Chapter 16, Problem 16.15P

(a)

Interpretation Introduction

Interpretation:

Equilibrium constant for the reaction has to be determined.

  ${\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}\rightleftharpoons 2{\text{I}}^{-}$

Concept Introduction:

For any spontaneous redox reaction the relation between ${\text{E}}_{\text{Cell}}^{\text{0}}$ and $\Delta {\text{G}}^o$ is as follows:

    $\Delta {\text{G}}^0=-{\text{nFE}}_{\text{cell}}^{\text{0}}$

Here, $\Delta {\text{G}}^o$ is the standard Gibbs free energy change of the reaction.

${\text{E}}_{\text{Cell}}^{\text{0}}$ is cell potential of overall reaction.

$\text{n}$ is number of electron transfered.

$\text{F}$ is Faraday constant.

The relation between equilibrium constant $\left(\text{K}\right)$ and $\Delta {\text{G}}^{\text{o}}$ is as follows:

    $\Delta {\text{G}}^0=-\text{RTlnK}$

Here, $\Delta {\text{G}}^{\text{o}}$ is standard Gibbs free energy change of the reaction.

$\text{K}$ is equilibrium constant of the reaction.

$\text{R}$ is gas constant.

$\text{T}$ is temperature in Kelvin scale.

Hence the relation between ${\text{E}}_{\text{Cell}}^{\text{0}}$ and equilibrium constant $\left(\text{K}\right)$ is as follows:

    $\text{K}={\text{e}}^{\frac{{\text{nFE}}_{\text{cell}}^{\text{0}}}{\text{RT}}}$

(b)

Interpretation Introduction

Interpretation:

Equilibrium constant of the reaction has to be determined.

  ${\text{I}}_{\text{2}}\left(\text{s}\right)+2{\text{e}}^{-}\rightleftharpoons 2{\text{I}}^{-}$

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

Solubility of ${\text{I}}_{\text{2}}$ in water has to be determined.

Concept Introduction:

For any spontaneous redox reaction the relation between ${\text{E}}_{\text{Cell}}^{\text{0}}$ and $\Delta {\text{G}}^{\text{o}}$ is as follows:

    $\Delta {\text{G}}^{\text{o}}=-{\text{nFE}}_{\text{cell}}^{\text{0}}$

Here, $\Delta {\text{G}}^{\text{o}}$ is the Gibbs free energy change of the reaction.

${\text{E}}_{\text{Cell}}^{\text{0}}$ is cell potential of overall reaction.

$\text{n}$ is number of electron transferred.

$\text{F}$ is Faraday constant.

The relation between equilibrium constant $\left(\text{K}\right)$ and $\Delta {\text{G}}^{\text{o}}$ is as follows:

    $\Delta {\text{G}}^{\text{o}}=-\text{RTlnK}$

Here, $\Delta {\text{G}}^{\text{o}}$ is the Gibbs free energy change of the reaction.

$\text{K}$ is equilibrium constant of overall reaction.

$\text{R}$ is universal gas constant.

$\text{T}$ is temperature in Kelvin scale.

Hence the relation between ${\text{E}}_{\text{Cell}}^{\text{0}}$ and equilibrium constant $\left(\text{K}\right)$ is as follows:

    $\text{K}={\text{e}}^{\frac{{\text{nFE}}_{\text{cell}}^{\text{0}}}{\text{RT}}}$

For solubility of any element, equilibrium constant $\left(\text{K}\right)$ of solubility reaction is equal to the solubility of that element.