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 pentance (g) respectively. The value `E_(cell)^(@)` for the pentance-oxygen fuel cell is:

Options

A. `1.968V`
B. `2.0968V`
C. `1.0968V`
D. `0.0968V`

Answer

C

Explanation

Writing the equation for pentane-oxygen fuel cell at respective electrodes and overall reaction, we get At Anode:
`underset(("pentace"))(CH_(5)H_(12))+10H_(2)O to 5CO_(2)+32H^(+)+32e ^(-)`
Al Cathode:
`(8O_(2)+ 32H^(+) +32e^(-) to 16H_(2)O)/("Overall:" C_(5) H_(12) +8O _(2) to 5CO_(2) +6H _(2) O)`
Calculation of `DeltaG ^(@)` for the above reaction
`DeltaG^(@)=[5xx(-394.4) +6xx(-237.2)] -[-8.2]`
`=-1972.0 -1423.2+8.2 =-3387.0kJ`
`=-3387000` Joules.
From the equation we find `n=32`
Using the relation, `Delta G^(@)=-nFE_(cell)^(@)` and substituting various values, we get
`-3387000=-32xx96500xxE_(cell)^(@) (F=96500C)`
or `E_(cell) ^(@) =(3387000)/(32xx96500)`
`=(3387000)/(3088000)or (3387)/(3088)V-1.0968V`
Thus option (c) is correct answer.