Chapter 19, Problem 19.87SP

Interpretation Introduction

Interpretation:

The balanced cell reaction, $\text{E°}$ and $\Delta \text{G°}$ for the cell needs to be determined that is composed of by combining two of the given cell reactions with the smallest $\text{E°}$.

$\begin{array}{l}{\text{Co}}_{}^{\text{2+}}\left(aq\right){\text{ + 2 e}}^{\text{-}}\stackrel{}{\to }\text{Co}\left(s\right)\text{E° = -0}\text{.28 V}\\ {\text{I}}_{\text{2}}\left(s\right){\text{ + 2 e}}^{\text{-}}\stackrel{}{\to }{\text{ 2 I}}_{}^{-}\left(aq\right)\text{ E° = 0}\text{.54 V}\\ {\text{Cu}}_{}^{\text{2+}}\left(aq\right){\text{+ 2 e}}^{\text{-}}\stackrel{}{\to }\text{Cu}\left(s\right)\text{ E° = 0}\text{.34 V}\end{array}$

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 the help of electrode reduction potential values. The reaction at each electrode is called half-reaction and the 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.