Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 18, Problem 32QAP
Interpretation Introduction

(a)

Interpretation:

The atoms being oxidized and reduced should be determined.

Concept Introduction:

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

The atom which gains electrons in the reaction that is which gets reduced in the reaction is said to be the oxidizing agent also known as the electron acceptor whereas the atom which loses electrons in the reaction that is which gets oxidized in the reaction is said to be the reducing agent also known as the electron donor.

Expert Solution
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Answer to Problem 32QAP

C being oxidized and Cl being reduced.

Explanation of Solution

Given:

4KClO3(s) + C6H12O6(s)  4KCl(s)+6H2O(l)+6CO2(g)

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

While determining the oxidation state of compound, the element with greater electronegativity is assigned with negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

The oxidation states are determined as:

The oxidation state for KClO3, C6H12O6(s), KCl, andCO2 the oxidation is determined as:

For KClO3 :

The oxidation state of K is assigned as + 1 and O is assigned as -2 and the oxidation state of Cl is assigned as x:

Since, KClO3 has no charge so; the sum of oxidation state must be zero.

so:

1 - x + 3(-2) = 0

x = + 5

For C6H12O6 :

The oxidation state of O is assigned as -2, of H as + 1 and the oxidation state of C is assigned as x:

Since, C6H12O6 has no charge so; the sum of oxidation state must be zero.

so:

6x + 12(1) + 6(-2) = 0

x = 0

For KCl :

The oxidation state of K is assigned as + 1 and the oxidation state of Cl is assigned as x:

Since, KCl has no charge so; the sum of oxidation state must be zero.

so:

1 + (-x) = 0

x = + 1

For CO2 :

The oxidation state of O is assigned as -2 and the oxidation state of C is assigned as x:

Since, CO2 has no charge so; the sum of oxidation state must be zero.

so:

x + 2(-2) = 0

x = + 4

4KClO3(s) + C6H12O6(s)  4KCl(s)+6H2O(l)+6CO2(g)

Oxidation states:+50+1+4

Since, the oxidation state of C increases from 0 to + 4 so, it undergoes oxidation and the oxidation state of Cl decreases from + 5 to + 1 so it undergoes reduction.

Interpretation Introduction

(b)

Interpretation:

The atoms being oxidized and reduced should be determined.

Concept Introduction:

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

The atom which gains electrons in the reaction that is which gets reduced in the reaction is said to be the oxidizing agent also known as the electron acceptor whereas the atom which loses electrons in the reaction that is which gets oxidized in the reaction is said to be the reducing agent also known as the electron donor.

Expert Solution
Check Mark

Answer to Problem 32QAP

C being oxidized and O being reduced.

Explanation of Solution

Given:

2 C8H18(l) +25O2 16CO2(g)+18H2O(l)

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

While determining the oxidation state of compound, the element with greater electronegativity is assigned with negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

The oxidation states are determined as:

The oxidation state for O2 is zero and for C8H18, CO2andH2O the oxidation is determined as:

For C8H18 :

The oxidation state of H is assigned as + 1 and the oxidation state of C is assigned as x:

Since, C8H18 has no charge so; the sum of oxidation state must be zero.

so:

8 x + 18(-1) = 0

x = -2.25

For H2O :

The oxidation state of H is assigned as + 1 and the oxidation state of O is assigned as x:

Since, H2O has no charge so; the sum of oxidation state must be zero.

so:

1(2) + x = 0

x = -2

For CO2 :

The oxidation state of O is assigned as -2 and the oxidation state of C is assigned as x:

Since, CO2 has no charge so; the sum of oxidation state must be zero.

so:

x + 2(-2) = 0

x = + 4

2 C8H18(l) +25O2 16CO2(g)+18H2O(l)

Oxidation states:-2.250+4-2

Since, the oxidation state of C increases from -2.25 to + 4 so, it undergoes oxidation and the oxidation state of O decreases from 0 to -2 so it undergoes reduction.

Interpretation Introduction

(c)

Interpretation:

The atoms being oxidized and reduced should be determined.

Concept Introduction:

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

The atom which gains electrons in the reaction that is which gets reduced in the reaction is said to be the oxidizing agent also known as the electron acceptor whereas the atom which loses electrons in the reaction that is which gets oxidized in the reaction is said to be the reducing agent also known as the electron donor.

Expert Solution
Check Mark

Answer to Problem 32QAP

P being oxidized and Cl being reduced.

Explanation of Solution

Given:

PCl3(g) +Cl2(g) PCl5(g)

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

While determining the oxidation state of compound, the element with greater electronegativity is assigned with negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

The oxidation states are determined as:

The oxidation state for Cl2 is zero and for PCl3 and PCl5 the oxidation is determined as:

For PCl3 :

The oxidation state of Cl is assigned as -1 and the oxidation state of P is assigned as x:

Since, PCl3 has no charge so; the sum of oxidation state must be zero.

so:

x + 3(-1) = 0

x = + 3

For PCl5 :

The oxidation state of Cl is assigned as -1 and the oxidation state of P is assigned as x:

Since, PCl5 has no charge so; the sum of oxidation state must be zero.

so:

x + 5(-1) = 0

x = + 5

PCl3(g) +Cl2(g) PCl5(g)

Oxidation states: + 30 + 5(for P) -1(for O)

Since, the oxidation state of P increases from + 3 to + 5 so, it undergoes oxidation and the oxidation state of Cl decreases from 0 to -1 so it undergoes reduction.

Interpretation Introduction

(d)

Interpretation:

The atoms being oxidized and reduced should be determined.

Concept Introduction:

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

The atom which gains electrons in the reaction that is which gets reduced in the reaction is said to be the oxidizing agent also known as the electron acceptor whereas the atom which loses electrons in the reaction that is which gets oxidized in the reaction is said to be the reducing agent also known as the electron donor.

Expert Solution
Check Mark

Answer to Problem 32QAP

Ca being oxidized and H being reduced.

Explanation of Solution

Given:

Ca(s) +H2(g) CaH2(g)

The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

While determining the oxidation state of compound, the element with greater electronegativity is assigned with negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

The oxidation states are determined as:

The oxidation state for CaandH2 is zero and for CaH2 the oxidation is determined as:

For CaH2 :

The oxidation state of Ca is assigned as + 2 and the oxidation state of H is assigned as x:

Since, CaH2 has no charge so; the sum of oxidation state must be zero.

so:

+2 + 2(x) = 0

x = -1

Ca(s) +H2(g) CaH2(g)

Oxidation states: 0 0 + 2(for Ca) -1(for H)

Since, the oxidation state of Ca increases from 0 to + 2 so, it undergoes oxidation and the oxidation state of H decreases from 0 to -1 so it undergoes reduction.

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Chapter 18 Solutions

Introductory Chemistry: A Foundation

Ch. 18 - Prob. 5ALQCh. 18 - Prob. 6ALQCh. 18 - In balancing oxidation-reduction equations, why is...Ch. 18 - What does it mean for a substance to be oxidized?...Ch. 18 - Label the following parts of the galvanic cell....Ch. 18 - Prob. 1QAPCh. 18 - Prob. 2QAPCh. 18 - For each of the following oxidation-reduction...Ch. 18 - For each of the following oxidation-reduction...Ch. 18 - For each of the following oxidation-reduction...Ch. 18 - Prob. 6QAPCh. 18 - Prob. 7QAPCh. 18 - Prob. 8QAPCh. 18 - Explain why, although it is not an ionic compound,...Ch. 18 - Prob. 10QAPCh. 18 - Prob. 11QAPCh. 18 - Prob. 12QAPCh. 18 - Prob. 13QAPCh. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 15QAPCh. 18 - Prob. 16QAPCh. 18 - . What is the oxidation state of chlorine in each...Ch. 18 - . What is the oxidation state of manganese in each...Ch. 18 - Prob. 19QAPCh. 18 - Prob. 20QAPCh. 18 - Prob. 21QAPCh. 18 - Prob. 22QAPCh. 18 - Prob. 23QAPCh. 18 - Prob. 24QAPCh. 18 - Prob. 25QAPCh. 18 - Prob. 26QAPCh. 18 - . Does an oxidizing agent donate or accept...Ch. 18 - Prob. 28QAPCh. 18 - Prob. 29QAPCh. 18 - Prob. 30QAPCh. 18 - Prob. 31QAPCh. 18 - Prob. 32QAPCh. 18 - Prob. 33QAPCh. 18 - Prob. 34QAPCh. 18 - Prob. 35QAPCh. 18 - Prob. 36QAPCh. 18 - Prob. 37QAPCh. 18 - Prob. 38QAPCh. 18 - Prob. 39QAPCh. 18 - Prob. 40QAPCh. 18 - Prob. 41QAPCh. 18 - Prob. 42QAPCh. 18 - Prob. 43QAPCh. 18 - Prob. 44QAPCh. 18 - . Balance each of the following...Ch. 18 - Prob. 46QAPCh. 18 - . Iodide ion, I- , is one of the most easily...Ch. 18 - Prob. 48QAPCh. 18 - Prob. 49QAPCh. 18 - Prob. 50QAPCh. 18 - . In which direction do electrons flow in a...Ch. 18 - Prob. 52QAPCh. 18 - . Consider the oxidation-reduction reaction...Ch. 18 - . Consider the oxidation—reduction reaction...Ch. 18 - Prob. 55QAPCh. 18 - Prob. 56QAPCh. 18 - Prob. 57QAPCh. 18 - Prob. 58QAPCh. 18 - Prob. 59QAPCh. 18 - Prob. 60QAPCh. 18 - Prob. 61QAPCh. 18 - . What are some important uses of electrolysis?Ch. 18 - . Although aluminum is one of the most abundant...Ch. 18 - . The “Chemistry in Focus” segment Water-Powered...Ch. 18 - Prob. 65APCh. 18 - Prob. 66APCh. 18 - Prob. 67APCh. 18 - Prob. 68APCh. 18 - Prob. 69APCh. 18 - Prob. 70APCh. 18 - Prob. 71APCh. 18 - Prob. 72APCh. 18 - Prob. 73APCh. 18 - . To obtain useful electrical energy from an...Ch. 18 - Prob. 75APCh. 18 - Prob. 76APCh. 18 - Prob. 77APCh. 18 - Prob. 78APCh. 18 - . The “pressure” on electrons to flow from one...Ch. 18 - Prob. 80APCh. 18 - Prob. 81APCh. 18 - Prob. 82APCh. 18 - Prob. 83APCh. 18 - . For each of the following unbalanced...Ch. 18 - Prob. 85APCh. 18 - Prob. 86APCh. 18 - Prob. 87APCh. 18 - . Balance each of the following...Ch. 18 - . Balance each of the following...Ch. 18 - . For each of the following oxidation-reduction...Ch. 18 - . For each of the following oxidation-reduction...Ch. 18 - . Assign oxidation sates to all of the atoms in...Ch. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 94APCh. 18 - Prob. 95APCh. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 97APCh. 18 - . In each of the following reactions, identify...Ch. 18 - . Balance each of the following half-reactions....Ch. 18 - Prob. 100APCh. 18 - Prob. 101APCh. 18 - Prob. 102APCh. 18 - . Consider the oxidation—reduction reaction...Ch. 18 - Prob. 104APCh. 18 - Prob. 105CP
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