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Standard free energies of formation (in kJ/mol) at 298 K are $-237.2$, $-394.4$ and $-8.2$ for $H_2O(l)$, $CO_2(g)$ and pentane$(g)$ respectively. The value of $E^\circ_{cell}$ for the pentane-oxygen fuel cell is:
A
1.0968 V
B
0.0968 V
C
1.968 V
D
2.0968 V
Detailed Solution
Cell reaction: $C_5H_{12}(g) + 8O_2(g) \rightarrow 5CO_2(g) + 6H_2O(l)$
$\Delta G^\circ = [(-394.4 \times 5) + (-237.2 \times 6)] - [(-8.2) + (8 \times 0)]$
$= [-1972.0 - 1423.2] + 8.2$
$= -3387$ kJ (taken as $-3387.5$ kJ in the source)
The standard free energy of formation of an elementary substance ($O_2$) is taken as zero.
For the fuel cell, the complete cell reaction is the combination of the following two half reactions:
Anode: $C_5H_{12}(g) + 10H_2O(l) \rightarrow 5CO_2(g) + 32H^+ + 32e^-$
Cathode: $8O_2(g) + 32H^+ + 32e^- \rightarrow 16H_2O(l)$
As the number of electrons exchanged is 32 here, $n = 32$.
$\Delta G^\circ = -nFE^\circ$
$-3387.5 \times 10^{3}$ J $= -32 \times 96500 \times E^\circ$
On solving, we get $E^\circ = \dfrac{3387.5 \times 10^{3}}{32 \times 96500} = 1.0968$ V
$\Delta G^\circ = [(-394.4 \times 5) + (-237.2 \times 6)] - [(-8.2) + (8 \times 0)]$
$= [-1972.0 - 1423.2] + 8.2$
$= -3387$ kJ (taken as $-3387.5$ kJ in the source)
The standard free energy of formation of an elementary substance ($O_2$) is taken as zero.
For the fuel cell, the complete cell reaction is the combination of the following two half reactions:
Anode: $C_5H_{12}(g) + 10H_2O(l) \rightarrow 5CO_2(g) + 32H^+ + 32e^-$
Cathode: $8O_2(g) + 32H^+ + 32e^- \rightarrow 16H_2O(l)$
As the number of electrons exchanged is 32 here, $n = 32$.
$\Delta G^\circ = -nFE^\circ$
$-3387.5 \times 10^{3}$ J $= -32 \times 96500 \times E^\circ$
On solving, we get $E^\circ = \dfrac{3387.5 \times 10^{3}}{32 \times 96500} = 1.0968$ V