Chapter U5.17, Problem 4E

(a)

Interpretation Introduction

Interpretation: The oxidized and reduced atom in the following reaction needs to be explained.

  ${\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)\to \text{CO}\left(\text{g}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept Introduction: Whenever an atom or ion loses electrons, another atom gains electron. Loss of electrons is known as oxidation, and gain of electrons is known as reduction. Oxidation and reduction always occur together. This process is known as oxidation-reduction, or redox.

Answer to Problem 4E

Hydrogen is oxidized and Oxygen is reduced in this reaction.

Explanation of Solution

The given reaction is balanced that means the equation have equal number of atoms in both sides.

  ${\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)\to \text{CO}\left(\text{g}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In this reaction, oxidation state of H increases from 0 ( ${\text{H}}_{\text{2}}$ ) to +1( ${\text{H}}_{\text{2}}\text{O}$ ).

Oxidation state of C decreases from +4 ( ${\text{CO}}_{\text{2}}$ ) to +2( $\text{CO}$ ).

Therefore, in this reaction H is oxidized and O is reduced.

(b)

Interpretation Introduction

Interpretation: Which atom is oxidized and which is reduced in the following reaction needs to be explained.

  ${\text{SF}}_{\text{4}}\left(\text{g}\right){\text{+F}}_{\text{2}}\left(\text{g}\right)\to {\text{SF}}_{\text{6}}\left(\text{g}\right)$

Concept Introduction: Whenever an atom or ion loses electrons, another atom gains electron. Loss of electrons is known as oxidation, and gain of electrons is known as reduction. Oxidation and reduction always occur together. This process is known as oxidation-reduction, or redox.

Answer to Problem 4E

Sulfur is oxidized and fluorine is reduced in this reaction.

Explanation of Solution

The given reaction is balanced that means the equation have equal number of atoms in both sides.

  ${\text{SF}}_{\text{4}}\left(\text{g}\right){\text{+F}}_{\text{2}}\left(\text{g}\right)\to {\text{SF}}_{\text{6}}\left(\text{g}\right)$

In this reaction, oxidation state of S increases from +4 ( ${\text{SF}}_4$ ) to +6( ${\text{SF}}_{\text{6}}$ ).

Oxidation state of F decreases from 0 ( ${\text{F}}_{\text{2}}$ ) to $-1$ ( ${\text{SF}}_{\text{6}}$ ).

Therefore, in this reaction S is oxidized and F is reduced.

(c)

Interpretation Introduction

Interpretation: Which atom is oxidized and which is reduced in the following reaction needs to be explained $\text{4Ag}\left(\text{s}\right){\text{+2H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{s}\right)\to {\text{2Ag}}_{\text{2}}\text{S}\left(\text{s}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{g}\right)$ .

Concept Introduction: Whenever an atom or ion loses electrons, another atom gains electron. Loss of electrons is known as oxidation, and gain of electrons is known as reduction. Oxidation and reduction always occur together. This process is known as oxidation-reduction, or redox.

Answer to Problem 4E

Silver is oxidized and Oxygen is reduced in this reaction.

Explanation of Solution

The given reaction is balanced that means the equation have equal number of atoms in both sides.

  $\text{4Ag}\left(\text{s}\right){\text{+2H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{s}\right)\to {\text{2Ag}}_{\text{2}}\text{S}\left(\text{s}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In this reaction, oxidation state of Ag increases from 0 (in $\text{Ag}$ ) to +1(in ${\text{Ag}}_{\text{2}}\text{S}$ ).

Oxidation state of O decreases from 0 (in ${\text{O}}_{\text{2}}$ ) to $-2$ ( ${\text{H}}_{\text{2}}\text{O}$ ).

Therefore, in this reaction Ag is oxidized and O is reduced.

(d)

Interpretation Introduction

Interpretation: Which atom is oxidized and which is reduced in the following reaction needs to be explained ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(\text{aq}\right){\text{+6O}}_{\text{2}}\left(\text{g}\right)\to 6{\text{CO}}_2\left(\text{g}\right){\text{+6H}}_{\text{2}}\text{O}\left(\text{l}\right)$ .

Concept Introduction: Whenever an atom or ion loses electrons, another atom gains electron. Loss of electrons is known as oxidation, and gain of electrons is known as reduction. Oxidation and reduction always occur together. This process is known as oxidation-reduction, or redox.

Answer to Problem 4E

Carbon is oxidized and Oxygen is reduced in this reaction.

Explanation of Solution

The given reaction is balanced that means the equation have equal number of atoms in both sides.

  ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(\text{aq}\right){\text{+6O}}_{\text{2}}\left(\text{g}\right)\to 6{\text{CO}}_2\left(\text{g}\right){\text{+6H}}_{\text{2}}\text{O}\left(\text{l}\right)$

In this reaction, oxidation state of C increases from 0 ( ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ ) to +4(in ${\text{CO}}_{\text{2}}$ ).

Oxidation state of O decreases from 0 (in ${\text{O}}_{\text{2}}$ ) to $-2$ in products.

Therefore, in this reaction C is oxidized and O is reduced.