Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2for H_(2)O(l), CO_(2)(g) and pentance (g) respectively. The value E_(cell)^(@) for the pentance-oxygen fuel cell is:

Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -

1.	Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2for H_(2)O(l), CO_(2)(g) and pentance (g) respectively. The value E_(cell)^(@) for the pentance-oxygen fuel cell is:
Answer» `1.968V` `2.0968V` `1.0968V` `0.0968V` Solution :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.