Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394…

Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394.4$ and -8.2 for $\mathrm{H}_2 \mathrm{O}(l) \mathrm{CO}_2(g)$ and pentane $(g)$, respectively. The value of $E_{\text {cell }}^0$ for the pentane-oxygen fuel cell is
  1. $1.968 \mathrm{~V}$
  2. $2.0968 \mathrm{~V}$
  3. $1.0968 \mathrm{~V}$
  4. $0.0968 \mathrm{~V}$

Solution

\(\begin{aligned}
& \mathrm{C}_5 \mathrm{H}_{12(g)}+8 \mathrm{O}_{2(9)} \longrightarrow 5 \mathrm{CO}_{2(g)}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{e})} \\
& \therefore \Delta \mathrm{G}_r=5\left(\Delta \mathrm{G}_r\right)_{\mathrm{CO}_2}+6\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\mathrm{H}_2 \mathrm{O}}-\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\text {Pentane }} \\
& =5(-5394.4)+6(-237.2)-(-8.2) \\
& =-1972-1423.2+8.2 \\
& \Delta \mathrm{G}_{\mathrm{R}}=-3387 \mathrm{~kJ} / \mathrm{mole} \\
& \mathrm{C}_5^{\frac{-12}{5}} \mathrm{H}_{12} \longrightarrow 5 \stackrel{4}{\mathrm{C}_2}
\end{aligned}\)
For 5 carbon atoms, there is a change of electrons from - 12 to 20
\(\begin{aligned}
& \therefore \text { No. of electrons involved }=32 \mathrm{e}^{-} \\
& \Delta G_r=-3387 \times 10^3=-32 \times E^{\circ} \times 96500 \\
& E^{\circ}=1.097 V
\end{aligned}\)

Asked in: NEET 2008 (Screening)

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