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Chapter 21: Problem 10

\text { Which of the following is not oxidized by } \mathrm{O}_{3} \text { ? }(a) KI (b) \(\mathrm{FeSO}_{4}\) (c) \(\mathrm{KMnO}_{4}\) (d) \(\mathrm{K}_{2} \mathrm{MnO}_{4}\)

Short Answer

Expert verified
(c) KMnO4 cannot be oxidized by ozone because it already has its manganese in the highest oxidation state.

Step by step solution

01

Understand Oxidation by Ozone

Ozone \((\mathrm{O}_3)\) is a powerful oxidizing agent. It can oxidize substances by adding oxygen or removing electrons from them. Identifying which substances can be oxidized involves knowing their chemical nature and their reaction capabilities with oxidants.
02

Analyze Each Option

We need to understand the chemical nature of each compound to determine its likelihood of being oxidized by ozone. (a) \(\text{KI}\) - Potassium iodide is a salt that can be oxidized because iodine has an intermediate oxidation state.(b) \(\mathrm{FeSO}_4\) - Ferrous sulfate contains iron in the +2 oxidation state, which can be further oxidized to the +3 state.(c) \(\mathrm{KMnO}_4\) - Potassium permanganate already has manganese in the +7 oxidation state, which is the highest oxidation state for manganese.(d) \(\mathrm{K}_2 \mathrm{MnO}_4\) - Potassium manganate has manganese in the +6 oxidation state and can be oxidized to a higher state.
03

Identify Which Cannot Be Oxidized by Ozone

Since \(\mathrm{KMnO}_4\) already has manganese in the highest possible oxidation state (+7) for manganese, it cannot be oxidized further. However, \(\text{KI}\), \(\mathrm{FeSO}_4\), and \(\mathrm{K}_2 \mathrm{MnO}_4\) have the potential to be oxidized by ozone.

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Key Concepts

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

Ozone as an oxidizing agent
Ozone Ozone Indirect Effects - Control pollutants: Ozone reacts with VOCs and nitrogen oxides in the air to control pollution. - Industrial synthesis: Ozone is used to synthesize chemicals and purify water due to its reactivity.
Potassium permanganate oxidation state
The oxidation state is a concept used to describe the degree of oxidation of an element within a compound. Specifically, it indicates how many electrons have been lost (or gained) by an atom compared to its elemental form. In the case of potassium permanganate Manganese, within the permanganate ion This higher oxidation state implies that KMnO_4 is already maximally oxidized, making it highly unlikely, if not impossible, to be further oxidized by an oxidizing agent such as ozone. This property makes potassium permanganate a very effective oxidizer itself in various chemical reactions. - KMnO_4 has manganese at the +7 oxidation state. - It is a stable compound because manganese can't lose more electrons. - KMnO_4 is often used as an oxidizing agent in industrial and laboratory settings.
Chemical reactivity with ozone
The chemical reactivity of a substance with ozone depends on its ability to either donate electrons or accept oxygen atoms. In chemical terms, reactions involving ozone are focused on two main actions: - Electron transfer: In some cases, a substance can be oxidized by losing electrons to the strongly oxidizing ozone molecule. Ozone can easily accept these electrons as it breaks apart to form oxygen gas and atomic oxygen. - Oxygen transfer: Ozone can also oxidize substances by transferring an oxygen atom to them. Examples of chemical reactivity with ozone: - Potassium iodide (KI): Iodine in KI can be oxidized from its -1 oxidation state to a +5 oxidation state by ozone, forming iodine oxide. - Ferrous sulfate ( It can be oxidized to ferric sulfate, where iron is in the +3 oxidation state. - Potassium manganate (K_2MnO_4): Although the manganese in K_2MnO_4 is at a high oxidation state (+6), it can still be oxidized by ozone to form permanganate, where manganese is at +7. Thus, substances that can either lose electrons or gain additional oxygen atoms tend to be reactive towards ozone, unlike potassium permanganate which cannot be further oxidized.

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Most popular questions from this chapter

Among the following, the correct statement is (a) Phosphates have no biological significance in humans (b) Between nitrates and phosphates, phosphates are less abundant in earth's crust (c) Between nitrates and phosphates, nitrates are less abundant in earth's crust (d) Oxidation of nitrates is possible in soil

The order of the oxidation state of the phosphorus atom in \(\mathrm{H}_{3} \mathrm{PO}_{2}\), \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) is [Adv. 2017] (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}\) (d) \(\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{A} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{2} \mathrm{PO}_{4}\)

Nitrogen(I) oxide is produced by:(a) thermal decomposition of ammonium nitrate (b) disproportionation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (c) thermal decomposition of ammonium nitrite (d) interaction of hydroxylamine and nitrous acid.

A metal ' \(\mathrm{M}\) ' reacts with nitrogen gas to afford ' \(\mathrm{M}_{3} \mathrm{~N}^{\prime}\). ' \(\mathrm{M}_{3} \mathrm{~N}\) ' on heating at high temperature gives back ' \(\mathrm{M}\) ' and on reaction with water produces a gas ' \(\mathrm{B}\) '. Gas ' \(\mathrm{B}\) ' reacts with aqueous solution of \(\mathrm{CuSO}_{4}\) to form a deep blue compound. 'M' and 'B' respectively are : [Main Online April 8, 2017] (a) \(\mathrm{Li}\) and \(\mathrm{NH}_{3}\) (b) \(\mathrm{Ba}\) and \(\mathrm{N}_{2}\) (c) \(\mathrm{Na}\) and \(\mathrm{NH}_{3}\) (d) \(\mathrm{Al}\) and \(\mathrm{N}_{2}\)

A colorless aqueous solution contains nitrates of two metals, \(\mathbf{X}\) and \(\mathbf{Y}\). When it was added to an aqueous solution of \(\mathrm{NaCl}\), a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue \(\mathbf{P}\) and a solution \(\mathbf{Q}\). The residue \(\mathbf{P}\) was soluble in aq. \(\mathrm{NH}_{3}\) and also in excess sodium thiosulfate. The hot solution \(\mathbf{Q}\) gave a yellow precipitate with KI. The metals \(\mathbf{X}\) and \(\mathbf{Y}\), respectively, are [Adv. 2020](a) \(\mathrm{Ag}\) and \(\mathrm{Pb}\) (b) \(\mathrm{Ag}\) and \(\mathrm{Cd}\) (c) \(\mathrm{Cd}\) and \(\mathrm{Pb}\) (d) \(\mathrm{Cd}\) and \(\mathrm{Zn}\)
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