/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Consider the following reaction:... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 9

Consider the following reaction: \(\mathrm{Mg}(s)+\mathrm{SnSO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\mathrm{Sn}(s)\) (a) Which species is oxidized? (b) Which species is reduced? (c) What is the oxidizing agent? (d) What is the reducing agent? (e) Draw a molecular-level picture of what happens at the surface of the magnesium metal.

Short Answer

Expert verified
a) Mg is oxidized. b) Sn is reduced. c) SnSO_4 is the oxidizing agent. d) Mg is the reducing agent. e) Mg atoms on the surface lose electrons and become Mg^2+ ions in the solution, while Sn^2+ ions in the solution gain electrons and deposit as Sn atoms on the Mg surface.

Step by step solution

01

Determine the Oxidation States

Before the reaction, Mg is in its elemental state, so its oxidation number is 0. In SnSO_4, S is in state +6, O is in its usual state of -2, so Sn must be in state +2 to balance the charges. In the products, MgSO_4 contains Mg in state +2, and the elemental Sn is back to a state of 0.
02

Identify Oxidized and Reduced Species

Here, Mg goes from 0 to +2, so it loses electrons and is oxidized. Sn goes from +2 to 0, so it gains electrons and is reduced.
03

Identify Oxidizing and Reducing Agents

The oxidizing agent is the species that is reduced, thus SnSO_4 is the oxidizing agent. The reducing agent is the species that is oxidized, which makes Mg the reducing agent.
04

Draw a Molecular-Level Picture

At the surface of the Mg metal, atoms of Mg are losing electrons to become Mg^2+ ions. These Mg^2+ ions are then leaving the surface of the metal and moving into the solution. As they do this, Sn^2+ ions in the solution are gaining electrons and are depositing as Sn atoms on the surface of the Mg metal.

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 91影视!

Key Concepts

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

Oxidation States
When looking at redox reactions, oxidation states help us track the transfer of electrons. An oxidation state is a number assigned to an element in a compound to signify how many electrons are gained or lost. For elements in their pure form, like magnesium (Mg) in this exercise, the oxidation state is 0. This is because no electrons are gained or lost in pure elements.

In the reaction given, magnesium starts with an oxidation state of 0 and ends with an oxidation state of +2 in magnesium sulfate (MgSO鈧�). This change indicates that magnesium loses two electrons during the reaction. Similarly, tin (Sn) in tin sulfate (SnSO鈧�) starts with an oxidation state of +2 and changes to 0 in the tin product, signifying a gain of two electrons.

Understanding oxidation states is crucial for identifying redox reactions and determining which species are oxidized or reduced. In essence, the oxidation state helps clarify the movement of electrons in a chemical reaction, making the analysis of complex reactions more manageable.
Oxidizing Agent
In redox reactions, oxidizing agents play an important role by facilitating the oxidation of another substance. An oxidizing agent helps another chemical species lose electrons while it itself gets reduced, gaining electrons. The concept can be understood better by examining its function in a real-life redox reaction.

In the provided reaction, tin sulfate (SnSO鈧�) is the oxidizing agent. This is because tin (Sn) gains electrons as its oxidation state changes from +2 in SnSO鈧� to 0 in elemental tin (Sn). By accepting electrons, SnSO鈧� allows magnesium (Mg) to lose electrons, thus oxidizing the Mg.

These electron transfers are what make the oxidizing agent a vital part of any redox reaction. Without it, the process of oxidation would not be able to occur, highlighting the reciprocal nature of electron transfers in chemical reactions.
Reducing Agent
A reducing agent, on the other hand, is the opposite of an oxidizing agent. It donates electrons to another substance, leading to the reduction of that substance. In doing so, the reducing agent itself undergoes oxidation.

In our reaction example, magnesium (Mg) acts as the reducing agent. It donates two electrons, changing its oxidation state from 0 to +2 in magnesium sulfate (MgSO鈧�). Because of the magnesium's ability to lose electrons, tin (Sn) in the SnSO鈧� compound is able to gain these electrons and is reduced.

These characteristics of reducing agents are fundamental in many chemical reactions, including those that occur in biological systems and industrial processes. By understanding how reducing agents contribute to electron transfers, one can better appreciate the essential nature of redox reactions across various fields of science.

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

The zinc-silver oxide battery, although expensive, is used to power satellite systems because of its light weight. Unbalanced half-reactions that occur in this battery are $$ \begin{gathered} \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s) \\ 2 \mathrm{AgO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ag}_{2} \mathrm{O}(s)+2 \mathrm{OH}^{-}(a q) \end{gathered} $$ Which equation represents the oxidation half-reaction? Which represents the reduction half-reaction? Add electrons to the appropriate side of each equation to account for the change in oxidation number.

How do oxidation numbers change during oxidation and during reduction?

When iodine is oxidized in concentrated \(\mathrm{HNO}_{3}\), it produces the white solid iodic acid, \(\mathrm{HIO}_{3}\) : $$ \mathrm{I}_{2}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{HIO}_{3}(s)+\mathrm{NO}_{2}(g) $$ Balance this equation, adding \(\mathrm{H}^{+}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) as necessary.

The reaction that occurs in gas grills involves the burning of propane: $$ \mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Assign oxidation numbers to all of the reactants and identify the oxidizing and reducing agents.

When a strip of nickel metal is placed in an aqueous solution of lead(II) nitrate, elemental lead coats the surface of the nickel strip and aqueous nickel(II) nitrate forms. (a) Write a balanced equation for this reaction. (b) What happens to the charge on nickel as products form? How many electrons are transferred? Does nickel gain or lose electrons? (c) What happens to the charge on lead as products form? How many electrons are transferred? Does lead gain or lose electrons?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
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.