Chapter 19, Problem 19.5IA

(a)

Interpretation Introduction

Interpretation:

The thermodynamic limit to the zero-current potential of fuel cells operating on hydrogen and oxygen has to be calculated.

Explanation of Solution

The zero-current potential of fuel cell operating on hydrogen and oxygen is calculated as,

    $\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ {\text{H}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\underset{}{\overset{}{\to }}{\text{CO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}\text{=}\text{-237KJ/mol}\\ \\ \text{ν}\text{=}\text{2}\\ \\ {\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ \text{=}\frac{\text{-(-237kJ/mol)}}{\text{2}\text{×}\text{96}\text{.48}\text{kC/mol}}\\ \\ \text{=}\text{+1}\text{.23V}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The thermodynamic limit to the zero-current potential of fuel cells operating on methane and air has to be calculated.

Explanation of Solution

The zero-current potential of fuel cell operating on methane and air is calculated as,

    $\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ {\text{CH}}_{\text{4}}\text{+}{\text{2O}}_{\text{2}}\underset{}{\overset{}{\to }}{\text{CO}}_{\text{2}}\text{+}{\text{2H}}_{\text{2}}\text{O}\\ \\ {\text{Δ}}_{\text{r}}{\text{G}}^{\text{0}}\text{=}\text{2}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(H}}_{\text{2}}\text{O)}\text{+}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(CO}}_{\text{2}}\text{)}\text{-}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(CH}}_{\text{4}}\text{)}\\ \\ \text{=}\text{[(2)}\times \text{(-237}\text{.1)}\text{+ (-394}\text{.4)}\text{-}\text{(-50}\text{.7)]KJ/mol}=-817.9\text{KJ/mol}\\ \\ \text{ν}\text{=}8\\ \\ {\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ \text{=}\frac{\text{-(}-817.9\text{KJ/mol)}}{8\text{×}\text{96}\text{.48}\text{kC/mol}}\\ \\ \text{=}\text{+1}\text{.06V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The thermodynamic limit to the zero-current potential of fuel cells operating on propane and air has to be calculated.

Explanation of Solution

The zero-current potential of fuel cell operating on propane and air is calculated as,

    $\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{+}{\text{5O}}_{\text{2}}\underset{}{\overset{}{\to }}{\text{3CO}}_{\text{2}}\text{+}{\text{4H}}_{\text{2}}\text{O}\\ \\ {\text{Δ}}_{\text{r}}{\text{G}}^{\text{0}}\text{=}\text{4}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(H}}_{\text{2}}\text{O)}\text{+}{\text{3Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(CO}}_{\text{2}}\text{)}\text{-}{\text{Δ}}_{\text{f}}{\text{G}}^{\text{0}}{\text{(C}}_{\text{3}}{\text{H}}_{\text{8}}\text{)}\\ \\ \text{=}\text{[(4)}\text{(-237}\text{.1)}\text{+ 3(-394}\text{.4)}\text{-}\text{(-23}\text{.4)]KJ/mol}\text{=}\text{-2108}\text{.2KJ/mol}\\ \\ \text{ν}\text{=}\text{16}\\ \\ {\text{E}}^{\text{0}}\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{G}}^{\text{0}}}{\text{νF}}\\ \\ \text{=}\frac{\text{-(-2108}\text{.2KJ/mol)}}{\text{16}\text{×}\text{96}\text{.48}\text{kC/mol}}\\ \\ \text{=}\text{+1}\text{.36V}\text{.}\end{array}$