Chapter F, Problem K.20E

(a)

Interpretation Introduction

Interpretation:

Given chemical equation for redox reaction has to be balanced.

    ${\text{SiCl}}_{\text{4}}\text{(l)}+{\text{H}}_{\text{2}}(\text{g})\to \text{Si}(\text{s})+\text{HCl}(\text{g})$

Explanation of Solution

The given reaction is written as follows;

    ${\text{SiCl}}_{\text{4}}\text{(l)}+{\text{H}}_{\text{2}}(\text{g})\to \text{Si}(\text{s})+\text{HCl}(\text{g})$

Balancing chlorine atom:  In the above equation, there are four chlorine atoms on the left side while there is one chlorine atom on the right side.  Adding coefficient $4$ before $\text{HCl}$ balances the chlorine atom on both sides of the equation.  The chemical equation obtained is given below;

    ${\text{SiCl}}_{\text{4}}\text{(l)}+{\text{H}}_{\text{2}}(\text{g})\to \text{Si}(\text{s})+4\text{HCl}(\text{g})$

Balancing hydrogen atom:  In the above equation, there are two hydrogen atoms on the left side while there are four hydrogen atoms on the right side.  Adding coefficient $2$ before ${\text{H}}_{\text{2}}$ balances the hydrogen atom on both sides of the equation.  The balanced chemical equation obtained is given below;

    ${\text{SiCl}}_{\text{4}}\text{(l)}+2{\text{H}}_{\text{2}}(\text{g})\to \text{Si}(\text{s})+4\text{HCl}(\text{g})$

Name of ${\text{SiCl}}_{\text{4}}$:

The compound is named using the rules for binary compounds.  There is one silicon atom and four chlorine atom.  Therefore, the name of the compound is given as silicon tetrachloride.

Oxidation state of Silicon in ${\text{SiCl}}_{\text{4}}$:

The total charge on ${\text{SiCl}}_{\text{4}}$ is zero.  The oxidation state of silicon is calculated as shown below;

    $\begin{array}{c}x+4(-1)=0\\ x-4=0\\ x=+4\end{array}$

The oxidation state of silicon is $+4$.

(b)

Interpretation Introduction

Interpretation:

Given chemical equation for redox reaction has to be balanced.

    ${\text{SnO}}_{\text{2}}\text{(s)}+\text{C}(\text{s})\to \text{Sn}(\text{l})+{\text{CO}}_{\text{2}}$

Explanation of Solution

The given reaction is written as follows;

    ${\text{SnO}}_{\text{2}}\text{(s)}+\text{C}(\text{s})\to \text{Sn}(\text{l})+{\text{CO}}_{\text{2}}$

Given equation has the same number of atoms for each kind of elements equal on both sides.  Therefore, this itself is a balanced equation.

Name of ${\text{SnO}}_{\text{2}}$:

The compound is named using the rules for ionic compounds.  There is one tin atom and two oxygen atom.  As tin is a transition metal, the oxidation state of tin is shown using Roman numeral.  Therefore, the name of the compound is given as tin(IV) oxide.

Oxidation state of tin in ${\text{SnO}}_{\text{2}}$:

The total charge on ${\text{SnO}}_{\text{2}}$ is zero.  The oxidation state of tin is calculated as shown below;

    $\begin{array}{c}x+2(-2)=0\\ x-4=0\\ x=+4\end{array}$

The oxidation state of tin is $+4$.

(c)

Interpretation Introduction

Interpretation:

Given chemical equation for redox reaction has to be balanced.

    ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{(s)}+\text{Ca}(\text{l})\to \text{V}(\text{s})+\text{CaO(s)}$

Explanation of Solution

The given reaction is written as follows;

    ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{(s)}+\text{Ca}(\text{l})\to \text{V}(\text{s})+\text{CaO(s)}$

Balancing oxygen atom:  In the above equation, there are five oxygen atoms on the left side of the equation, while there is only one oxygen atom in the right side of the equation.  Adding coefficient $5$ before $\text{CaO}$ balances the oxygen atoms.  The chemical equation obtained is given as follows;

    ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{(s)}+\text{Ca}(\text{l})\to \text{V}(\text{s})+5\text{CaO(s)}$

Balancing vanadium atom:  In the above equation, there are two vanadium atoms on the left side of the equation, while there is only one vanadium atom in the right side of the equation.  Adding coefficient $2$ before $\text{V}$ balances the vanadium atoms.  The chemical equation obtained is given as follows;

    ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{(s)}+\text{Ca}(\text{l})\to 2\text{V}(\text{s})+5\text{CaO(s)}$

Balancing calcium atom:  In the above equation, there is one calcium atom on the left side of the equation, while there are five calcium atoms in the right side of the equation.  Adding coefficient $5$ before $\text{Ca}$ balances the calcium atoms.  The balanced chemical equation obtained is given as follows;

    ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{(s)}+5\text{Ca}(\text{l})\to 2\text{V}(\text{s})+5\text{CaO(s)}$

Name of ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}$:

The compound is named using the rules for ionic compounds.  There are two vanadium atoms and five oxygen atoms.  As vanadium is a transition metal, the oxidation state of tin is shown using Roman numeral.  Therefore, the name of the compound is given as vanadium pentoxide.

Oxidation state of vanadium in ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}$:

The total charge on ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}$ is zero.  The oxidation state of tin is calculated as shown below;

    $\begin{array}{c}2x+5(-2)=0\\ 2x-10=0\\ x=+5\end{array}$

The oxidation state of vanadium is $+5$.

(d)

Interpretation Introduction

Interpretation:

Given chemical equation for redox reaction has to be balanced.

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{Mg}(\text{s})\to \text{B}(\text{s})+\text{MgO(s)}$

Explanation of Solution

The given reaction is written as follows;

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{Mg}(\text{s})\to \text{B}(\text{s})+\text{MgO(s)}$

Balancing oxygen atom:  In the above equation, there are three oxygen atoms on the left side of the equation, while there is only one oxygen atom in the right side of the equation.  Adding coefficient $3$ before $\text{MgO}$ balances the oxygen atoms.  The chemical equation obtained is given as follows;

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{Mg}(\text{s})\to \text{B}(\text{s})+3\text{MgO(s)}$

Balancing boron atom:  In the above equation, there are two boron atoms on the left side of the equation, while there is only one boron atom in the right side of the equation.  Adding coefficient $2$ before $\text{B}$ balances the boron atoms.  The chemical equation obtained is given as follows;

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+\text{Mg}(\text{s})\to 2\text{B}(\text{s})+3\text{MgO(s)}$

Balancing magnesium atom:  In the above equation, there is one magnesium atom on the left side of the equation, while there are three magnesium atoms in the right side of the equation.  Adding coefficient $3$ before $\text{Mg}$ balances the magnesium atoms.  The balanced chemical equation obtained is given as follows;

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}+3\text{Mg}(\text{s})\to 2\text{B}(\text{s})+3\text{MgO(s)}$

Name of ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$:

The compound is named using the rules for binary compounds.  There are two boron atoms and three oxygen atoms.  Therefore, the name of the compound is given as boron trioxide.

Oxidation state of boron in ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$:

The total charge on ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$ is zero.  The oxidation state of tin is calculated as shown below;

    $\begin{array}{c}2x+3(-2)=0\\ 2x-6=0\\ x=+3\end{array}$

The oxidation state of boron is $+3$.