/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 An aqueous solution of \(\mathrm... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 44

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.

Short Answer

Expert verified
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 by step solution

01

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:\[5SO_2 + 2KMnO_4 + 2H_2O \rightarrow 2MnSO_4 + K_2SO_4 + 2H_2SO_4\]
02

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:\[3SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O\]
03

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:\[SO_2 + 2Hg_2(NO_3)_2 + 2H_2O \rightarrow 4Hg + 2HNO_3 + 2H_2SO_4\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈦) 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 SO鈧 reducing KMnO鈧 where,
  • The oxidation of S in SO鈧 is from +4 to +6 (becoming SO鈧劼测伝).
  • The reduction of Mn in MnO鈧勨伝 is from +7 to +2 (becoming Mn虏鈦).
By applying these principles, we achieve the equation: 5SO鈧 + 2KMnO鈧 + 2H鈧侽 鈫 2MnSO鈧 + K鈧係O鈧 + 2H鈧係O鈧. This balanced form assures that all elements and charges align on both sides of the equation.
Sulfur Dioxide Reducing Agent
Sulfur dioxide ( SO鈧 ) 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 SO鈧 to SO鈧劼测伝 . 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 KMnO鈧, K鈧侰r鈧侽鈧, and metal cations like mercury,
  • SO鈧 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 ( KMnO鈧 ), MnO鈧勨伝 ions act as strong oxidizing agents due to manganese's high oxidation state of +7 . When SO鈧 reduces KMnO鈧, the manganese ions are reduced to Mn虏鈦 in MnSO鈧, 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 MnO鈧勨伝 diminishes as it forms pale pink-colored Mn虏鈦 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: 5SO鈧 + 2KMnO鈧 + 2H鈧侽 鈫 2MnSO鈧 + K鈧係O鈧 + 2H鈧係O鈧.
Potassium Dichromate Reduction
K鈧侰r鈧侽鈧, known as potassium dichromate, is another potent oxidizing agent in acidic redox reactions. When reduced by SO鈧, the dichromate ions (Cr鈧侽鈧嚶测伝) convert to Cr鲁鈦 ions. The chromium oxidation changes from +6 in Cr鈧侽鈧嚶测伝 to +3 in Cr鲁鈦, signaling a reduction. Here鈥檚 how this plays out:
  • SO鈧 provides the electrons needed by Cr鈧侽鈧嚶测伝 to convert into Cr鲁鈦.
  • 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: 3SO鈧 + K鈧侰r鈧侽鈧 + H鈧係O鈧 鈫 Cr鈧(SO鈧)鈧 + K鈧係O鈧 + H鈧侽. Both elemental and charge balancing are maintained, completing the redox process.
Mercury Reduction
The reduction of mercury(I) nitrate ( Hg鈧(NO鈧)鈧 ) using SO鈧 features an interesting transformation. Here, the mercury ion Hg鈧偮测伜 converts to metallic mercury (Hg鈦), a reduction of oxidation state from +1 (per Hg atom in Hg鈧偮测伜) to 0 . This happens as:
  • Each Hg atom in Hg鈧偮测伜 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: SO鈧 + 2Hg鈧(NO鈧)鈧 + 2H鈧侽 鈫 4Hg + 2HNO鈧 + 2H鈧係O鈧. This showcases mercury's transition to a stable, lower oxidation state.

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Most popular questions from this chapter

Identify the following hydrides as ionic, metallic, or molecular: (a) \(\mathrm{H}_{2} \mathrm{~S},(\mathbf{b}) \mathrm{LiH},(\mathbf{c}) \mathrm{VH}_{0.56}\)

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Write a balanced chemical reaction for the condensation reaction between \(\mathrm{H}_{3} \mathrm{PO}_{4}\) molecules to form \(\mathrm{H}_{6} \mathrm{P}_{4} \mathrm{O}_{13}\).

Account for the following observations: (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) are quite basic.

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\). (e) Hydrazine reduces \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}\) in base (hydrazine is oxidized to \(\mathrm{N}_{2}\) ).
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