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 .

Explanation

Step I : Calculate of `DeltaG^(@)` for the fuel cell
The combustion reaction taking place in the fuel cell is :
`C_(5)H_(12)(g)+8O_(2)(g) to 5CO_(2)(g)+6H_(2)O(l)`
`DeltaG^(@)=[5xxDelta_(f)G^(@)CO_(2)(g)+6Delta_(f)G^(@)H_(2)O(l)]-[Delta_(f)G^(@)C_(5)H_(12)(g)+8Delta_(f)G^(@)O_(2)(g)]`
`=[5xx(-394.4)+6xx(-237.2)]-[-8.2+"zero"]`
`=[-1972-1423.2]+8.2=-3387kJ`
Step II : Calculate of `E_(cell)^(@)`
The combustion reaction taking place in the fuel cell is a redox reaction. The oxidation and reduction half reactions may be represented as :
`C_(5)H_(12)(g)+10H_(2)O(l)to5CO_(2)(g)+32H^(+)(aq)+32e^(-)`
`8O_(2)(g)+32H^(+)(aq)+32e^(-) to 16H_(2)O(l)`.
No.of electrons (n) exchanged in the reaction =32
`DeltaG^(@)=-"nF "E_(cell)^(@)" or "E
_(cell)^(@)=-(DeltaG^(@))/(nF)`
`E_(cell)^(@)=-((-3387xx10^(3)J))/(32xx36500 "Jvolt"^(-1))=1.0968 V`