Question

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is

$\mathbf{2}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

The two half-cell reactions are

$$\begin{gathered} \mathbf{CO}\mathbf{+}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}\mathbf{\to }{\mathbf{CO}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}} \\ {\mathbf{O}}_{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{O}}^{\mathbf{2}\mathbf{-}} \end{gathered}$$

The two half-reactions are carried out in separate compartments connected with a solid mixture of CeO₂ and Gd₂O₃. Oxide ions can move through this solid at high temperatures (about 800°C). ∆Gfor the overall reaction at 800°C under certain concentration conditions is -380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Short answer

Cell potential is 0.984 V.

Step-by-step solution

Analyzing the given data

The overall reaction, $2\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Two half reactions;$$\begin{gathered} \mathrm{CO}+{\mathrm{O}}^{2-}\to {\mathrm{CO}}_2+2{\mathrm{e}}^{-} \\ {\mathrm{O}}_2+4{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-} \end{gathered}$$

From the above two equations, we get 4 mol of electrons.

Calculating the cell potential of the cell

We know that

$\begin{array}{rcl}\mathrm{\Delta G}& =& -{\mathrm{nFE}}_{\mathrm{cell}}\\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& -\frac{{\mathrm{\Delta G}}^{\mathrm{o}}}{\mathrm{nF}}\\ & =& \frac{-(-380 \mathrm{kJ})}{4(96485 \mathrm{C}/\mathrm{mol})}\left(\frac{{10}^3 \mathrm{J}}{1 \mathrm{kJ}}\right)\\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& 0.984 \mathrm{V}\end{array}$

Hence, cell potential is 0.984 V.