Question

Direct methanol fuel cells (DMFCs) have shown somepromise as a viable option for providing “green” energyto small electrical devices. Calculate E° for the reactionthat takes place in DMFCS:
$C{H}_{3}OH\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
Use values of$\Delta {G^o}_f$ from Appendix 4.

Short answer

The standard potential of the reaction is 1.21 V.

Step-by-step solution

Formula for the calculation of Eo

The formula to calculate,

$\frac{\Delta G^o}{-nF}=E^o$

Calculation of standard Gibbs free energy for the reaction

The standard Gibbs free energy for the reaction can be calculated as:

$$\begin{gathered} \Delta {G^0}_r=\Delta {G^0}_f\left[C{O}_2\right(g\left)\right]+2\Delta {G^0}_f\left[H_{2}O\right(l\left)\right]-\Delta {G^0}_f\left[C{H}_{3}OH\right(l)-\frac{3}{2}\Delta {G^0}_f[O_2\left(g\right)] \\ =1\times (-394 kJ mo{l}^{-1})+2\times (-237 kJ mo{l}^{-1})-1\times (-166 kJ mo{l}^{-1})-(\frac{3}{2}\times 0 kJ mo{l}^{-1}\left)\right] \\ =-702 kJ mo{l}^{-1} \end{gathered}$$

Calculation to find the standard potential of the reaction

The balanced half-cell reaction can be denoted as:

$$\begin{gathered} C{H}_{3}OH\left(l\right)+H_{2}O\left(l\right)\to C{O}_2\left(g\right)+6H^{+}+6e^{-1} \\ \frac{3}{2}{O}_2\left(g\right)+6e^{-}+6H^{+}\to 3H_{2}O\left(l\right) \end{gathered}$$

To calculate the electrode potential,

$$\begin{gathered} \Delta G^0=-nF{E^o}_{cell} \\ \Delta G^0=-702 kJ \\ =-702000 J \\ -702000 J=-6\times 9.6485\times {10}^4\times {E^0}_{cell} \\ {E^0}_{cell}=\frac{-702000 J}{-6\times 9.6485\times {10}^4} \\ =1.21 J{C}^{-1} \\ {E^0}_{cell}=1.21 V \end{gathered}$$