Chapter 18, Problem 26A

(a)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{MnO}}_4^{-}\left(\text{aq}\right)+\text{Zn}\left(\text{s}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 26A

The balance reduction half-reaction is-

  ${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}$

The balance oxidation half-reaction is-

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{MnO}}_4^{-}\left(\text{aq}\right)+\text{Zn}\left(\text{s}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$

Let oxidation state of Mn is x. Then, in ${\text{MnO}}_4^{-}$ its oxidation state is-

$\begin{array}{c}x+4\left(-2\right)=-1\\ x=8-1\\ =7\end{array}$

Oxidation state of O is $-2$ . Zn is in free state so, its oxidation state is zero. Oxidation state of both Mn and Zn is $+2$ .

Oxidation state of Mn is decreases from $+7$ to $+2$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)$

Other element except O is balanced. Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ . Now, add electrons to balance the charge. To balance charge on both side, add five electrons to the left. Hence, the balanced equation is-

${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of Zn is increases from $0$ to $+2$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$

Both the elements are balanced. Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the right. Hence, the balanced equation is-

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (1) by 2 and equation (2) by 5. After that, add both the equation to get the net balance equation.

  $\begin{array}{c}{\text{2MnO}}_4^{-}\left(\text{aq}\right)+16{\text{H}}^{+}+10{\text{e}}^{-}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\\ \text{5Zn}\left(\text{s}\right)\to 5{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+10{\text{e}}^{-}\\ {\text{2MnO}}_4^{-}\left(\text{aq}\right)+\text{5Zn}\left(\text{s}\right)+16{\text{H}}^{+}+10{\text{e}}^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}+5{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge is balanced in both sides.

(b)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+{\text{H}}^{\text{+}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 26A

The balance reduction half-reaction is-

  ${\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)$

The balance oxidation half-reaction is-

${\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+{\text{H}}^{\text{+}}\left(\text{aq}\right)$

Oxidation state of Sn is $4$ . H is in free state so, its oxidation state is zero. Oxidation state of Sn and H is $+2$ and $+1$ respectively.

Oxidation state of Sn decreases from $+4$ to $+2$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{Sn}}^{\text{4+}}\left(\text{aq}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)$

Both Sn is balanced. Now, add electrons to balance charge. To balance charge on both side add two electrons to the left. Hence, the balanced equation is-

${\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)$ …… (1)

Oxidation state of H increases from $0$ to $+1$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)$

Both the elements are balanced. Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the right. Hence, the balanced equation is-

${\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Add equation (1) and (2) to get the net balance equation.

$\begin{array}{c}{\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)\\ {\text{H}}_{\text{2}}\left(\text{g}\right)\to 2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{Sn}}^{\text{4+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to {\text{Sn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge is balanced in both sides.

(c)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

$\text{Zn}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NO}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 26A

The balance reduction half-reaction is-

  ${\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+{\text{e}}^{-}\to {\text{NO}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

The balance oxidation half-reaction is-

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

$\text{Zn}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NO}}_{\text{2}}\left(\text{g}\right)$

Zn is in free state so, its oxidation state is zero. Oxidation state of N in ${\text{NO}}_3^{-}$ is $+5$ , O is $-2$ . In product side, oxidation state of N is +4, O is $-2$ and Zn is $+2$ .

Oxidation state of N decreases from $+5$ to $+4$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{NO}}_{\text{2}}\left(\text{aq}\right)$

Other element except O is balanced. Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ . Now, add electrons to balance the charge. To balance charge on both sides, add one electron to the left. Hence, the balanced equation is-

${\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+{\text{e}}^{-}\to {\text{NO}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of Znincreases from $0$ to $+2$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$

Both the elements are balanced. Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the right. Hence, the balanced equation is-

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (1) by 2. After that, add both the equation to get the net balance equation.

$\begin{array}{c}{\text{2NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{\text{+}}+2{\text{e}}^{-}\to 2{\text{NO}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\\ \text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{2NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{\text{+}}+\text{Zn}\left(\text{s}\right)\to 2{\text{NO}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}{\text{O+Zn}}^{\text{2+}}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge is balanced in both sides.

(d)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{Br}}_{\text{2}}\left(\text{l}\right)\to \text{S}\left(\text{s}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 26A

The balance reduction half-reaction is-

  ${\text{Br}}_{\text{2}}\left(\text{l}\right)+2{\text{e}}^{-}\to 2{\text{Br}}^{-}\left(\text{aq}\right)$

The balance oxidation half-reaction is-

${\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\to \text{S}\left(\text{s}\right)+2{\text{H}}^{+}+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)+{\text{Br}}_{\text{2}}\left(\text{l}\right)\to \text{S}\left(\text{s}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

${\text{Br}}_{\text{2}}$ is in free state so, its oxidation state is zero. Oxidation state of S in ${\text{H}}_{\text{2}}\text{S}$ is $-2$ , H is $+1$ . In product side, oxidation state of S is 0, and ${\text{Br}}^{-}$ is $-1$ .

Oxidation state of Br decreases from $0$ to $-1$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{Br}}_{\text{2}}\left(\text{l}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)$

To balance both the elements, multiply reactant side Br by 2.

${\text{Br}}_{\text{2}}\left(\text{l}\right)\to 2{\text{Br}}^{-}\left(\text{aq}\right)$

Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the left. Hence, the balanced equation is-

${\text{Br}}_{\text{2}}\left(\text{l}\right)+2{\text{e}}^{-}\to 2{\text{Br}}^{-}\left(\text{aq}\right)$ …… (1)

Oxidation state of S is increases from $-2$ to 0. Hence, it is oxidized. The oxidation half-reaction is as follows:

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

Balance H by adding ${\text{H}}^{+}$ in right side.

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

Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the right. Hence, the balanced equation is-

${\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\to \text{S}\left(\text{s}\right)+2{\text{H}}^{+}+2{\text{e}}^{-}$ …… (1)

Add both the equation (1) and (2) to get the net balance equation.

$\begin{array}{c}{\text{Br}}_{\text{2}}\left(\text{l}\right)+2{\text{e}}^{-}\to 2{\text{Br}}^{-}\left(\text{aq}\right)\\ {\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\to \text{S}\left(\text{s}\right)+2{\text{H}}^{+}+2{\text{e}}^{-}\\ {\text{Br}}_{\text{2}}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{S}\left(\text{g}\right)\to 2{\text{Br}}^{-}\left(\text{aq}\right)+\text{S}\left(\text{s}\right)\end{array}$

In the overall reaction, charge is balanced in both sides.