Problem 44 - Chemistry The Central Science In Si Units

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

Balanced reactions are: \(5SO_2 + 2KMnO_4 + 2H_2O \rightarrow 2MnSO_4 + K_2SO_4 + 2H_2SO_4\), \(3SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O\), \(SO_2 + 2Hg_2(NO_3)_2 + 2H_2O \rightarrow 4Hg + 2HNO_3 + 2H_2SO_4\).

Step 1: Understanding the Redox Reaction with KMnO4

In this reaction, sulfur dioxide (SO_2) acts as a reducing agent and reduces potassium permanganate (KMnO_4) to manganese sulfate (MnSO_4). The unbalanced reaction is:\[SO_2(aq) + KMnO_4(aq) + H_2O(l) \rightarrow MnSO_4(aq) + K_2SO_4(aq) + H_2SO_4(aq)\]Balancing the reaction in acidic medium, considering the oxidation changes and charges, we get:\[5SO_2 + 2KMnO_4 + 2H_2O \rightarrow 2MnSO_4 + K_2SO_4 + 2H_2SO_4\]

Step 2: Balancing the Reaction with K2Cr2O7

In the second reaction, SO_2 reduces potassium dichromate (K_2Cr_2O_7) to chromium ions (Cr^{3+}) in an acidic solution. First, we write the unbalanced equation:\[SO_2(aq) + K_2Cr_2O_7(aq) + H_2SO_4(aq) \rightarrow Cr^{3+}(aq) + SO_3(aq) + K_2SO_4(aq) + H_2O(l)\]Balancing the redox equation under acidic conditions, it becomes:\[3SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O\]

Step 3: Balancing the Reaction with Hg2(NO3)2

For the third reaction, SO_2 reduces mercury(I) nitrate (Hg_2(NO_3)_2) to mercury metal. The unbalanced equation:\[SO_2(aq) + Hg_2(NO_3)_2(aq) + H_2O(l) \rightarrow Hg(l) + HNO_3(aq) + H_2SO_4(aq)\]Balancing the equation considering the change in oxidation number:\[SO_2 + 2Hg_2(NO_3)_2 + 2H_2O \rightarrow 4Hg + 2HNO_3 + 2H_2SO_4\]

Key Concepts

Acidic Medium Balancing

When balancing redox reactions conducted in acidic medium, the key is to ensure that both mass and charge are conserved. This usually involves

• Adding water molecules to balance oxygen atoms.
• Introducing hydrogen ions (H⁺) for hydrogen atom balance.
• Ensuring charge balance by adding electrons where necessary.

These steps turn unbalanced reactions into balanced equations that depict the underlying chemistry accurately. Consider the example of SO₂ reducing KMnO₄ where,

• The oxidation of S in SO₂ is from +4 to +6 (becoming SO₄²⁻).
• The reduction of Mn in MnO₄⁻ is from +7 to +2 (becoming Mn²⁺).

By applying these principles, we achieve the equation: 5SO₂ + 2KMnO₄ + 2H₂O → 2MnSO₄ + K₂SO₄ + 2H₂SO₄. This balanced form assures that all elements and charges align on both sides of the equation.

Sulfur Dioxide Reducing Agent

Sulfur dioxide ( SO₂ ) is a common reducing agent in chemical reactions. It donates electrons to other species, causing them to reduce. The sulfur typically undergoes oxidation itself, in this case transforming from SO₂ to SO₄²⁻ . This change from +4 to +6 oxidation state involves a loss of electrons by sulfur, referred to as oxidation. As it interacts with powerful oxidizing agents like KMnO₄, K₂Cr₂O₇, and metal cations like mercury,

• SO₂ helps other elements achieve reduction (gain of electrons).
• It undergoes its own oxidation simultaneously.

This dual role is vital in the stoichiometry and balancing of redox reactions, where the reducing agent facilitates the reduction of other components.

KMnO4 Reduction

Within redox reactions involving potassium permanganate ( KMnO₄ ), MnO₄⁻ ions act as strong oxidizing agents due to manganese's high oxidation state of +7 . When SO₂ reduces KMnO₄, the manganese ions are reduced to Mn²⁺ in MnSO₄, shifting the manganese oxidation state from +7 to +2. This process is detailed as follows:

• Manganese gains electrons during the reaction, thus undergoing reduction.
• The purple color of MnO₄⁻ diminishes as it forms pale pink-colored Mn²⁺ ions.

This color change is often indicative of the reaction's progress. By balancing all components in an acidic medium, we get the full balanced reaction equation: 5SO₂ + 2KMnO₄ + 2H₂O → 2MnSO₄ + K₂SO₄ + 2H₂SO₄.

Potassium Dichromate Reduction

K₂Cr₂O₇, known as potassium dichromate, is another potent oxidizing agent in acidic redox reactions. When reduced by SO₂, the dichromate ions (Cr₂O₇²⁻) convert to Cr³⁺ ions. The chromium oxidation changes from +6 in Cr₂O₇²⁻ to +3 in Cr³⁺, signaling a reduction. Here’s how this plays out:

• SO₂ provides the electrons needed by Cr₂O₇²⁻ to convert into Cr³⁺.
• As it gains electrons, the solution may shift in color due to the formation of trivalent chromium.

Following balance under acidic conditions, the equation culminates to: 3SO₂ + K₂Cr₂O₇ + H₂SO₄ → Cr₂(SO₄)₃ + K₂SO₄ + H₂O. Both elemental and charge balancing are maintained, completing the redox process.

Mercury Reduction

The reduction of mercury(I) nitrate ( Hg₂(NO₃)₂ ) using SO₂ features an interesting transformation. Here, the mercury ion Hg₂²⁺ converts to metallic mercury (Hg⁰), a reduction of oxidation state from +1 (per Hg atom in Hg₂²⁺) to 0 . This happens as:

• Each Hg atom in Hg₂²⁺ gains electrons.
• Metallic mercury forms, appearing as small beads or droplets in solution.

Sulfur dioxide acts as the reducing agent, undergoing oxidation, while assisting other elements in reduction. Balancing reveals: SO₂ + 2Hg₂(NO₃)₂ + 2H₂O → 4Hg + 2HNO₃ + 2H₂SO₄. This showcases mercury's transition to a stable, lower oxidation state.