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 $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{CO}_{2}(\mathrm{~g})$ and pentane $(\mathrm{g})$ respectively. The value $\mathrm{E}_{\text {cell }}^{\circ}$ for the pentane-oxygen fuel cell is Answer upto 3 decimal places without rounding off.
  1. 1.906
  2. 1.960
  3. 1.069
  4. 1.096

Solution

Writing the equation for pentane-oxygen fuel cell at respective electrodes and overall reaction, we get
At Anode:
$\mathrm{C}_{5} \mathrm{H}_{12}+10 \mathrm{H}_{2} \mathrm{O} ightarrow 5 \mathrm{CO}_{2}+32 \mathrm{H}^{+}+32 \mathrm{e}^{-}$
(pentane)
At Cathode:
$\frac{8 \mathrm{O}_{2}+32 \mathrm{H}^{+}+32 \mathrm{e}^{-} ightarrow 16 \mathrm{H}_{2} \mathrm{O}}{\text { Overall }: \mathrm{C}_{5} \mathrm{H}_{12}+8 \mathrm{O}_{2} ightarrow 5 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O}}$
Calculation of $\Delta G^{\circ}$ for the above reaction $\Delta G^{\circ}=[5 \times(-394.4)+6 \times(-237.2)] -[-8.2]$
$=-1972.0-1423.2+8.2=-3387.0 \mathrm{~kJ}$
$=-3387000$ Joules. From the equation we find $\mathrm{n}=32$ Using the relation, $\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$ and substituting
various values, we get
$-3387000=-32 \times 96500 \times E_{\text {cell }}^{\circ}(\mathrm{F}=96500 \mathrm{C})$
or $E_{\text {cell }}^{\circ}=\frac{3387000}{32 \times 96500}$
$=\frac{3387000}{3088000}$ or $\frac{3387}{3088} \mathrm{~V}=1.0968 \mathrm{~V}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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