Chapter 20, Problem 46A

Interpretation Introduction

Interpretation:

The cell potential of biofuel cell is to be calculated.

Concept introduction:

Anode is that electrode where oxidation is occurs and cathode is that electrode where reduction is occurs.

Answer to Problem 46A

The cell potential of the reaction is

$\begin{array}{l}{\text{E°}}_{\text{cathode}}{\text{= E°}}_{{\text{Fe}}^{\text{+3}}|{\text{Fe}}^{\text{+2}}}{\text{-E°}}_{\begin{array}{l}\text{NaDH}|{\text{NaD}}^{\text{+}}\end{array}}\\ \text{=0}\text{.36-}\left(\text{-0}\text{.32}\right)\\ \text{=0}\text{.68V}\end{array}$

The cell potential of the cell is $\text{0}\text{.68V}$

Explanation of Solution

Given information:

The standard reduction potential ${\text{Fe}}^{\text{+3}}|{\text{Fe}}^{\text{+2}}$ = $\text{0}\text{.36}$ V

${\text{E°}}_{\text{anode}}\text{=}\text{-0}\text{.32}\text{V}$

The cell potential of the given cell reaction is:

$\begin{array}{l}{\text{E°}}_{\text{cathode}}{\text{= E°}}_{{\text{Fe}}^{\text{+3}}|{\text{Fe}}^{\text{+2}}}{\text{-E°}}_{\begin{array}{l}\text{NaDH}|{\text{NaD}}^{\text{+}}\end{array}}\\ \text{=0}\text{.36-}\left(\text{-0}\text{.32}\right)\\ \text{=0}\text{.68V}\end{array}$

The cell potential of the cell is $\text{0}\text{.68V}$

Conclusion

The cell potential of the reaction is

$\begin{array}{l}{\text{E°}}_{\text{cathode}}{\text{= E°}}_{{\text{Fe}}^{\text{+3}}|{\text{Fe}}^{\text{+2}}}{\text{-E°}}_{\begin{array}{l}\text{NaDH}|{\text{NaD}}^{\text{+}}\end{array}}\\ \text{=0}\text{.36-}\left(\text{-0}\text{.32}\right)\\ \text{=0}\text{.68V}\end{array}$

The cell potential of the cell is $\text{0}\text{.68V}$