Chapter 14, Problem 14.DE

(a)

Interpretation Introduction

Interpretation:

The Nernst equation for given cell has to be written.

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 write: The Nernst equation for given cell.

$\begin{array}{l}\text{Right hand cell:}{\text{H}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{0}{\text{.5H}}^{\text{+}}{}_{\text{(g)}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{0V}\\ \text{-}\\ \text{Left hand cell:}{\text{Ag}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Ag}}_{\text{(s)}}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{0}\text{.799V}\\ \\ \text{E}\text{=}{\text{E}}_{\text{+}}{}^{\text{0}}\text{-}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{-0}\text{.799V}\\ \\ \text{E}\text{=}\left(\text{0-}\text{0}\text{.05916}\text{log}\frac{{\text{P}}_{{\text{H}}_{\text{2}}}{}^{\text{1/2}}}{{\text{[H}}^{\text{+}}\text{]}}\right)\text{-}\left(\text{0}\text{.799 - 0}\text{.059 16 1og}\frac{\text{1}}{{\text{[Ag}}^{\text{+}}\text{]}}\right)\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cell voltage for the given cell and direction of flow of electrons have 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 for the given cell and direction of flow of electrons.

The concentration of silver ion is calculated using solubility product

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[I}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{8}\text{.3}\text{×}{\text{10}}^{\text{-17}}}{\text{0}\text{.10}}\text{=}\text{8}\text{.3}\text{×}{\text{10}}^{\text{-16}}\text{M}\end{array}$

The cell voltage is determined as

$\begin{array}{l}\text{E}\text{=}\left(\text{0-}\text{0}\text{.05916}\text{log}\frac{{\text{P}}_{{\text{H}}_{\text{2}}}{}^{\text{1/2}}}{{\text{[H}}^{\text{+}}\text{]}}\right)\text{-}\left(\text{0}\text{.799 - 0}\text{.059 16 1og}\frac{\text{1}}{{\text{[Ag}}^{\text{+}}\text{]}}\right)\\ \\ \text{E}\text{=}\left(\text{0-}\text{0}\text{.05916}\text{log}\frac{\sqrt[]{\text{0}\text{.20}}}{\text{0}\text{.10}}\right)\text{-}\left(\text{0}\text{.799 - 0}\text{.059 16 1og}\frac{\text{1}}{\text{8}\text{.3}\text{×}{\text{10}}^{\text{-16}}}\right)\\ \\ \text{= -0}\text{.038-(-0}\text{.093) = 0}\text{.055 V}\end{array}$

The flow of electron from right hand electrode silver to left hand electrode platinum because the cell voltage is positive.

(c)

Interpretation Introduction

Interpretation:

The standard reduction potential by deriving Nernst equation 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 standard reduction potential by deriving Nernst equation.

$\begin{array}{l}\text{Right hand cell:}{\text{H}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{0}{\text{.5H}}_{\text{2}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{0}\text{V}\\ \text{-}\\ \text{Left hand cell:}{\text{AgI}}_{\text{(s)}}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Ag}}_{\text{(s)}}\text{+}{\text{I}}^{\text{-}}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{?}\end{array}$

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

$\begin{array}{l}\text{0}\text{.055}\text{=}\left(\text{0}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{\sqrt[]{\text{0}\text{.20}}}{\text{0}\text{.10}}\right)\text{-}\left({\text{E}}^{\text{0}}\text{-}\text{0}\text{.05916}\text{log}\text{(0}\text{.10)}\right)\\ \\ {\text{E}}^{\text{0}}\text{=}\text{-0}\text{.153 V}\end{array}$

The appendix value ${\text{E}}^{\text{0}}\text{=}\text{-0}\text{.152 V}$ is near to calculated value ${\text{E}}^{\text{0}}\text{=}\text{-0}\text{.153 V}$.