Chapter 22, Problem 22.165MP

Interpretation Introduction

Interpretation:

The standard cell potential E◦ and ΔG◦needs to be determined for the given reactions and the spontaneity of disproportionation reactions needs to be determined.

${\text{3 M}}_{\text{aq}}^{\text{+}}\stackrel{}{\to }{\text{M}}_{\text{aq}}^{\text{+}}{\text{+ 2 M}}_{\text{(s) }}\text{ M=InorTl}$

Concept introduction:

In the electrochemical cell, the reactions at cathode and anode occur due to the difference in their reduction electrode potential value. The EMF of the cell can be calculated with help of electrode reduction potential values. The reaction at each electrode is called as half-reaction and combination of both half-reactions gives the cell reaction of given electrochemical cell. The standard cell potential for an electrochemical cell can be calculated as:

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{cathode}}^{\text{°}}\text{}{\text{ - E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{reduction}}^{\text{°}}\text{}{\text{ - E}}_{\text{oxidation}}^{\text{°}}\end{array}$

Cathode involves the reduction process whereas oxidation occurs at the anode. The spontaneity of the reaction can be determined with the help of $\Delta {\text{G}}_r^{°}$ and ${\text{E}}_{\text{cell}}^{\text{°}}$ can be written as:

${\text{ΔG}}_{\text{r}}^{\text{°}}\text{= -n }\times \text{F }\times {\text{ E}}_{\text{cell}}^{\text{°}}$

Here n is the number of e- involve in half-reaction and F is 96458 C/mol e-. Hence the negative value of ${\text{E}}_{\text{cell}}^{\text{°}}$ indicates the non-spontaneity of reaction.