Question

Calculate E° for the reaction

$C{H}_{3}OH\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
using values of $\Delta {G_i}^o$in Appendix 4. Will E° increase ordecrease with an increase in temperature? (See Exercise 45 for the dependence of E° on temperature.)

Short answer

Thevalue of $E^o=1.21 V$. It increases with temperature when the electrode cell potential decreases.

Step-by-step solution

Calculating the standard free energy

The standard free energy can be calculated as:

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

The free energy is -702000 J.

Calculating the electrode cell potential

The free energy and negative cell potential are proportional to each other.

$$\begin{gathered} \Delta {G^0}_f=-nF{E}^o \\ {E}^o=\frac{\Delta {G^0}_f}{-nF} \\ {E}^o=\frac{702000 J}{-(6 mol)(96485 C mo{l}^{-1})} \\ {E}^o=1.21 V \end{gathered}$$

When the temperature increases, there is a decrease in electrode cell potential.