Chapter 11, Problem 21E

(a)

Interpretation Introduction

Interpretation:

The sketch of galvanic cell along with cathode and anode and the direction of electron flow, the direction of flow of ions through salt bridge, the balanced chemical equation and the E⁰cell should be predicted.

Concept Introduction:

In galvanic cell chemical energy converted into electrical energy.

At anode oxidation takes place which means loss of electrons.

At cathode reduction takes place which means gain of electrons.

Answer to Problem 21E

The E⁰cell is0.27 V

Reaction at anode:

  ${\text{2Br}}^{\text{-}}\to {\text{Br}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}$

Reaction at cathode:

  ${\text{Cl}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}$

Overall reaction:

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

Explanation of Solution

Given information:

  $\begin{array}{l}{\text{Cl}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{E}}^{\text{0}}\text{=}\text{1}\text{.36}\text{V}\\ {\text{Br}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{E}}^{\text{0}}\text{=}\text{1}\text{.09}\text{V}\end{array}$

The diagram of cell is shown below:

  

The direction of flow of electrons is from anode to cathode.

Negative ions flow towards anode.

At anode oxidation takes place

At cathode reduction takes place.

The cell representation is shown below:

The oxidation half-cell reaction is shown below:

  ${\text{2Br}}^{\text{-}}\to {\text{Br}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}$

The reduction half-cell reaction is shown below:

  ${\text{Cl}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}$

The overall reaction is shown below:

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

The calculation of E⁰cell is shown below:

E⁰cell = E⁰cathode − E⁰anode

  = 1.36 − (-1.09)

  = 0.27 V

(b)

Interpretation Introduction

Interpretation:

The sketch of galvanic cell along with cathode and anode and the direction of electron flow, the direction of flow of ions through salt bridge, the balanced chemical equation and the E⁰cell should be predicted.

Concept Introduction:

In galvanic cell chemical energy converted into electrical energy.

At anode oxidation takes place which means loss of electrons.

At cathode reduction takes place which means gain of electrons.

Answer to Problem 21E

The E⁰cell is 0.09 V

Reaction at anode:

  ${\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\to {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\text{+}{\text{5e}}^{\text{-}}$

Reaction at cathode:

  ${\text{IO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{IO}}_{\text{3}}{}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

Overall reaction:

  ${\text{5IO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{2Mn}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{5IO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right){\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{6H}}^{\text{+}}\left(\text{aq}\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}{\text{E}}^{\text{0}}\text{=}\text{1}\text{.51}\text{V}\\ {\text{IO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{IO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}{\text{E}}^{\text{0}}\text{=}\text{1}\text{.60}\text{V}\end{array}$

The diagram is shown below:

  

The direction of flow of electrons is from anode to cathode.

Negative ions flow towards anode.

At anode oxidation takes place

At cathode reduction takes place.

The oxidation half-cell reaction is shown below:

  ${\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\to {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\text{+}{\text{5e}}^{\text{-}}$

The reduction half-cell reaction is shown below:

  ${\text{IO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{IO}}_{\text{3}}{}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

The overall reaction is shown below:

  ${\text{5IO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{2Mn}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{5IO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right){\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{6H}}^{\text{+}}\left(\text{aq}\right)$

The calculation of E⁰cell is shown below:

E⁰cell = E⁰cathode − E⁰anode

  = 1.60 − 1.51

  = 0.09 V

(c)

Interpretation Introduction

Interpretation:

The sketch of galvanic cell along with cathode and anode and the direction of electron flow, the direction of flow of ions through salt bridge, the balanced chemical equation and the E⁰cell should be predicted.

Concept Introduction:

In galvanic cell chemical energy converted into electrical energy.

At anode oxidation takes place which means loss of electrons.

At cathode reduction takes place which means gain of electrons.

Answer to Problem 21E

The E⁰cell is 1.10 V

Reaction at anode:

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to {\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}$

Reaction at cathode:

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2H}}_{\text{2}}\text{O}$

Overall reaction:

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

Explanation of Solution

Given information:

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2H}}_{\text{2}}\text{O}{\text{E}}^{\text{0}}\text{=}\text{1}\text{.78}\text{V}\\ {\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{E}}^{\text{0}}\text{=}\text{0}\text{.68}\text{V}\end{array}$

The diagram is shown below:

  

The direction of flow of electrons is from anode to cathode.

Negative ions flow towards anode.

At anode oxidation takes place

At cathode reduction takes place.

The oxidation half-cell reaction is shown below:

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to {\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}$

The reduction half-cell reaction is shown below:

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{2H}}_{\text{2}}\text{O}$

The overall reaction is shown below:

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

The calculation of E⁰cell is shown below:

E⁰cell = E⁰cathode − E⁰anode

  = 1.78 − 0.68

  = 1.10 V

(d)

Interpretation Introduction

Interpretation:

The sketch of galvanic cell along with cathode and anode and the direction of electron flow, the direction of flow of ions through salt bridge, the balanced chemical equation and the E⁰cell should be predicted.

Concept Introduction:

In galvanic cell chemical energy converted into electrical energy.

At anode oxidation takes place which means loss of electrons.

At cathode reduction takes place which means gain of electrons.

Answer to Problem 21E

The E⁰cell is 1.14 V

Reaction at anode:

  $\text{Mn}\to {\text{Mn}}^{\text{2+}}\text{+}2{\text{e}}^{\text{-}}$

Reaction at cathode:

  ${\text{Fe}}^{3+}\text{+}3{\text{e}}^{\text{-}}\to \text{Fe}$

Overall reaction:

  $\text{3Mn}\left(\text{s}\right)\text{+}{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\to {\text{3Mn}}^{\text{2+}}\left(\text{aq}\right)\text{+}\text{2Fe}\left(\text{s}\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}{\text{Mn}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\to \text{Mn}{\text{E}}^{\text{0}}\text{=}\text{-1}\text{.18}\text{V}\\ {\text{Fe}}^{\text{3+}}\text{+}{\text{3e}}^{\text{-}}\to \text{Fe}{\text{E}}^{\text{0}}\text{=}\text{-}\text{0}\text{.036}\text{V}\end{array}$

The diagram is shown below:

  

The direction of flow of electrons is from anode to cathode.

Negative ions flow towards anode.

At anode oxidation takes place

At cathode reduction takes place.

The oxidation half-cell reaction is shown below:

  $\text{Mn}\to {\text{Mn}}^{\text{2+}}\text{+}2{\text{e}}^{\text{-}}$

The reduction half-cell reaction is shown below:

  ${\text{Fe}}^{3+}\text{+}3{\text{e}}^{\text{-}}\to \text{Fe}$

The overall reaction is shown below:

  $\text{3Mn}\left(\text{s}\right)\text{+}{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\to {\text{3Mn}}^{\text{2+}}\left(\text{aq}\right)\text{+}\text{2Fe}\left(\text{s}\right)$

The calculation of E⁰cell is shown below:

E⁰cell = E⁰cathode − E⁰anode

  = - 0.036- (-1.18)

  = 1.14 V