/*! 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 50 An aqueous solution of \(\mathrm... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 50

An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q)\), (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
The balanced equations for the given reactions are: (a) \(5 \mathrm{SO}_{2}+2 \mathrm{KMnO}_{4}+14 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O}\) (b) \(3\mathrm{SO}_{2}+\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}+16\mathrm{H}^{+} \rightarrow 3\mathrm{SO}_{3}^{2-}+2\mathrm{Cr}^{3+} +7\mathrm{H}_{2}\mathrm{O}\) (c) \(2\mathrm{SO}_{2}+\mathrm{Hg}_{2}\mathrm{(NO}_{3)\mathrm{)}_{2}+2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{Hg}+2\mathrm{SO}_{3}^{2-}+4\mathrm{H}^{+}+4\mathrm{NO}_{3}^{-}\)

Step by step solution

01

Identify the reactants in each reaction

(a) Aqueous SO鈧� and KMnO鈧� (b) Aqueous SO鈧� and acidic K鈧侰r鈧侽鈧� (c) Aqueous SO鈧� and Hg鈧�(NO鈧�)鈧�
02

Write unbalanced equations for each reaction

(a) SO鈧� + KMnO鈧� 鈫� MnSO鈧� (b) SO鈧� + K鈧侰r鈧侽鈧� 鈫� Cr鲁鈦� (c) SO鈧� + Hg鈧�(NO鈧�)鈧� 鈫� Hg
03

Balance each reaction by adjusting the stoichiometric coefficients and adding appropriate H鈧侽, H鈦�, and OH鈦� species to balance the atoms and charge

(a) Overall half-reactions for KMnO鈧� and SO鈧� are \(5 \mathrm{SO}_{2}+14 \mathrm{H}^{+} +2 \mathrm{MnO}_{4}^{-} \rightarrow 5 \mathrm{SO}_{3}^{2-}+2 \mathrm{Mn}^{2+} +8 \mathrm{H}_{2} \mathrm{O}\) Adding eight water molecules and balancing the charge with fourteen H鈦�, the balanced equation for the reaction between KMnO鈧� and SO鈧� is \[5 \mathrm{SO}_{2}+2 \mathrm{KMnO}_{4}+14 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O}\] (b) Overall half-reactions for K鈧侰r鈧侽鈧� and SO鈧� are \(3\mathrm{SO}_{2}+16\mathrm{H}^{+}+\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow 3\mathrm{SO}_{3}^{2-}+2\mathrm{Cr}^{3+} +7\mathrm{H}_{2}\mathrm{O}\) Adding seven water molecules and balancing the charge with sixteen H鈦�, the balanced equation for the reaction between K鈧侰r鈧侽鈧� and acidic SO鈧� is \[3\mathrm{SO}_{2}+\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}+16\mathrm{H}^{+} \rightarrow 3\mathrm{SO}_{3}^{2-}+2\mathrm{Cr}^{3+} +7\mathrm{H}_{2}\mathrm{O}\] (c) Overall half-reactions for Hg鈧�(NO鈧�)鈧� and SO鈧� are \(2\mathrm{SO}_{2}+2\mathrm{H}_{2}\mathrm{O}+4\mathrm{H}^{+}+\mathrm{Hg}_{2}\mathrm{(NO}_{3)\mathrm{)}_{2} \rightarrow 2\mathrm{Hg}+3\mathrm{SO}_{3}^{2-}+4\mathrm{H}^{+}\) Adding two water molecules and balancing the charge with four H鈦�, the balanced equation for the reaction between Hg鈧�(NO鈧�)鈧� and SO鈧� is \[2\mathrm{SO}_{2}+\mathrm{Hg}_{2}\mathrm{(NO}_{3)\mathrm{)}_{2}+2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{Hg}+2\mathrm{SO}_{3}^{2-}+4\mathrm{H}^{+}+4\mathrm{NO}_{3}^{-}\]

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is crucial in chemistry because it ensures that the number of atoms for each element is equal on both sides of the equation. This reflects the law of conservation of mass. When you balance a chemical equation, you adjust the stoichiometric coefficients, which are the numbers in front of molecules or compounds, to ensure that the atoms balance on both the reactant and product sides.
In the context of redox reactions, balancing becomes a bit more involved because you also need to balance the charges. For example, consider the reaction of sulfur dioxide (\( \mathrm{SO}_2 \)) with potassium permanganate (\( \mathrm{KMnO}_4 \)). Here, you start by writing the unbalanced equation, then add water (\( \mathrm{H}_2\mathrm{O} \)) to balance oxygen atoms, and hydrogen ions (\( \mathrm{H}^+ \)) to balance hydrogen atoms. Finally, you may need to add electrons to balance the charges.
Using these methods accurately balances the equation for all elements involved, ensuring that all atoms and charges are accounted for.
Half-Reactions
Half-reactions are a way of splitting the overall redox reaction into two separate equations 鈥� one for oxidation and one for reduction. This division helps in understanding and balancing redox reactions better.
Oxidation involves the loss of electrons, while reduction involves the gain of electrons. For instance, in the redox reaction between \( \mathrm{SO}_2 \) and \( \mathrm{KMnO}_4 \), sulfur dioxide undergoes oxidation, losing electrons, while \( \mathrm{MnO}_4^- \) is reduced as it gains electrons. By writing separate half-reactions, you get to see visually where electrons are transferred between reactants.
The electrons are then adjusted so that both half-reactions have the same number of electrons to ensure a balanced combined equation. Finally, these two balanced half-reactions are combined to form the full redox equation.
Aqueous Solutions
In many chemical reactions, particularly redox reactions, the reactants and products are often in aqueous solutions. An aqueous solution is simply a solution where water is the solvent. The molecules or ions dissolve in water and become surrounded by water molecules, enabling them to react more readily.
In the given reactions, substances like \( \mathrm{KMnO}_4 \) and \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \) are initially added to water. Once dissolved, they disassociate into their respective ions and can undergo redox reactions with reagents like \( \mathrm{SO}_2 \).
The presence of water also facilitates the balancing of reactions that involve ions. When balancing a reaction in an aqueous environment, water, \( \mathrm{H}^+ \), and \( \mathrm{OH}^- \) ions are often used to balance the equation as they are abundant in such a medium.
Oxidation-Reduction Reactions
Redox reactions, short for oxidation-reduction reactions, are a type of chemical reaction that involves the transfer of electrons between two species. These reactions are characterized by changes in oxidation states of the involved atoms.
In any redox reaction, there is always an oxidizing agent that gains electrons and a reducing agent that loses electrons. For example, in the reaction involving \( \mathrm{SO}_2 \) and \( \mathrm{KMnO}_4 \), \( \mathrm{KMnO}_4 \) acts as the oxidizing agent as it gains electrons, while \( \mathrm{SO}_2 \) is the reducing agent as it loses electrons.
Redox reactions are crucial in various natural and industrial processes, such as metabolism in biological systems and the corrosion of metals. Understanding these reactions requires recognizing changes in electron distribution and balancing the entire process in terms of both mass and charge.

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Most popular questions from this chapter

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) chlorate ion, (b) hydroiodic acid, (c) iodine trichloride, (d) sodium hypochlorite, (e) perchloric acid, (f) xenon tetrafluoride.

Carbon forms an unusual, unstable oxide of formula \(\mathrm{C}_{3} \mathrm{O}_{2}\) called carbon suboxide. Carbon suboxide is made by using \(\mathrm{P}_{2} \mathrm{O}_{5}\) to dehydrate the dicarboxylic acid called malonic acid, which has the formula \(\mathrm{HOOC}-\mathrm{CH}_{2}-\mathrm{COOH}\). (a) Write a balanced reaction for the production of carbon suboxide from malonic acid. (b) Suggest a Lewis structure for \(\mathrm{C}_{3} \mathrm{O}_{2}\). (Hint: The Lewis structure of malonic acid suggests which atoms are connected to which.) (c) By using the information in Table \(8.5\), predict the \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{C}_{3} \mathrm{O}_{2}\). (d) Sketch the Lewis structure of a product that could result by the addition of \(2 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to 1 mol of \(\mathrm{C}_{3} \mathrm{O}_{2}\)

Removal of perchlorate from water supplies is difficult. Naturally occurring microorganisms are, however, capable of destroying perchlorate in solution in minutes. What do you think might be the type of reaction occurring in the microorganisms, and what do you predict might be the fate of the perchlorate ion in the reaction?

Write a balanced equation for each of the following reactions: (a) preparation of white phosphorus from calcium phosphate, (b) hydrolysis of \(\mathrm{PBr}_{3}\), (c) reduction of \(\mathrm{PBr}_{3}\) to \(\mathrm{P}_{4}\) in the gas phase, using \(\mathrm{H}_{2}\).

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(\mathrm{s})\), reacts with ozone to form \(\mathrm{PbSO}_{4}(\mathrm{~s})\) and \(\mathrm{O}_{2}(\mathrm{~g}) .\) (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to ZnO. (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\).
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