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Chapter 5: Problem 100

Briefly describe (a) half-equation method of balancing redox equations; (b) disproportionation reaction; (c) titration; (d) standardization of a solution.

Short Answer

Expert verified
A: The half-equation method breaks a redox reaction into oxidation and reduction half-reactions, which are individually balanced before getting combined into a final balanced reaction. B: A disproportionation reaction is one where an element in a molecule gets both oxidized and reduced, forming two or more different products. C: Titration is an analysis method used to determine the concentration of an unknown solution using a solution of known concentration. D: Standardization of a solution is a process used to determine the exact concentration of a solution through titration.

Step by step solution

01

A: Half-equation Method of Balancing Redox Equations

In this method, the redox equation is split into two half-reactions - one for oxidation, and one for reduction. Each of these half-reactions is balanced separately, and then combined to give the balanced redox equation. Steps include: 1. Divide into half-reactions, 2. Balance atoms other than O and H in each half-reaction, 3. Balance O atoms by adding H2O, 4. Balance H by adding H+, 5. Balance charge by adding electrons, 6. Make electron loss equal electron gain, 7. Combine half-reactions and simplify.
02

B: Disproportionation Reaction

Disproportionation Reaction is a chemical reaction where a molecule is transformed into two or more different products, with the molecule itself getting both oxidised and reduced. Here, the same element is simultaneously oxidized and reduced. For example, the reaction 2 H2O2 --> 2 H2O + O2 is a disproportionation reaction, where oxygen in hydrogen peroxide (H2O2) is both reduced to H2O and oxidized to O2.
03

C: Titration

Titration is a procedure in which a solution – the titrant – is used to analyse another solution – the analyte. Usually, one uses a buret to add the titrant to the analyte until the reaction between them is complete; the point at which this occurs is called the end point. An indicator is used to visually signal the end point of the reaction.
04

D: Standardization of a Solution

Standardization of a solution in chemistry refers to the process of determining the exact concentration of a solution. This is accomplished by titrating a solution of known concentration with the solution of unknown concentration until the reaction is complete, as signaled by a color change from an appropriate indicator.

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Most popular questions from this chapter

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) \(\mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow\) (b) \(\mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow\) (c) \(\operatorname{FeS}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow\) (d) \(\mathrm{K}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow\) (e) \(\mathrm{Mg}(\mathrm{s})+\mathrm{H}^{+} \longrightarrow\)

Which of the following aqueous solutions has the highest concentration of \(\mathrm{K}^{+}\) ? (a) \(0.0850 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4};\) (b) a solution containing \(1.25 \mathrm{g} \mathrm{KBr} / 100 \mathrm{mL} ;\) (c) a solution having \(8.1 \mathrm{mg} \mathrm{K}^{+} / \mathrm{mL}\).
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