Chapter 17, Problem 47E

(a)

Interpretation Introduction

Interpretation:

The reaction taking place in two different galvanic cells is given. The $E°$ of the given galvanic cells, the spontaneity of the reactions and balanced chemical equation for each cell is to be stated.

Concept introduction:

The galvanic cell converts chemical energy into electrical energy while the electrolytic cell converts electrical energy into chemical energy.

The species at anode undergoes oxidation while the species at cathode undergoes reduction and the electrons generated at the anode are transferred through wire to the cathode.

To determine: The set up of the cell given in the chemical equation and the direction of electron flow, identification of the cathode and anode and balanced chemical equation.

Answer to Problem 47E

1. a. The value of $E{°}_{\text{cell}}$ is calculated as $\underset{\_}{0.97V}$.

The cell reaction is spontaneous in nature.

The balanced chemical equation for given reaction (a) is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

Explanation of Solution

The value of $E{°}_{\text{cell}}$ is calculated as $\underset{\_}{0.97V}$.

The reaction taking place at cathode is,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)E{°}_{\text{red}}=1.51\text{V}$

The reaction taking place at anode is,

${\text{2I}}^{-}\left(aq\right)\to {\text{I}}_2\left(aq\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=-0.54\text{V}$

Multiply reduction half with a coefficient of $2$ and oxidation half with a coefficient of $5$ and add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\\ {\text{10I}}^{-}\left(aq\right)\to 5{\text{I}}_2\left(aq\right)+10{\text{e}}^{-}\end{array}$

The final equation is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.54\text{V}+\text{1}\text{.51}\text{V}\\ =\underset{\_}{0.97V}\end{array}$

The value of $E{°}_{\text{cell}}$ is $\underset{\_}{0.97V}$.

The cell reaction is spontaneous in nature.

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

Where,

• $\Delta G°$ is the Gibbs free energy change at the standard conditions.
• $n$ is the number of electrons involved in the reaction.
• $F$ is the Faraday’s constant.
• $E{°}_{\text{cell}}$ is the cell potential at the standard condition.

When the value of $E{°}_{\text{cell}}$ comes out to be positive then $\Delta G°$ becomes negative. Therefore it leads to a spontaneous reaction.

The balanced chemical equation for given reaction (a) is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

The reduction half cell reaction is,

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

The change in the oxidation number of manganese is from $+7$ to $+2$.

The oxidation half reaction is,

${\text{I}}^{-}\left(aq\right)\to {\text{I}}_2\left(aq\right)$

The change in oxidation number of iodine is from $-1$ to zero.

All the atoms except oxygen and hydrogen in the reduction half-reaction are already balanced. So, directly balance the oxygen atoms by adding water molecules to the right hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

Balance the hydrogen atom by adding ${\text{H}}^{+}$ to the left hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$ (1)

Balance all the atoms other than hydrogen and oxygen in the oxidation half reaction,

${\text{2I}}^{-}\left(aq\right)\to {\text{I}}_2\left(aq\right)$

As there are no hydrogen or oxygen atoms. So, directly balance the charge by adding electrons at the appropriate side,

${\text{2I}}^{-}\left(aq\right)\to {\text{I}}_2\left(aq\right)+2{\text{e}}^{-}$ (2)

Multiply equation (1) by $2$ and equation (2) by $5$ to cancel out electrons and add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\\ {\text{10I}}^{-}\left(aq\right)\to 5{\text{I}}_2\left(aq\right)+10{\text{e}}^{-}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

(b)

Interpretation Introduction

Interpretation:

The reaction taking place in two different galvanic cells is given. The $E°$ of the given galvanic cells, the spontaneity of the reactions and balanced chemical equation for each cell is to be stated.

Concept introduction:

The galvanic cell converts chemical energy into electrical energy while the electrolytic cell converts electrical energy into chemical energy.

The species at anode undergoes oxidation while the species at cathode undergoes reduction and the electrons generated at the anode are transferred through wire to the cathode.

To determine: The set up of the cell given in the chemical equation and the direction of electron flow, identification of the cathode and anode and the balanced chemical equation.

Answer to Problem 47E

1. a. The value of $E{°}_{\text{cell}}$ is calculated as $\underset{\_}{0.97V}$.

The cell reaction is spontaneous in nature.

The balanced chemical equation for given reaction (a) is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

Explanation of Solution

The value of $E{°}_{\text{cell}}$ is calculated as $\underset{\_}{-1.36V}$.

The reaction taking place at cathode is,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)E{°}_{\text{red}}=1.51\text{V}$

The reaction taking place at anode is,

${\text{2F}}^{-}\left(aq\right)\to {\text{F}}_2\left(aq\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=-2.87\text{V}$

Multiply reduction half with a coefficient of $2$ and oxidation half with a coefficient of $5$ and add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\\ {\text{10F}}^{-}\left(aq\right)\to 5{\text{F}}_2\left(aq\right)+10{\text{e}}^{-}\end{array}$

The final equation is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{I}}_2\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-2.87\text{V+1}\text{.51}\text{V}\\ \text{=}\underset{\_}{-1.36V}\end{array}$

The value of $E{°}_{\text{cell}}$ is $\underset{\_}{-1.36V}$.

The cell reaction is non-spontaneous in nature.

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

Where,

• $\Delta G°$ is the Gibbs free energy change at the standard conditions.
• $n$ is the number of electrons involved in the reaction.
• $F$ is the Faraday’s constant.
• $E{°}_{\text{cell}}$ is the cell potential at the standard condition.

When the value of $E{°}_{\text{cell}}$ comes out to be negative then $\Delta G°$ becomes positive. Therefore it leads to a non-spontaneous reaction.

The balanced chemical equation for given reaction (a) is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{F}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{F}}_2\left(aq\right)$

The reduction half cell reaction is,

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

The change in the oxidation number of manganese is from $+7$ to $+2$.

The oxidation half reaction is,

${\text{F}}^{-}\left(aq\right)\to {\text{F}}_2\left(aq\right)$

The change in oxidation number of fluorine is from $-1$ to zero.

All the atoms except oxygen and hydrogen in the reduction half-reaction are already balanced. So, directly balance the oxygen atoms by adding water molecules to the right hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

Balance the hydrogen atom by adding ${\text{H}}^{+}$ to the left hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$ (1)

Balance all the atoms other than hydrogen and oxygen in the oxidation half reaction,

${\text{2F}}^{-}\left(aq\right)\to {\text{F}}_2\left(aq\right)$

As there are no hydrogen or oxygen atoms. So, directly balance the charge by adding electrons at the appropriate side,

${\text{2F}}^{-}\left(aq\right)\to {\text{F}}_2\left(aq\right)+2{\text{e}}^{-}$ (2)

Multiply equation (1) by $2$ and equation (2) by $5$ to cancel out electrons and add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\\ {\text{10F}}^{-}\left(aq\right)\to 5{\text{F}}_2\left(aq\right)+10{\text{e}}^{-}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{2MnO}}_4{}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{F}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{F}}_2\left(aq\right)$