Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Challenge Identify the oxidizing agent and the reducing agent in each reaction. a. \(M g+I_{2} \rightarrow M g l_{2}\) b. \(H_{2} S+C l_{2} \rightarrow S+2 H C l\)

Short Answer

Expert verified
For reaction a (\(Mg + I_2 \rightarrow MgI_2\)), the reducing agent is \(Mg\) and the oxidizing agent is \(I_2\). For reaction b (\(H_2S + Cl_2 \rightarrow S + 2HCl\)), the reducing agent is \(H_2S\) and the oxidizing agent is \(Cl_2\).

Step by step solution

01

Determine the oxidation states of the elements in the reactants and the products

Determine the oxidation states of the different elements in the reactants and products: \(Mg\) is in its elemental state and has an oxidation state of 0. \(I_2\) is in its elemental state and has an oxidation state of 0. In \(MgI_2\), \(Mg\) has an oxidation state of +2 and each \(I\) has an oxidation state of -1. Now let's compare the changes in oxidation state for each element.
02

Identify the changes in oxidation states and their roles as oxidizing or reducing agents

Analyze the changes in oxidation states for each element: \(Mg\): 0 → +2 (oxidation) \(I\): 0 → -1 (reduction) \(Mg\) loses electrons and undergoes oxidation, which means that \(Mg\) is the reducing agent in this reaction. On the other hand, \(I_2\) gains electrons and undergoes reduction, so \(I_2\) is the oxidizing agent. For reaction b: \(H_2S + Cl_2 \rightarrow S + 2HCl\)
03

Determine the oxidation states of the elements in the reactants and the products

Determine the oxidation states of the different elements in the reactants and products: In \(H_2S\), each \(H\) has an oxidation state of +1 and \(S\) has an oxidation state of -2. In \(Cl_2\), both \(Cl\) atoms have an oxidation state of 0. In the products, \(S\) is in its elemental state and has an oxidation state of 0. In \(HCl\), each \(H\) has an oxidation state of +1 and each \(Cl\) has an oxidation state of -1. Now let's compare the changes in oxidation state for each element.
04

Identify the changes in oxidation states and their roles as oxidizing or reducing agents

Analyze the changes in oxidation states for each element: \(S\): -2 → 0 (oxidation) \(Cl\): 0 → -1 (reduction) \(S\) loses electrons and undergoes oxidation, which means that \(H_2S\) is the reducing agent in this reaction. On the other hand, \(Cl_2\) gains electrons and undergoes reduction, so \(Cl_2\) is the oxidizing agent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Oxidation States
Understanding oxidation states is fundamental in identifying the roles substances play in redox reactions. An oxidation state, often referred to as the oxidation number, is a measure of the degree of oxidation of an atom in a chemical compound. It represents the number of electrons that an atom either gains or loses when it forms a chemical bond.

In essence, it’s like the atom’s 'charge' if all the bonds to it were completely ionic. For example, in a molecule like water (H2O), oxygen has an oxidation state of -2, while hydrogen has an oxidation state of +1. These numbers help us track electrons during a reaction which is crucial for understanding redox processes.

Oxidation State Changes

When a substance moves from a neutral element to a part of a compound, as we saw in the exercises with magnesium (Mg) and iodine (I2), their oxidation states change. Magnesium starts at 0 and goes to +2, losing electrons, and iodine goes from 0 to -1, each atom gaining one electron. This gives us a clear picture of the reaction’s flow of electrons.
Oxidizing and Reducing Agents
In the dance of redox reactions, oxidizing and reducing agents are the key partners. An oxidizing agent takes electrons from another substance, therefore it is 'reduced'. Conversely, a reducing agent donates electrons to another substance, and thus it is 'oxidized'. These agents are always found in pairs; one cannot exist without the other in a redox reaction.

Role Identification

To spot these agents, look at the changes in oxidation states. For instance, with reaction (a), magnesium (Mg) loses two electrons and increases its oxidation state from 0 to +2 which makes it the reducing agent. Iodine (I2), on the other hand, accepts those electrons, decreasing its oxidation number from 0 to -1, becoming the oxidizing agent. In our second example, sulfur (S) is part of hydrogen sulfide (H2S) and it loses electrons, thus H2S acts as the reducing agent, while chlorine (Cl2) gains those electrons, marking it as the oxidizing agent.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds, leading to new substances. Redox reactions are a special type of chemical reaction where oxidation and reduction occur simultaneously.

Every chemical reaction follows the law of conservation of mass, where the mass of the reactants equals the mass of the products. In a redox process, it's critical to ensure that the number of atoms and the charge are balanced on both sides of the reaction equation. This allows us to accurately represent the physical transactions occurring at the atomic level.

In the textbook examples, when magnesium reacts with iodine, the product is magnesium iodide, and sulfur reacting with chlorine gas produces elemental sulfur and hydrogen chloride. These transformations showcase the principles of redox reactions, with the transfer of electrons leading to significant chemical change.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

State what an oxidation half-reaction shows. What does a reduction half- reaction show?

Identify the oxidizing agent and the reducing agent in each of these redox equations. a. \(N_{2}+3 H_{2} \rightarrow 2 \mathrm{NH}_{3}\) b. \(2 \mathrm{Na}+\mathrm{I}_{2} \rightarrow 2 \mathrm{NaI}\)

Explain why not all oxidation reactions involve oxygen.

Apply The following equations show redox reactions that are sometimes used in the laboratory to generate pure nitrogen gas and pure dinitrogen monoxide gas (nitrous oxide, \(\mathrm{N}_{2} \mathrm{O} )\) $$\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ $$\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ a. Determine the oxidation number of each element in the two equations, and then make diagrams showing the changes in oxidation numbers that occur in each reaction. b. Identify the atom that is oxidized and the atom that is reduced in each of the two reactions. c. Identify the oxidizing and reducing agents in each of the two reactions. d. Write a sentence telling how the electron transfer taking place in these two reactions differs from that taking place here $$2 \mathrm{AgNO}_{3}+\mathrm{Zn} \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}$$

Use the half-reaction method to balance the redox equations. Begin by writing the oxidation and reduction half-reactions. Leave the balanced equation in ionic form. \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{BiO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{Bi}^{2+}(\mathrm{aq})\) (in acid solution)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free