Chapter 11, Problem 23E

(a)

Interpretation Introduction

Interpretation: The standard line notation for the cell having following anode and cathode reactions needs to be determined.

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

Concept Introduction : Standard line notation is way to express a reaction a reaction in the electrochemical cell. In this notation, the cathode is separarted from the anode by using a salt bridge. In this cell, the anode is placed on the left and cathode on the right

Oxidation half cell: In this half-cell oxidation of metal occurs at anode

Reduction half cell: In this half-cell reduction of ion occurs at cathode

Flow of electron in the cell is from anode to cathode.

Answer to Problem 23E

The cell notation is,

  $\text{Pt|}{\text{Br}}^{\text{-}}\text{(1}\text{M)}{\text{Br}}_{\text{2}}\text{(1}\text{atm)||}{\text{Cl}}_{\text{2}}\text{(1}\text{atm)}\text{|}{\text{Cl}}^{\text{-}}\text{(1}\text{M)}\text{|}\text{Pt}$

Explanation of Solution

First cell is

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

From the above reaction it can be seen that ${\text{Cl}}_{\text{2}}$ have more reduction potential hence get reduced and ${\text{Br}}_{\text{2}}$ get oxidized so the cell notation will be,

  $\text{Pt|}{\text{Br}}^{\text{-}}\text{(1}\text{M)}{\text{Br}}_{\text{2}}\text{(1}\text{atm)||}{\text{Cl}}_{\text{2}}\text{(1}\text{atm)}\text{|}{\text{Cl}}^{\text{-}}\text{(1}\text{M)}\text{|}\text{Pt}$

The reaction is occurring at platinum electrode.

Then the oxidation half-cell is:

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

And the reduction half-cell is:

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

(b)

Interpretation Introduction

Interpretation: The standard line notation for the cell having following anode and cathode reactions needs to be determined.

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

Concept Introduction : Standard line notation is way to express a reaction a reaction in the electrochemical cell. In this notation, the cathode is separarted from the anode by using a salt bridge. In this cell, the anode is placed on the left and cathode on the right

Oxidation half cell: In this half-cell oxidation of metal occurs at anode

Reduction half cell: In this half-cell reduction of ion occurs at cathode

Answer to Problem 23E

The cell notation is,

  $\text{Pt|}{\text{Mn}}^{\text{2+}}\text{(1}\text{M),}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{||}{\text{IO}}_{\text{4}}{}^{\text{-}}\text{(1}{\text{M),IO}}_{\text{3}}{}^{\text{-}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}\text{Pt}$

Explanation of Solution

The second cell is

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

From the above reaction it can be seen that ${\text{IO}}_{\text{4}}{}^{\text{-}}$ have more reduction potential hence get reduced and ${\text{MnO}}_{\text{4}}{}^{\text{-}}$ get oxidized so the cell notation will be,

  $\text{Pt|}{\text{Mn}}^{\text{2+}}\text{(1}\text{M),}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{||}{\text{IO}}_{\text{4}}{}^{\text{-}}\text{(1}{\text{M),IO}}_{\text{3}}{}^{\text{-}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}\text{Pt}$

The reaction is occurring at platinum electrode.

Then the oxidation half-cell is:

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

And the reduction half-cell is:

  ${\text{IO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}\to {\text{IO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}$

(c)

Interpretation Introduction

Interpretation: The standard line notation for the cell having following anode and cathode reactions needs to be determined.

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

Concept Introduction : Standard line notation is way to express a reaction a reaction in the electrochemical cell. In this notation, the cathode is separarted from the anode by using a salt bridge. In this cell, the anode is placed on the left and cathode on the right

Oxidation half cell: In this half-cell oxidation of metal occurs at anode

Reduction half cell: In this half-cell reduction of ion occurs at cathode

Answer to Problem 23E

The cell notation is:

  $\text{Pt|}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}{\text{O}}_{\text{2}}\text{(1}\text{atm)}\text{||}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}\text{Pt}$

Explanation of Solution

The third cell is

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

From the above reaction it can be seen that ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ have more reduction potential hence get reduced easily and in ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ ,oxygen is represented as ${\text{O}}_{\text{2}}{}^{2-}$ and it gets reduced to ${\text{O}}^{\text{2-}}$ so the cell notation will be,

  $\text{Pt|}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}{\text{O}}_{\text{2}}\text{(1}\text{atm)}\text{||}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(1}\text{M),}{\text{H}}^{\text{+}}\text{(1}\text{M)}\text{|}\text{Pt}$

The reaction is occurring at platinum electrode.

Then the oxidation half-cell is:

  ${\text{2H}}_{\text{2}}{\text{O}}_2\to {\text{O}}_{\text{2}}\text{+}{\text{2H}}^{\text{+}}\text{+}{\text{2e}}^{\text{-}}$

And the reduction half-cell is:

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

(d)

Interpretation Introduction

Interpretation: The standard line notation for the cell having following anode and cathode reactions needs to be determined.

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

Concept Introduction:Standard line notation is way to express a reaction a reaction in the electrochemical cell. In this notation, the cathode is separarted from the anode by using a salt bridge. In this cell, the anode is placed on the left and cathode on the right

Oxidation half cell: In this half-cell oxidation of metal occurs at anode

Reduction half cell: In this half-cell reduction of ion occurs at cathode

Answer to Problem 23E

The cell notation is,

  $\text{Mn}\text{|}{\text{Mn}}^{\text{2+}}\text{(1}\text{M)}\text{||}{\text{Fe}}^{\text{3+}}\text{(1}\text{M)}\text{|}\text{Fe}$

Explanation of Solution

The fourth cell is

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

In the above reaction it can be seen that ${\text{Fe}}^{\text{3+}}$ has negative reduction potential hence get easily reduced and ${\text{Mn}}^{\text{2+}}$ have large negative reduction potential hence get easily oxidized, so the cell notation will be,

  $\text{Mn}\text{|}{\text{Mn}}^{\text{2+}}\text{(1}\text{M)}\text{||}{\text{Fe}}^{\text{3+}}\text{(1}\text{M)}\text{|}\text{Fe}$

The reaction is occurring at platinum electrode.

Then the oxidation half-cell is:

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

And the reduction half-cell is:

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