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Of the following, the two solids that will liberate \(\mathrm{Cl}_{2}(\mathrm{g})\) when heated with \(\mathrm{HCl}(\mathrm{aq})\) are \((\mathrm{a}) \mathrm{NaCl}(\mathrm{s})\) (b) \(\mathrm{ZnCl}_{2}(\mathrm{s}) ;(\mathrm{c}) \mathrm{MnO}_{2}(\mathrm{s}) ;(\mathrm{d}) \mathrm{CuO}(\mathrm{s}) ;(\mathrm{e}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s})\) (f) \(\mathrm{NaOH}(\mathrm{s})\)

Short Answer

Expert verified
The two solids that will liberate \(Cl_{2}(g)\) when heated with \(HCl(aq)\) are \(MnO_{2}(s)\) and \(K_{2}Cr_{2}O_{7}(s)\).

Step by step solution

01

Use Reactivity Series

Firstly, we need to use the reactivity series or electrochemical series. In the electrochemical series, the displacement of Chlorine would occur by elements that are above it in the series. So, from the given options, \(MnO_{2}(s)\) and \(K_{2}Cr_{2}O_{7}(s)\) can displace Chlorine as Manganese and Chromium are above Chlorine in the reactivity series.
02

Formulate the Reactions

Let's illustrate this with balanced chemical equations. For \(MnO_{2}(s)\), the reaction with \(HCl(aq)\) can be written as: \(4HCl(aq) + MnO_{2}(s) \rightarrow MnCl_{2}(aq) + 2H_{2}O(l) + Cl_{2}(g)\). Similarly, for \(K_{2}Cr_{2}O_{7}(s)\), the reaction with \(HCl(aq)\) can be written as: \(14HCl(aq) + K_{2}Cr_{2}O_{7}(s) \rightarrow 2KCl(aq) + 2CrCl_{3}(s) + 7H_{2}O(l) + 3Cl_{2}(g)\). In both reactions, \(Cl_{2}(g)\) is liberated and therefore, the solids \(MnO_{2}(s)\) and \(K_{2}Cr_{2}O_{7}(s)\) meet the condition imposed by the question.

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Key Concepts

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

Chemical Reactions
Chemical reactions describe how substances interact to form new compounds. In this exercise, we see a couple of reactions with hydrochloric acid (HCl) where chlorine gas (Cl\((g)\)) is liberated. Let's break this down. A chemical reaction involves reactants, like manganese(IV) oxide (MnO\(_2\)) or potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)), and products, like manganese(II) chloride or chromium(III) chloride alongside chlorine gas and water.
  • A balanced reaction ensures mass conservation, meaning the number of atoms for each element stays the same before and after the reaction.
  • Manganese and chromium, from the reactants, transform during the interaction, producing chlorine gas, a significant change.
When you combine MnO\(_2\) with HCl, manganese chloride, water, and chlorine gas form. Similarly, potassium dichromate with HCl yields potassium chloride, chromium chloride, water, and chlorine gas. These reactions are redox in nature, involving oxidation of Cl\(^-\) to Cl\(_2\) and reduction of manganese or chromium.
Chlorine Liberation
Chlorine liberation happens when Cl\(_2\) gas is released during a chemical reaction. This phenomenon is linked to the substances' placement within the reactivity series. Here, both MnO\(_2\) and K\(_2\)Cr\(_2\)O\(_7\) have the potential to liberate chlorine from hydrochloric acid due to their relative reactivity positions.
  • Manganese and chromium are more reactive compared to chlorine in these reactions.
  • This higher reactivity allows them to help displace and release chlorine gas.
Understanding chlorine liberation is crucial for numerous industrial and chemical processes. For example, chlorine is essential in water purification and the production of plastics. When MnO\(_2\) and K\(_2\)Cr\(_2\)O\(_7\) interact with HCl, they create an environment where the chlorine ions (Cl\(^-\)) are oxidized to chlorine gas (Cl\(_2\)). This transition marks the liberation of chlorine, mainly observed when Cl\(_2\) bubbles form and escape into the atmosphere.
Electrochemical Series
The electrochemical series, or reactivity series, is a chart of metals based on their reactivity. This concept predicts the outcome of displacement reactions, such as those needed to determine which compounds can liberate chlorine gas. The series orders substances from most reactive to least reactive.
  • Reactivity indicates how easily an element will lose or gain electrons in a reaction.
  • Highly reactive elements, like manganese and chromium, displace less reactive ones, like chlorine.
In our scenario, MnO\(_2\) and K\(_2\)Cr\(_2\)O\(_7\) are capable of liberating chlorine from HCl because manganese and chromium lie above chlorine in the series. As a general rule, elements higher up in the series can displace those below them. This principle helps chemists predict reaction behavior and conditions for chlorine liberation. By considering the elemental rankings, they can choose suitable reactants for specific processes, such as bleaching or disinfecting, where chlorine generation is desired. Understanding the electrochemical series is key to mastering control over chemical reactions in various fields.

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Most popular questions from this chapter

Write plausible half-equations to represent each of the following in basic solution. (a) oxidation of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) to \(\mathrm{FeO}_{4}^{2-}\) (b) reduction of \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) to silver metal

What products are obtained when \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}^{3+}(\mathrm{aq})\) are each treated with a limited amount of NaOH(aq)? With an excess of \(\mathrm{NaOH}(\) aq)? Why are the results different in these two cases?A certain steel is to be analyzed for \(\mathrm{Cr}\) and \(\mathrm{Mn}\). By suitable treatment, the Cr in the steel is oxidized to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and the \(\mathrm{Mn}\) to \(\mathrm{MnO}_{4}(\mathrm{aq}) . \mathrm{A} 10.000 \mathrm{g}\) sample of steel is used to produce \(250.0 \mathrm{mL}\) of a solution containing \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) . \mathrm{A} 10.00 \mathrm{mL}\) portion of this solution is added to \(\mathrm{BaCl}_{2}(\mathrm{aq}),\) and by proper adjustment of the \(\mathrm{pH}\), the chromium is completely precipitated as \(\mathrm{BaCrO}_{4}(\mathrm{s}) ; 0.549 \mathrm{g}\) is obtained. A second \(10.00 \mathrm{mL}\) portion of the solution requires exactly \(15.95 \mathrm{mL}\) of \(0.0750 \mathrm{M} \mathrm{Fe}^{2+}(\mathrm{aq})\) for its titration in acidic solution. Calculate the \(\%\) Cr and \% \(\mathrm{Mn}\) in the steel sample. [Hint: In the titration \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) is reduced to \(\mathrm{Mn}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) is reduced to \(\left.\mathrm{Cr}^{3+}(\mathrm{aq}) ; \text { the } \mathrm{Fe}^{2+}(\mathrm{aq}) \text { is oxidized to } \mathrm{Fe}^{3+}(\mathrm{aq}) \cdot\right]\)

Complete and balance the following equations. If no reaction occurs, so state. (a) \(\operatorname{TiCl}_{4}(\mathrm{g})+\mathrm{Na}(1) \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{Ag}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) (d) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+\mathrm{KOH}(\mathrm{aq}) \longrightarrow\) (e) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\)

You are given these three reducing agents: \(\mathrm{Zn}(\mathrm{s})\) \(\mathrm{Sn}^{2+}(\mathrm{aq}),\) and \(\mathrm{I}^{-}(\mathrm{aq}) .\) Use data from Appendix \(\mathrm{D}\) to determine which of them can, under standard-state conditions in acidic solution, reduce (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) to \(\mathrm{Cr}^{3+}(\mathrm{aq})\) (b) \(\mathrm{Cr}^{3+}(\text { aq })\) to \(\mathrm{Cr}^{2+}(\mathrm{aq})\) (c) \(\mathrm{SO}_{4}^{2-}(\text { aq })\) to \(\mathrm{SO}_{2}(\mathrm{g})\)

By means of a chemical equation, give an example to represent the reaction of (a) a transition metal with a nonoxidizing acid; (b) a transition metal oxide with \(\mathrm{NaOH}(\mathrm{aq}) ;(\mathrm{c})\) an inner transition metal with \(\mathrm{HCl}(\mathrm{aq})\).

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