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

Identify the oxidized reactant, the reduced reactant, the oxidizing agent, and the reducing agent in the following reactions: (a) \(\mathrm{Fe}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Cu}(s)\) (b) \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MgCl}_{2}(s)\) (c) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Cr}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\)

Short Answer

Expert verified
(a) Oxidized: Fe, Reduced: Cu²⁺, Oxidizing agent: Cu²⁺, Reducing agent: Fe. (b) Oxidized: Mg, Reduced: Cl₂, Oxidizing agent: Cl₂, Reducing agent: Mg. (c) Oxidized: Al, Reduced: Cr₂O₃, Oxidizing agent: Cr₂O₃, Reducing agent: Al.

Step by step solution

01

Determine oxidation states for Reaction (a)

In reaction (a), \(\mathrm{Fe}(s) + \mathrm{Cu}^{2+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{Cu}(s)\): - Solid iron \((\mathrm{Fe})\) has an oxidation state of 0.- Copper \((\mathrm{Cu}^{2+})\) has an oxidation state of +2.- \(\mathrm{Fe}^{2+}\) becomes +2, meaning it has lost 2 electrons (oxidized).- \(\mathrm{Cu}\) goes from +2 in \(\mathrm{Cu}^{2+}\) to 0, meaning it gains 2 electrons (reduced).
02

Identify components for Reaction (a)

In reaction (a): - **Oxidized reactant** is \(\mathrm{Fe}\) because its oxidation state increases from 0 to +2.- **Reduced reactant** is \(\mathrm{Cu}^{2+}\) because its oxidation state decreases from +2 to 0.- **Oxidizing agent** is \(\mathrm{Cu}^{2+}\) as it gains electrons.- **Reducing agent** is \(\mathrm{Fe}\) as it loses electrons.
03

Determine oxidation states for Reaction (b)

In reaction (b), \(\mathrm{Mg}(s) + \mathrm{Cl_2}(g) \rightarrow \mathrm{MgCl_2}(s)\): - Magnesium \((\mathrm{Mg})\) has an oxidation state of 0.- Chlorine \((\mathrm{Cl_2})\) has an oxidation state of 0.- In \(\mathrm{MgCl_2}\), \(\mathrm{Mg}\) becomes +2 and each \(\mathrm{Cl}\) becomes -1.- \(\mathrm{Mg}\) loses 2 electrons and \(\mathrm{Cl_2}\) gains electrons.
04

Identify components for Reaction (b)

In reaction (b):- **Oxidized reactant** is \(\mathrm{Mg}\) because its oxidation state increases from 0 to +2.- **Reduced reactant** is \(\mathrm{Cl_2}\) because its oxidation state decreases from 0 to -1.- **Oxidizing agent** is \(\mathrm{Cl_2}\) as it gains electrons.- **Reducing agent** is \(\mathrm{Mg}\) as it loses electrons.
05

Determine oxidation states for Reaction (c)

In reaction (c), \(2 \mathrm{Al}(s) + \mathrm{Cr_2 O_3}(s) \rightarrow 2 \mathrm{Cr}(s) + \mathrm{Al_2 O_3}(s)\): - Aluminum \((\mathrm{Al})\) has an oxidation state of 0.- Chromium \((\mathrm{Cr})\) in \(\mathrm{Cr_2O_3}\) is +3 and the oxide ion \((\mathrm{O})\) is -2.- In \(\mathrm{Al_2O_3}\), \(\mathrm{Al}\) is +3.- \(\mathrm{Cr}\) in \(\mathrm{Cr_2O_3}\) becomes 0 in \(\mathrm{Cr}\).
06

Identify components for Reaction (c)

In reaction (c):- **Oxidized reactant** is \(\mathrm{Al}\) because its oxidation state increases from 0 to +3.- **Reduced reactant** is \(\mathrm{Cr_2O_3}\) because \(\mathrm{Cr}\) reduces from +3 to 0.- **Oxidizing agent** is \(\mathrm{Cr_2O_3}\) as it gains electrons.- **Reducing agent** is \(\mathrm{Al}\) as it loses electrons.

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
Oxidation states, also known as oxidation numbers, are theoretical charges assigned to atoms within a molecule to indicate the degree of oxidation or reduction. They provide a way to keep track of electrons during chemical reactions. By determining the oxidation states of atoms in both the reactants and products, one can identify which atoms have gained or lost electrons. This helps to clarify the changes occurring in redox reactions.
  • Oxidation refers to an increase in oxidation state, which corresponds to the loss of electrons.
  • Reduction refers to a decrease in oxidation state, indicating the gain of electrons.
Understanding oxidation states is crucial for correctly identifying the processes and agents in redox reactions.
Oxidizing Agent
An oxidizing agent is a substance that promotes the oxidation of another species during a chemical reaction by gaining electrons itself. In doing so, it undergoes reduction.
In redox reactions, identifying the oxidizing agent involves finding the molecule or ion that has gained electrons and, therefore, experienced a reduction in its oxidation state.
  • Example: In reaction (a), \({\text{Cu}^{2+}}\) serves as the oxidizing agent because it gains 2 electrons, reducing its oxidation state from +2 to 0.
The ability to identify the oxidizing agent is essential for understanding how electron transfer manifests in chemical processes.
Reducing Agent
A reducing agent, or reductant, is the opposite of the oxidizing agent. It donates electrons to another species, facilitating reduction, while itself undergoing oxidation.
In redox reactions, the reducing agent can be recognized as the species whose oxidation state increases, as it loses electrons.
  • For example, in reaction (b), magnesium (Mg) acts as the reducing agent since it loses 2 electrons, increasing its oxidation state from 0 to +2.
Identifying the reducing agent helps to comprehend which components are donating electrons and undergoing oxidation within a reaction.
Electron Transfer
Electron transfer is at the heart of redox reactions. It involves the movement of electrons from one atom or molecule to another, leading to changes in oxidation states and the reconfiguration of bonds.
  • An oxidation process involves the loss of electrons by a species, increasing its oxidation state.
  • A reduction process, on the other hand, involves the gain of electrons, decreasing its oxidation state.
For instance, during the electron transfer in reaction (c), aluminum (Al) loses electrons, while chromium (Cr) gains electrons. This transfer forms the basis of identifying oxidized and reduced species, as well as oxidizing and reducing agents.
Chemical Reactions
Chemical reactions involve the rearrangement of atoms and bonds to form new substances. Redox reactions, a specific type of chemical reaction, focus on the exchange of electrons between reactants.
  • These reactions are essential in processes such as combustion, corrosion, and energy production.
  • Balancing redox reactions includes ensuring that the number of electrons lost equals the number of electrons gained, maintaining charge balance.
Understanding the fundamental principles of chemical reactions, like balancing and electron transfer, offers insights into how substances interact and transform.

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

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l)\) (b) \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \longrightarrow \mathrm{Br}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (c) \(\mathrm{SbCl}_{3}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{5}(s)\)

Iodine, used as an antiseptic agent, can be prepared in the laboratory by the following reaction: \(\begin{array}{l} 2 \mathrm{NaI}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnSO}_{4}(a q)+\mathrm{I}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array}\) (a) Determine the oxidation number for the Mn and I on both sides of the equation. (b) Identify the oxidizing and reducing agents.

Write balanced ionic equations and net ionic equations for the following reactions: (a) Aqueous sulfuric acid is neutralized by aqueous potassium hydroxide. (b) Aqueous magnesium hydroxide is neutralized by aqueous hydrochloric acid.

Which of the following substances are likely to be soluble in water? (a) \(\mathrm{ZnSO}_{4}\) (b) \(\mathrm{NiCO}_{3}\) (c) \(\mathrm{PbCl}_{2}\) (d) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Pepto-Bismol, an antacid and antidiarrheal, contains bismuth subsalicylate, \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}\). Some users of this product can experience a condition known as "black tongue," which is caused by the reaction of bismuth(III) ions with trace amounts of \(\mathrm{S}^{2-}\) in saliva to form a black precipitate. Write the balanced net ionic equation for this precipitation reaction.

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