half-reaction by 1 (no need to change) and the reduction half-reaction by 2 to equalize the number of electrons, and then add them together: \(SO_{2} \rightarrow SO_{4}^{2-} + 2 e^{-}\) \(2(Hg^{2+} + 2 e^{-} \rightarrow Hg)\) The final balanced equation is: \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\) The balanced equations for the three reactions are: 1. \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\) 2. \(3 SO_{2} + 2 Cr_{2}O_{7}^{2-}+ 14 H^{+} \rightarrow 3 SO_{4}^{2-} + 4 Cr^{3+} + 7 H_{2}O\) 3. \(SO_{2} + 2 Hg^{2+} + 2 H^{+} \rightarrow SO_{4}^{2-} + 2 Hg\)

Key Concepts

Oxidation-Reduction

Redox reactions, short for oxidation-reduction reactions, involve the transfer of electrons between substances. These reactions are pivotal in chemistry. They occur when one substance loses electrons (oxidation) and another gains electrons (reduction). Identifying these changes involves determining the oxidation states of elements in the reactants and products. For instance, in the reduction of

• KMnO₄, manganese transitions from an oxidation state of +7 to +2, signifying electron gain, or reduction.
• SO₂ acts as a reducing agent, moving from +4 to +6, indicating electron loss, or oxidation.

Recognizing these transformations helps balance the electrons in the overall reaction, essential for understanding redox processes.

Balancing Chemical Equations

Balancing chemical equations is crucial for representing chemical reactions accurately. In a balanced equation, the number of atoms for each element is the same on both sides of the equation. This balance reflects the conservation of mass.
To balance equations in redox reactions, you need to:

• Write the oxidation and reduction half-reactions separately.
• Equalize the number of electrons lost in oxidation with those gained in reduction, often by multiplying the half-reactions by appropriate factors.
• Combine the adjusted half-reactions to form the complete balanced equation.

For example, in balancing \(5 SO_{2} + 2 MnO_{4}^{-} + 2 H^{+} \rightarrow 5 SO_{4}^{2-} + 2 Mn^{2+} + H_{2}O\), you ensure that electron count is balanced, reflecting the accurate depiction of electron transfer during the reaction.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to transform reactants into products. A chemical equation symbolizes this process, depicting the substances consumed and those produced. In the provided examples, solutions of chemicals interact in the aqueous phase.
Redox reactions are a type of chemical reaction where oxidation and reduction processes occur simultaneously. This is seen when:

• SO₂ reduces KMnO₄ to MnSO₄, resulting in a change of oxidation state and product formation.
• Similarly, in acidic medium, it reduces K₂Cr₂O₇ to Cr³⁺, illustrating complex interplays between different atoms and ions.

Understanding the role of solvents and reaction conditions, like acidity, is key in studying how these processes occur, paving the way for advanced topics like reaction kinetics and equilibria.