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In the reaction : \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (a) \(\mathrm{I}_{2}\) is reducing agent. (b) \(I_{2}\) is oxidising agent and \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is reducing agent. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is oxidising agent. (d) \(\mathrm{I}_{2}\) is reducing agent and \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is oxidising agent.

Short Answer

Expert verified
Iodine (\(I_{2}\) is the oxidizing agent and thiosulfate (\(S_{2} O_{3}^{2-}\) is the reducing agent, so the correct answer is (b).

Step by step solution

01

Understand the Concept of Oxidizing and Reducing Agents

An oxidizing agent is a substance that gains electrons in a chemical reaction and is reduced. A reducing agent is a substance that loses electrons in a chemical reaction and is oxidized.
02

Determine the oxidation changes for each element

Compare the oxidation states of iodine (I) and sulfur (S) in reactants and products to determine any changes in oxidation state. In the reactant \(\mathrm{I}_{2}\), the oxidation state of iodine is 0, and in the product \(2 \mathrm{I}^{-}\), it is -1. For sulfur in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), the average oxidation state is +2, and in \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\), it is also +2.
03

Identify the Oxidizing and Reducing Agents

Since iodine is going from 0 to -1, it is gaining electrons and is acting as an oxidizing agent. The sulfur's oxidation state doesn't change, meaning it is losing electrons and acting as a reducing agent.
04

Evaluate the Options

Based on the changes in oxidation states, \(\mathrm{I}_{2}\) acts as an oxidizing agent because it gains electrons, and \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) acts as a reducing agent because it loses electrons. We can disregard options (a) and (c) because \(\mathrm{I}_{2}\) is not reducing, and \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is not oxidizing. We can also disregard option (d) because it incorrectly identifies \(\mathrm{I}_{2}\) as reducing. Therefore, option (b) is the correct answer.

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes that transform one set of chemical substances into another. They occur when atoms and molecules, known as reactants, interact to form new compounds, known as products. Reactions can be simple, involving the exchange or sharing of electrons between atoms, or complex, involving multiple steps and intermediate compounds.

Understanding chemical reactions is essential for grasping the various changes that substances undergo. For instance, combustion is a type of chemical reaction where a substance reacts with oxygen to release energy in the form of heat and light. Synthesis, decomposition, single replacement, and double replacement are other common types of chemical reactions. Each type follows specific patterns of reactants and products, showcasing how atoms and molecules rearrange during these transformations.

The study of chemical reactions also includes analyzing reaction rates, which is how fast a reaction proceeds, and reaction mechanisms, which are the step-by-step sequences of elementary steps that describe how a reaction occurs at the molecular level.
Oxidation States
Oxidation states, often referred to as oxidation numbers, are theoretical numbers assigned to an element in a chemical compound that represent the number of electrons lost or gained by an atom of that element in the compound. The oxidation state is used to keep track of how the distribution of electrons among the atoms changes in a reaction.

In an elemental form, atoms have an oxidation state of 0 because they have not gained or lost electrons. When elements form compounds, their oxidation states can change. Metals, for instance, often have positive oxidation states because they tend to lose electrons. Non-metals may have negative oxidation states, reflecting their tendency to gain electrons.

Identifying oxidation states is critical in the study of chemical reactions, particularly redox reactions. It helps chemists determine which atoms are oxidized and which are reduced during reactions. To calculate the oxidation state, one must apply specific rules, including the idea that the sum of the oxidation states for all atoms in a neutral molecule must equal zero, and in an ion, it must equal the charge of the ion.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between two species. These reactions are crucial in many natural and industrial processes, such as cellular respiration, photosynthesis, batteries, and corrosion.

An oxidizing agent, or oxidant, accepts electrons in a redox reaction and by doing so, becomes reduced. Conversely, a reducing agent, or reductant, donates electrons and becomes oxidized during the reaction. Understanding these agents is fundamental for solving redox reactions, as they determine the flow of electrons.

Oxidation and reduction occur simultaneously in a redox reaction, which means that one substance cannot be oxidized unless another is reduced. An easy way to remember this concept is the mnemonic 'LEO the lion says GER': 'Lose Electrons Oxidize, Gain Electrons Reduce.' Recognizing the changes in oxidation states helps explain the outcomes of redox processes, making it possible to predict which substance will be oxidized and which will be reduced.

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Most popular questions from this chapter

If \(\mathrm{Mn}^{3+}\) ions are unstable in solution and undergo disproportionation to give \(\mathrm{Mn}^{2+}, \mathrm{MnO}_{2}\) and \(\mathrm{H}^{+}\) ions. What will be the balanced equation for the reaction? (a) \(3 \mathrm{Mn}^{3+}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{Mn}^{2+}+8 \mathrm{H}^{+}\) (b) \(\mathrm{Mn}^{3+}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+4 \mathrm{H}^{+}\) (c) \(\mathrm{Mn}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+4 \mathrm{H}^{+}\) (d) \(2 \mathrm{Mn}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}^{+}\)

Oxidation numbers of \(\mathrm{Mn}\) in its compounds \(\mathrm{MnCl}_{2}\) \(\mathrm{Mn}(\mathrm{OH})_{3}, \mathrm{MnO}_{2}\) and \(\mathrm{KMnO}_{4}\) respectively are (a) \(+2,+4,+7,+3\) (b) \(+2,+3,+4,+7\) (c) \(+7,+3,+2,+4\) (d) \(+7,+4,+3,+2\)

What will be the products of electrolysis of an aqueous solution of \(\mathrm{AgNO}_{3}\) with silver electrodes? (a) Ag from Ag anode dissolves while \(\mathrm{Ag}^{+}\)from solution gets deposited on cathode. (b) \(\mathrm{Ag}\) is liberated at cathode and \(\mathrm{O}_{2}\) is liberated at anode. (c) \(\mathrm{Ag}\) at cathode and nitric acid at anode is liberated. (d) No reaction takes place.

Which of the following is true about the given redox reaction? $$ \mathrm{SnCl}_{2}+2 \mathrm{FeCl}_{3} \rightarrow \mathrm{SnCl}_{4}+2 \mathrm{FeCl}_{2} $$ (a) \(\mathrm{SnCl}_{2}\) is oxidised and \(\mathrm{FeCl}_{3}\) acts as oxidising agent. (b) \(\mathrm{FeCl}_{3}\) is oxidised and acts as oxidising agent. (c) \(\mathrm{SnCl}_{2}\) is reduced and acts as oxidising agent. (d) \(\mathrm{FeCl}_{3}\) is oxidised and \(\mathrm{SnCl}_{2}\) acts as a oxidising agent.

Which of the following is the strongest oxidizing agent? (a) \(\mathrm{F}_{2}\) (b) \(\mathrm{Cl}_{2}\) (c) \(\mathrm{Br}_{2}\) (d) \(\mathrm{I}_{2}\)

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