Chapter 5, Problem 5I.1E

(a)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Ag}}_{\text{(s)}}{\text{|AgNO}}_{{\text{3(aq,m}}_{\text{L}}\text{)}}{\text{||AgNO}}_{{\text{3(aq,m}}_{\text{R}}\text{)}}{\text{|Ag}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Ag}}_{\text{(s)}}{\text{|AgNO}}_{{\text{3(aq,m}}_{\text{L}}\text{)}}{\text{||AgNO}}_{{\text{3(aq,m}}_{\text{R}}\text{)}}{\text{|Ag}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

  ${\text{AgNO}}_{{\text{3(aq,m}}_{\text{R}}\text{)}}{\text{+e}}^{\text{-}}\to {\text{Ag}}_{\text{(s)}}$

Left-hand half-reaction:

  ${\text{AgNO}}_{{\text{3(aq,m}}_{\text{L}}\text{)}}{\text{+e}}^{\text{-}}\to {\text{Ag}}_{\text{(s)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.

  ${\text{AgNO}}_{{\text{3(aq,m}}_{\text{R}}\text{)}}{\text{+ e}}^{\text{-}}{\text{+ Ag}}_{\text{(s)}}\to {\text{AgNO}}_{{\text{3(aq,m}}_{\text{L}}\text{)}}{\text{+ e}}^{\text{-}}+{\text{Ag}}_{\text{(s)}}$

The common components on both sides gets cancelled.  The cell reaction is given as

  ${\text{Ag}}^{\text{+}}{}_{{\text{(aq,m}}_{\text{R}}\text{)}}\to {\text{Ag}}^{\text{+}}{}_{{\text{(aq,m}}_{\text{L}}\text{)}}$

Nernst equation:

It is a one-electron reaction.  Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  $\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\text{m}}_{\text{L}}}{{\text{m}}_{\text{R}}}\\ \\ {\text{E}}_{\text{cell}}\text{ = -}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\text{m}}_{\text{L}}}{{\text{m}}_{\text{R}}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Pt}}_{\text{(s)}}{\text{|H}}_{{\text{2(g,p}}_{\text{L}}\text{)}}{\text{|HCl}}_{\text{(aq)}}{\text{|H}}_{\text{2}}{}_{{\text{(g, p}}_{\text{R}}\text{)}}{\text{|Pt}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Pt}}_{\text{(s)}}{\text{|H}}_{{\text{2(g,p}}_{\text{L}}\text{)}}{\text{|HCl}}_{\text{(aq)}}{\text{|H}}_{\text{2}}{}_{{\text{(g, p}}_{\text{R}}\text{)}}{\text{|Pt}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

  ${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+e}}^{\text{-}}\to \frac{\text{1}}{\text{2}}{\text{H}}_{{\text{2(g,p}}_{\text{R}}\text{)}}$

Left-hand half-reaction:

  ${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+e}}^{\text{-}}\to \frac{\text{1}}{\text{2}}{\text{H}}_{{\text{2(g,p}}_{\text{L}}\text{)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.

  $\frac{\text{1}}{\text{2}}{\text{H}}_{{\text{2(g,p}}_{\text{L}}\text{)}}\to \frac{\text{1}}{\text{2}}{\text{H}}_{{\text{2(g,p}}_{\text{R}}\text{)}}$

Nernst equation:

It is a one-electron reaction.  Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  $\begin{array}{l}{\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\text{p}}_{\text{L}}}{{\text{p}}_{\text{R}}}\\ \\ {\text{E}}_{\text{cell}}\text{ = -}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\text{p}}_{\text{L}}}{{\text{p}}_{\text{R}}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Pt}}_{\text{(s)}}{\text{|K}}_{\text{3}}{{\text{[Fe(CN)}}_{\text{6}}\text{]}}_{\text{(aq)}}{\text{,K}}_{\text{4}}{{\text{[Fe(CN)}}_{\text{6}}\text{]}}_{\text{(aq)}}{\text{||Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{,H}}^{\text{+}}{}_{\text{(aq)}}{\text{|MnO}}_{\text{2(s)}}{\text{|Pt}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Pt}}_{\text{(s)}}{\text{|K}}_{\text{3}}{{\text{[Fe(CN)}}_{\text{6}}\text{]}}_{\text{(aq)}}{\text{,K}}_{\text{4}}{{\text{[Fe(CN)}}_{\text{6}}\text{]}}_{\text{(aq)}}{\text{||Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{,H}}^{\text{+}}{}_{\text{(aq)}}{\text{|MnO}}_{\text{2(s)}}{\text{|Pt}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

It corresponds to the reduction of ${\text{MnO}}_{\text{2}}{\text{ to Mn}}^{\text{2+}}$.

  ${\text{MnO}}_{\text{2(s)}}{\text{+ 4H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Left-hand half-reaction:

It corresponds to the reduction of ${\text{Fe}}^{\text{3+}}{\text{ to Fe}}^{\text{2+}}$.

  ${\text{Fe}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\to {\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.  By reversing and doubling the stoichiometric coefficients of the left hand half reaction, the balanced equation can be written as

  ${\text{MnO}}_{\text{2(s)}}{\text{+ 4H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ 2Fe}}^{\text{3+}}{}_{\text{(aq)}}\to {\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+ 2Fe}}^{\text{2+}}{}_{\text{(aq)}}$

Nernst equation:

Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{c}}_{{\text{Mn}}^{\text{2+}}}{\text{c}}_{{}_{{\text{Fe}}^{\text{2+}}}}^{\text{2}}{\text{c}}^{{\sout{\text{ο}}}^{\text{3}}}}{{\text{c}}_{{\text{H}}^{\text{+}}}^{\text{4}}{\text{c}}_{{\text{Fe}}^{\text{3+}}}^{\text{2}}}$

(d)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Pt}}_{\text{(s)}}{\text{|Cl}}_{\text{2(g)}}{\text{|HCl}}_{\text{(aq)}}{\text{||HBr}}_{\text{(aq)}}{\text{|Br}}_{\text{2(l)}}{\text{|Pt}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Pt}}_{\text{(s)}}{\text{|Cl}}_{\text{2(g)}}{\text{|HCl}}_{\text{(aq)}}{\text{||HBr}}_{\text{(aq)}}{\text{|Br}}_{\text{2(l)}}{\text{|Pt}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

It corresponds to the reduction of ${\text{Br}}_{\text{2}}{\text{ to Br}}^{-}$.

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

Left-hand half-reaction:

It corresponds to the reduction of ${\text{Cl}}_2{\text{ to Cl}}^{-}$.

  $\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}{\text{+ e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}{}_{\text{(aq)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.

  $\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2(l)}}{\text{+ Cl}}^{\text{-}}{}_{\text{(aq)}}\to {\text{Br}}^{\text{-}}{}_{\text{(aq)}}\text{+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}$

Nernst equation:

Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\text{(p}}_{{\text{Cl}}_{\text{2}}}{\text{/p}}^{\sout{\text{ο}}}{\text{)c}}_{{\text{Br}}^{\text{-}}}}{{\text{c}}_{{\text{Cl}}^{\text{-}}}}$

(e)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Pt}}_{\text{(s)}}{\text{|Fe}}^{\text{3+}}{}_{\text{(aq)}}{\text{,Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{||Sn}}^{\text{4+}}{}_{\text{(aq)}}{\text{,Sn}}^{\text{2+}}{}_{\text{(aq)}}{\text{|Pt}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Pt}}_{\text{(s)}}{\text{|Fe}}^{\text{3+}}{}_{\text{(aq)}}{\text{,Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{||Sn}}^{\text{4+}}{}_{\text{(aq)}}{\text{,Sn}}^{\text{2+}}{}_{\text{(aq)}}{\text{|Pt}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

It corresponds to the reduction of ${\text{Sn}}^{\text{4+}}{\text{ to Sn}}^{\text{2+}}$.

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

Left-hand half-reaction:

It corresponds to the reduction of ${\text{Fe}}^{\text{3+}}{\text{ to Fe}}^{\text{2+}}$.

  ${\text{Fe}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\to {\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.

  ${\text{Sn}}^{\text{4+}}{}_{\text{(aq)}}{\text{+ 2Fe}}^{\text{2+}}{}_{\text{(aq)}}\to {\text{Sn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2Fe}}^{\text{3+}}{}_{\text{(aq)}}$

Nernst equation:

Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{c}}_{{\text{Sn}}^{\text{2+}}}{\text{c}}_{{}_{{\text{Fe}}^{\text{3+}}}}^{\text{2}}}{{\text{c}}_{{\text{Sn}}^{\text{4+}}}{\text{c}}_{{\text{Fe}}^{\text{2+}}}^{\text{2}}}$

(f)

Interpretation Introduction

Interpretation:

The cell reaction, electrode half-reaction and Nernst equation for the given cell has to be written.

  ${\text{Fe}}_{\text{(s)}}{\text{|Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{||Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{,H}}^{\text{+}}{}_{\text{(aq)}}{\text{|MnO}}_{\text{2(s)}}{\text{|Pt}}_{\text{(s)}}$

Concept Introduction:

Cell reaction:

The overall reaction which takes place in the cell, written on the assumption that the right hand electrode is the cathode, that is assuming that the spontaneous reaction is the reduction that occurs in the right hand compartment.

Electrode half-reaction:

On the anode half reaction, oxidation occurs.  For example,

  ${\text{Zn}}_{\text{(s)}}{\text{= Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}$

On the cathode half reaction, reduction occurs.  For example,

  ${\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}{\text{ = Cu}}_{\text{(s)}}$

Nernst equation:

It is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and concentrations of chemical species undergoing reduction and oxidation.

The Nernst equation for an electrochemical half-cell is

  $\begin{array}{l}{\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{lnQ}\\ \\ {\text{E}}_{\text{red}}{\text{= E}}_{\text{red}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}\end{array}$

The Nernst equation for an electrochemical reaction (full cell)

  ${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{-}}\text{-}\frac{\text{RT}}{\text{zF}}{\text{lnQ}}_{\text{r}}$

Explanation of Solution

The given cell is

  ${\text{Fe}}_{\text{(s)}}{\text{|Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{||Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{,H}}^{\text{+}}{}_{\text{(aq)}}{\text{|MnO}}_{\text{2(s)}}{\text{|Pt}}_{\text{(s)}}$

The half-cell reaction can be given as

Right-hand half-reaction:

It corresponds to the reduction of ${\text{MnO}}_{\text{2}}{\text{ to Mn}}^{2+}$.

  ${\text{MnO}}_{\text{2(s)}}{\text{+ 4H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Left-hand half-reaction:

It corresponds to the reduction of ${\text{Fe}}^{\text{2+}}\text{ to Fe}$.

  ${\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}\to {\text{Fe}}_{\text{(s)}}$

The combination of right-hand half-reaction and left-hand half-reaction gives an overall cell reaction.

  ${\text{MnO}}_{\text{2(s)}}{\text{+ Fe}}_{\text{(s)}}{\text{+ 4H}}^{\text{+}}{}_{\text{(aq)}}\to {\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Nernst equation:

Assuming that solutions are sufficiently dilute, the Nernst equation can be given as

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\sout{\text{ο}}}\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{c}}_{{\text{Mn}}^{\text{2+}}}{\text{c}}_{{}_{{\text{Fe}}^{\text{2+}}}}^{}{\text{c}}^{{\sout{\text{ο}}}^{\text{2}}}}{{\text{c}}_{{\text{H}}^{\text{+}}}^{\text{4}}}$