Chapter 14, Problem 14.30P

(a)

Interpretation Introduction

Interpretation:

The net reaction for given solutions has to be written.

Concept introduction:

Identifying relevant half-reaction:

Reduction reaction of each half cell have to be identified.  In order to identify it, element with two oxidation number should be taken.   From the elements reduction reaction for both right and left hand cell is given.

Explanation of Solution

To write: The net reaction for given solutions.

Elements with two oxidation states are ${\text{Ce}}^{\text{3+}}$, ${\text{Ce}}^{\text{4+}}$ and ${\text{MnO}}_{\text{4}}{}^{\text{-}}$, ${\text{Mn}}^{\text{2+}}$.  The reduction reaction for both half-cell is given as

$\begin{array}{l}\text{Right hand cell:}{\text{5Ce}}^{\text{4+}}\text{+}{\text{5e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{5Ce}}^{\text{3+}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{1}\text{.92}\text{V}\\ \text{-}\\ \text{Left hand cell:}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\text{+}{\text{5e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{1}\text{.229V}\\ \\ \text{Net}\text{reaction}{\text{5Ce}}^{\text{4+}}\text{+}{\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }{\text{5Ce}}^{\text{3+}}\text{+}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\\ \\ {\text{E}}^{\text{0}}\text{=}\text{0}\text{.193}\text{V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The standard free energy and equilibrium constant of given cell has to be determined.

Concept introduction:

Relation between ${\text{E}}^{\text{0}}\text{and}\text{K}$ is given as

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\text{K}\\ \\ \text{K}\text{=}{\text{10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}\\ \\ {\text{ΔG}}^{\text{0}}\text{=}{\text{-nFE}}^{\text{0}}\\ \\ {\text{E}}^{\text{0}}\text{is}\text{standard}\text{reduction}\text{potential}\\ \\ \text{K}\text{is}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

To determine: The standard free energy and equilibrium constant of given cell.

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}0.193\text{V}\\ \\ {\text{ΔG}}^{\text{0}}\text{=}{\text{-nFE}}^{\text{0}}\text{=}\text{-1(96485}\text{C/mol)(0}\text{.193}\text{V)}\text{=-93}\text{.1KJ}\\ \\ \text{K}\text{=}{\text{10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}\text{=}{\text{10}}^{\text{1(}0.193\text{V}\text{)/0}\text{.05916}}\text{=}2\text{×}{\text{10}}^{\text{16}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The cell voltage of given cell has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

${\text{aA+ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{bB}$

The Nernst equation results in the half-cell potential E as,

$\text{E=E°-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

Here,

$\text{E°}$ = standard reduction potential ( ${\text{A}}_{\text{A}}{\text{=A}}_{\text{B}}\text{=1}$ )

$\text{R}$ = gas constant $\left(\text{8}\text{.314}\text{J/}\left(\text{K}\text{.mol}\right)\right)\text{=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)$

T =Temperature (in K)

N = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$ )

A = Activity of species, i

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode ( ${\text{E}}_{\text{+}}$) – potential of left hand electrode( ${\text{E}}_{-}$).

Explanation of Solution

To determine: The cell voltage of given cell.

$\begin{array}{l}\text{Right hand cell:}{\text{5Ce}}^{\text{4+}}\text{+}{\text{5e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{5Ce}}^{\text{3+}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{1}\text{.92}\text{V}\\ \text{-}\\ \text{Left hand cell:}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\text{+}{\text{5e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{1}\text{.229V}\\ \\ \text{Net}\text{reaction}{\text{5Ce}}^{\text{4+}}\text{+}{\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }{\text{5Ce}}^{\text{3+}}\text{+}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{8H}}^{\text{+}}\end{array}$

Applying the electrode potential in Nernst equation, the cell voltage is calculated as

$\begin{array}{l}{\text{E}}_{\text{+}}\text{=}\left(\text{1}\text{.70}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{[0}\text{.100]}}^{\text{5}}}{{\text{[1}\text{×}{\text{10}}^{\text{-4}}\text{]}}^{\text{5}}}\right)\text{-}\left(\text{-0}\text{.44}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{\text{[1}\text{×}{\text{10}}^{\text{-4}}\text{]}}{\text{[0}\text{.1][1}{\text{.0]}}^{\text{8}}}\right)\\ \\ \text{=}\text{1}\text{.5223 - 1}\text{.5425 = -0}\text{.020 V}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The change in free energy of given cell has to be determined.

Concept introduction:

Relation between change in free energy and change in electric potential:

$\text{ΔG}\text{=}\text{-n}\text{.N}\text{.F}\text{.E }$

Here,

$\text{E}$ is volts

n = unit charge per molecule

N = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$ )

Explanation of Solution

To determine: The change in free energy of given cell.

The free energy difference is calculated as

$\begin{array}{l}\text{ΔG}\text{=}\text{-n}\text{.N}\text{.F}\text{.E }\\ \\ \text{=}\text{-5mol×}\text{96485}\text{C/mol}\text{×}\text{-0}\text{.020}\text{V}\\ \\ \text{=}\text{+10KJ}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The $\text{pH}$ at which given solution attains equilibrium has to be determined.

Concept introduction:

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

Relation between ${\text{E}}^{\text{0}}\text{and}\text{K}$ is given as

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\text{K}\\ \\ \text{K}\text{=}{\text{10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}\\ \\ {\text{ΔG}}^{\text{0}}\text{=}{\text{-nFE}}^{\text{0}}\\ \\ {\text{E}}^{\text{0}}\text{is}\text{standard}\text{reduction}\text{potential}\\ \\ \text{K}\text{is}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

To determine: The $\text{pH}$ at which given solution attains equilibrium.

When the solution is at equilibrium the cell voltage E becomes zero.

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\text{K}\\ \\ {\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{5}}\text{log}\frac{{{\text{[Ce}}^{\text{3+}}\text{]}}^{\text{5}}{\text{[MnO}}_{\text{4}}{}^{\text{-}}{\text{][H}}^{\text{+}}{\text{]}}^{\text{8}}}{{\text{[Ce}}^{\text{4+}}{\text{][Mn}}^{\text{2+}}\text{]}}\\ \\ {\text{[H}}^{\text{+}}\text{]}\text{=}\text{0}\text{.62}\\ \\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\text{=}\text{-log(0}\text{.62)}\\ \\ \text{=}\text{0}\text{.21}\end{array}$