Chapter 19, Problem 27E

Interpretation Introduction

(a)

Interpretation:

The value of E °_cell should be determined for the given reaction as follows: ${\text{MnO}}_4^{-}(\text{aq)}+{\text{8H}}^{\text{+}}(\text{aq) }+5{\text{Ce}}^{\text{2+}}(\text{aq)}\to 5{\text{Ce}}^{\text{4+}}(\text{aq)}+{\text{Mn}}^{\text{2+}}(\text{aq})+4{\text{H}}_{\text{2}}\text{O(l})$

Concept introduction:

The E°_cell value for a redox reaction gives by following equation.

$E{°}_{cell}=E{°}_{Reduction}-E{°}_{Oxidation}$

Interpretation Introduction

(b)

Interpretation:

The Δ_rG° should be determined for the reaction as follows: ${\text{MnO}}_4^{-}(\text{aq)}+{\text{8H}}^{\text{+}}(\text{aq) }+5{\text{Ce}}^{\text{2+}}(\text{aq)}\to 5{\text{Ce}}^{\text{4+}}(\text{aq)}+{\text{Mn}}^{\text{2+}}(\text{aq})+4{\text{H}}_{\text{2}}\text{O(l})$

Concept introduction:

The E°_cell value for a redox reaction gives by following equation.

$E{°}_{cell}=E{°}_{Reduction}-E{°}_{Oxidation}$

When the reactants and products are in their standard states,

$\begin{array}{l}{\Delta }_{\text{r}}{\text{G}}^{\circ }=-z\text{F}{E}_{cell}^{\circ }\\ \text{Where,}\\ z=\text{Electron number}\\ \text{F}=\text{Faraday constant}=96,485\text{C}{\text{mol}}^{-1}\\ E_{cell}^{\circ }=\text{Standard }\text{}\text{Electron cell potential}\end{array}$

The electron number, z, for a reaction depends on how the reaction is written.

Interpretation Introduction

(c)

Interpretation:

The K should be determined for the reaction as follows: ${\text{MnO}}_4^{-}(\text{aq)}+{\text{8H}}^{\text{+}}(\text{aq) }+5{\text{Ce}}^{\text{2+}}(\text{aq)}\to 5{\text{Ce}}^{\text{4+}}(\text{aq)}+{\text{Mn}}^{\text{2+}}(\text{aq})+4{\text{H}}_{\text{2}}\text{O(l})$.

Concept introduction:

The E°_cell value for a redox reaction gives by following equation.

$E{°}_{cell}=E{°}_{Reduction}-E{°}_{Oxidation}$

When the reactants and products are in their standard states,

$\begin{array}{l}{\Delta }_{\text{r}}{\text{G}}^{\circ }=-z\text{F}{E}_{cell}^{\circ }\\ \text{Where,}\\ z=\text{Electron number}\\ \text{F}=\text{Faraday constant}=96,485\text{C}{\text{mol}}^{-1}\\ E_{cell}^{\circ }=\text{Standard }\text{}\text{Electron cell potential}\end{array}$

The electron number, z, for a reaction depends on how the reaction is written.

Also,

$\begin{array}{l}{\Delta }_{\text{r}}{\text{G}}^{\circ }=-\text{RT}\ln \text{K}\\ -\text{RT}\ln \text{K}=-z\text{F}{E}_{cell}^{\circ }\\ \therefore \ln \text{K}=\frac{-z\text{F}{E}_{cell}^{\circ }}{-\text{R}T}\\ \ln \text{K}=11604\text{C}{\text{J}}^{\text{-1}}\frac{z{E}_{cell}^{\circ }}{T}\\ \text{K}=e^{11604\text{C}{\text{J}}^{\text{-1}}\frac{z{E}_{cell}^{\circ }}{T}}\end{array}$

Interpretation Introduction

(d)

Interpretation:

It should be determined, whether the given reaction completes or not. ${\text{MnO}}_4^{-}(\text{aq)}+{\text{8H}}^{\text{+}}(\text{aq) }+5{\text{Ce}}^{\text{2+}}(\text{aq)}\to 5{\text{Ce}}^{\text{4+}}(\text{aq)}+{\text{Mn}}^{\text{2+}}(\text{aq})+4{\text{H}}_{\text{2}}\text{O(l})$

Concept introduction:

The E°_cell value for a redox reaction gives by following equation.

$E{°}_{cell}=E{°}_{Reduction}-E{°}_{Oxidation}$

When the reactants and products are in their standard states,

$\begin{array}{l}{\Delta }_{\text{r}}{\text{G}}^{\circ }=-z\text{F}{E}_{cell}^{\circ }\\ \text{Where,}\\ z=\text{Electron number}\\ \text{F}=\text{Faraday constant}=96,485\text{C}{\text{mol}}^{-1}\\ E_{cell}^{\circ }=\text{Standard }\text{}\text{Electron cell potential}\end{array}$

The electron number, z, for a reaction depends on how the reaction is written.