Standard free energies of formation (I `kJ //`mol ) at `298 K` are ` -237 .2 , - 394 .4` and `- 8.2 ` for `H_2 O(1), CO_2 (g)` and pentange (g) , respectively . The value of `E_(cell)^@` for the pentane-oxygen fuel cell is .

Options

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

Answer

A

Explanation

`C_(5)H_(12)(g)+8O_(2)(g)rArr 5CO_(2)(g)+6H_(2)O(l)`
`DeltaG^(@)=[(-394.4xx5)+(-237.2xx6)]`
`-[(-8.2)+(8xx0)]`
`=-3387kJ`
The standard free energy change of elementary substances is taken as zero.
For the fual cell, the complete cell reaction is :
`C_(5)H_(12)(g)+8O_(2)(g)rarr 5CO_(2)(g)+6H_(2)O(l)`
This reaction is the combination of the following two half reactions. :
`C_(5)H_(12)(g)+10H_(2)O(l)rarr5CO_(2)(g)+32H^(+)+32e`
`8O_(2)(g)+32H^(+)+32erarr 16H_(2)O(l)`
As the number of electrons exchanged is 32 here. so n = 32
`DeltaG^(@)=-nFE^(@)`
`-3387.5xx10^(3)J=-32xx96500"J/vole"xxE^(@)`
On solving, we get
`E^(@)=1.09698V`