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.

Answer to Problem 32QAP

C being oxidized and Cl being reduced.

Explanation of Solution

Given:

$4\text{K}\text{C}\text{l}{\text{O}}_3\left(\text{s}\right)+{\text{ C}}_6{\text{H}}_{12}{\text{O}}_6\left(\text{s}\right)\to \text{ 4KCl}\left(\text{s}\right)+6{\text{H}}_2\text{O}(l)+6\text{C}{\text{O}}_2(g)$

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 $\text{K}\text{C}\text{l}{\text{O}}_3,{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(\text{s}\right),\text{KCl},\text{a}\text{n}\text{d}\text{C}{\text{O}}_2$ the oxidation is determined as:

For $\text{K}\text{C}\text{l}{\text{O}}_3$ :

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, $\text{K}\text{C}\text{l}{\text{O}}_3$ has no charge so; the sum of oxidation state must be zero.

so:

1 - x + 3(-2) = 0

x = + 5

For ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ :

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

Since, ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ has no charge so; the sum of oxidation state must be zero.

so:

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

x = 0

For $\text{KCl}$ :

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

Since, $\text{KCl}$ has no charge so; the sum of oxidation state must be zero.

so:

1 + (-x) = 0

x = + 1

For $\text{C}{\text{O}}_2$ :

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

Since, $\text{C}{\text{O}}_2$ has no charge so; the sum of oxidation state must be zero.

so:

x + 2(-2) = 0

x = + 4

$4\text{K}\text{C}\text{l}{\text{O}}_3\left(\text{s}\right)+{\text{ C}}_6{\text{H}}_{12}{\text{O}}_6\left(\text{s}\right)\to \text{ 4KCl}\left(\text{s}\right)+6{\text{H}}_2\text{O}(l)+6\text{C}{\text{O}}_2(g)$

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.

Answer to Problem 32QAP

C being oxidized and O being reduced.

Explanation of Solution

Given:

${\text{2 C}}_8{\text{H}}_{18}\left(\text{l}\right)\text{ +}{\text{25O}}_2\to \text{ 1}6\text{C}{\text{O}}_2(g)+18{\text{H}}_2\text{O}(l)$

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 ${\text{O}}_2$ is zero and for ${\text{C}}_8{\text{H}}_{18},\text{C}{\text{O}}_2\text{a}\text{n}\text{d}{\text{H}}_2\text{O}$ the oxidation is determined as:

For ${\text{C}}_8{\text{H}}_{18}$ :

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

Since, ${\text{C}}_8{\text{H}}_{18}$ has no charge so; the sum of oxidation state must be zero.

so:

8 x + 18(-1) = 0

x = -2.25

For ${\text{H}}_2\text{O}$ :

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

Since, ${\text{H}}_2\text{O}$ has no charge so; the sum of oxidation state must be zero.

so:

1(2) + x = 0

x = -2

For $\text{C}{\text{O}}_2$ :

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

Since, $\text{C}{\text{O}}_2$ has no charge so; the sum of oxidation state must be zero.

so:

x + 2(-2) = 0

x = + 4

${\text{2 C}}_8{\text{H}}_{18}\left(\text{l}\right)\text{ +}{\text{25O}}_2\to \text{ 1}6\text{C}{\text{O}}_2(g)+18{\text{H}}_2\text{O}(l)$

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.

Answer to Problem 32QAP

P being oxidized and Cl being reduced.

Explanation of Solution

Given:

$\text{P}\text{C}{\text{l}}_3\left(\text{g}\right)\text{ +}\text{C}{\text{l}}_2(g)\to \text{P}\text{C}{\text{l}}_5(g)$

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 $\text{C}{\text{l}}_2$ is zero and for $\text{P}\text{C}{\text{l}}_3\text{a}\text{n}\text{d}\text{P}\text{C}{\text{l}}_5$ the oxidation is determined as:

For $\text{P}\text{C}{\text{l}}_3$ :

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

Since, $\text{P}\text{C}{\text{l}}_3$ has no charge so; the sum of oxidation state must be zero.

so:

x + 3(-1) = 0

x = + 3

For $\text{P}\text{C}{\text{l}}_5$ :

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

Since, $\text{P}\text{C}{\text{l}}_5$ has no charge so; the sum of oxidation state must be zero.

so:

x + 5(-1) = 0

x = + 5

$\text{P}\text{C}{\text{l}}_3\left(\text{g}\right)\text{ +}\text{C}{\text{l}}_2(g)\to \text{P}\text{C}{\text{l}}_5(g)$

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.

Answer to Problem 32QAP

Ca being oxidized and H being reduced.

Explanation of Solution

Given:

$\text{C}\text{a}\left(\text{s}\right)\text{ +}{\text{H}}_2(g)\to \text{C}\text{a}{\text{H}}_2(g)$

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 $\text{C}\text{a}\text{a}\text{n}\text{d}{\text{H}}_2$ is zero and for $\text{C}\text{a}{\text{H}}_2$ the oxidation is determined as:

For $\text{C}\text{a}{\text{H}}_2$ :

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

Since, $\text{C}\text{a}{\text{H}}_2$ has no charge so; the sum of oxidation state must be zero.

so:

+2 + 2(x) = 0

x = -1

$\text{C}\text{a}\left(\text{s}\right)\text{ +}{\text{H}}_2(g)\to \text{C}\text{a}{\text{H}}_2(g)$

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.