Electrochemistry Ques 57
57. Standard free energies of formation (in $kJ / mol$ ) at $298 K$ are $-237.2,-394.4$ and -8.2 for $H_2 O(l)$, $CO_2(g)$ and pentane $(g)$, respectively. The value of $E^{\circ}$ cell for the pentane-oxygen fuel cell is :
[2008]
(a) $1.968 V$
(b) $2.0968 V$
(c) $1.0968 V$
(d) $0.0968 V$
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Answer:
Correct Answer: 57.(c)
Solution:
(c) Writing the equation for pentane-oxygen fuel cell at respective electrodes and over all reaction, we get
At Anode:
$\underset{\text{(pentane)}}{C_5 H _{12}}+10 H_2 O \to 5 CO_2+32 H^{+}+32 e^{-}$
At Cathode:
$\frac{8 O_2+32 H^{+}+32 e^{-} \to 16 H_2 O}{\text{ Over all : } C_5 H _{12}+8 O_2 \to 5 CO_2+6 H_2 O}$
Calculation of $\Delta G^{\circ}$ for the above reaction $\Delta G^{\circ}=[5 \times(-394.4)+6 \times(-237.2)]-[-8.2]$ $=-1972.0-1423.2+8.2=-3387.0 kJ$ $=-3387000$ Joules.
From the overall equation we find $n=32$
Using the relation, $\Delta G^{\circ}=-n F E _{\text{cell }}^{0}$ and substituting various values, we get
$-3387000=-32 \times 96500 \times E _{\text{cell }}^{0}(F=96500 C)$
or $\quad E _{\text{cell }}^{0}=\frac{3387000}{32 \times 96500}=\frac{3387000}{3088000}$
or $\frac{3387}{3088} V=1.0968 V$
BETA
