Chapter 6, Problem 6L.2E

(a)

Interpretation Introduction

Interpretation:

The Gibbs free energy of the reaction has to be written.

The given cell reaction is,

  ${\text{3Cr}}^{\text{3+}}\left(\text{aq}\right)\text{+Bi}\left(\text{s}\right)\stackrel{}{\to }{\text{3Cr}}^{\text{2+}}\left(\text{aq}\right){\text{+Bi}}^{\text{3+}}\left(\text{aq}\right){\text{E}}_{\text{cell}}{}^{\text{o}}\text{=-0}\text{.61}\text{V}$

Concept Introduction:

The standard Gibbs free energy is defined as the difference of the molar Gibbs free energies of the products and reactants in their standard states.

The Gibbs free energy of the reaction is calculated as,

  ${\text{ΔG=-nFE}}_{\text{cell}}$

Here, $\text{ΔG=}$ Gibbs free energy

    $\text{n=}$ Number of electrons transferred (in moles)

  $\text{F=}$ Faraday’s constant $\left(\text{9}{\text{.6485×10}}^{\text{4}}\text{C}\cdot {\text{mol}}^{\text{-1}}\right)$

  ${\text{E}}_{\text{cell}}\text{=}$ Cell potential

(b)

Interpretation Introduction

Interpretation:

The Gibbs free energy of the reaction has to be written.

The given cell reaction is,

  $\text{Mg}\left(\text{s}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{Mg}}^{\text{2+}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{aq}\right){\text{+2OH}}^{\text{+}}\left(\text{aq}\right){\text{E}}_{\text{cell}}{}^{\text{o}}\text{=2}\text{.36}\text{V}$

Concept Introduction:

Refer to part (a).