Chapter 8, Problem 8.89QA

Interpretation Introduction

To determine:

a) Oxidation numbers of all the elements in the given reaction.

b) The elements being oxidized and being reduced.

Answer to Problem 8.89QA

Solution:

a)  $3 Si{O}_{2\left(s\right)}+2 F{e}_3{O}_{4\left(s\right)}\to 3 F{e}_{2}Si{O}_{4\left(s\right)}+O_{2\left(g\right)}$

Elements	O.N. on reactant side	O.N. on product side
Si	+4	+4
O	-2	-2 (in $F{e}_{2}Si{O}_4$) and 0 (in $O_2$)
Fe	+3 (two Fe) and +2 (one Fe)	+2

In this reaction, iron is reduced $\left(+3 to+2\right)$ and oxygen is oxidized $(-2 to 0)$.

b) $Si{O}_{2\left(s\right)}+2 {Fe}_{\left(s\right)}+O_{2\left(g\right)}\to F{e}_{2}Si{O}_{4\left(s\right)}$

Elements	O.N. on reactant side	O.N. on product side
Si	+4	+4
O	-2 (in $Si{O}_2$) and 0 (in $O_2$)	-2
Fe	0	+2

In this reaction, iron is oxidized $\left(0 to+2\right)$ and oxygen is reduced $\left(0 to-2\right).$

c) $4 Fe{O}_{\left(S\right)}+O_{2\left(g\right)}+6 H_2{O}_{\left(l\right)}\to 4 Fe{\left(OH\right)}_{3\left(s\right)}$

Elements	O.N. on reactant side	O.N. on product side
H	+1	+1
O	-2 (in $FeO and H_{2}O$) and 0 (in $O_2$)	-2
Fe	+2	+3

In this reaction, iron is oxidized $(+2 to+3)$ and oxygen is reduced $\left(0 to-2\right).$

Explanation of Solution

1) Concept:

We are asked to assign oxidation numbers to each element in the given reactions. We will follow the rules for assigning oxidation numbers to atoms. Using the changes in oxidation number, we can find out oxidized and reduced elements in the reactions. Oxidation numbers of some elements are constant as, in case of O, the common oxidation number is -2 with some exceptions. Also, the oxidation number of H is +1 except when it is attached to metal (-1).

2) Given:

i) $3 Si{O}_{2\left(s\right)}+2 F{e}_3{O}_{4\left(s\right)}\to 3 F{e}_{2}Si{O}_{4\left(s\right)}+O_{2\left(g\right)}$

ii) $Si{O}_{2\left(s\right)}+2 {Fe}_{\left(s\right)}+O_{2\left(g\right)}\to F{e}_{2}Si{O}_{4\left(s\right)}$

iii) $4 Fe{O}_{\left(S\right)}+O_{2\left(g\right)}+6 H_2{O}_{\left(l\right)}\to 4 Fe{\left(OH\right)}_{3\left(s\right)}$

3) Calculations:

a) $3 Si{O}_{2\left(s\right)}+2 F{e}_3{O}_{4\left(s\right)}\to 3 F{e}_{2}Si{O}_{4\left(s\right)}+O_{2\left(g\right)}$

The sum of oxidation numbers of neutral compounds equals zero. Therefore, in $Si{O}_2$, sum of oxidation state of $Si$ and $O$ is zero.

Let the oxidation number of $Si$ be $x.$

$x+(2\times \left(O\right))=0$

$x+(2\times \left(-2\right))=0$

$x=+4$

The oxidation state of $Fe$ in $F{e}_3{O}_4$ is

$3x+4\left(O\right)=0$

$3x+4(-2)=0$

$3x=8$

$x=+ 2.66$

So, in $F{e}_3{O}_4$, two $Fe$ are in $+3$ oxidation state and one $Fe$ is in $+2$ oxidation state.

Oxidation state of $F{e}_{2}Si{O}_4$ is calculated as below by considering oxidation number of $Fe$ as $x$ and oxidation state of $Si$ as $y.$

$2x+y+4\left(O\right)=0$

$2x+y=+8$

Here, two $Fe$ are in $+2$ oxidation state, and $Si$ is in $+4$ oxidation state.

Here, in the reaction, as $Fe$ undergoes a change in the oxidation state from $+3 to+2,$ it means $Fe$ gets reduced. Further, $O$ goes from $-2 to 0$. Therefore, $O$ gets oxidized.

b) $Si{O}_{2\left(s\right)}+2 {Fe}_{\left(s\right)}+O_{2\left(g\right)}\to F{e}_{2}Si{O}_{4\left(s\right)}$

The sum of oxidation number of neutral compounds equals zero. So, in $Si{O}_2$, sum of oxidation state of $Si$ and $O$ is zero.

Let the oxidation number of $Si$ be $x.$

$x+(2\times \left(O\right))=0$

$x+(2\times \left(-2\right))=0$

$x=+4$

Oxidation number of compounds in its elemental form is zero. So oxidation state of $Fe$ and $O$ is zero.

Oxidation state in $F{e}_{2}Si{O}_4$ is calculated as below by considering oxidation number of $Fe$ as $x$ and oxidation state of $Si$ as $y.$

$2x+y+(4\times \left(O\right))=0$

$2x+y+(4\times \left(-2\right))=0$

$2x+y=+8$

So here, two $Fe$ are in $+2$ oxidation state and $Si$ is in $+4$ oxidation state.

Here, in the reaction, as $O$ undergoes a change in the oxidation state from $0 to-2,$ it means $O$ is reduced. Further, $Fe$ changes from $0 to+2,$ so it get oxidized.

c) $4 Fe{O}_{\left(S\right)}+O_{2\left(g\right)}+6 H_2{O}_{\left(l\right)}\to 4 Fe{\left(OH\right)}_{3\left(s\right)}$

The sum of oxidation number of neutral compounds equals zero. So, in $FeO$, the sum of oxidation state of $Fe$ and $O$ is zero.

Let the oxidation number of $Fe$ be $x.$

$x+(1\times \left(O\right))=0$

$x+(1\times \left(-2\right))=0$

$x=+2$

Oxidation number of compounds in its elemental form is zero. So oxidation state of $O$ is zero.

So, oxidation state of $Fe$ in $Fe({OH)}_3$ is

$x+(3\times \left(O\right))+(3\times \left(H\right))=0$

$x+(3\times \left(-2\right))+(3\times \left(+1\right))=0$

$x=+3$

So, in the reaction, $Fe$ undergoes from $+2$ oxidation state to $+3,$ so it is oxidized. Further, oxygen undergoes from $0$ to $-2,$ so it is reduced.

Conclusion:

Oxidation means loss of electrons, and reduction is gain of electrons. From the change in the oxidation state of elements, the oxidized and reduced species have been calculated.