Standard free energies of formation of (in `KJ mol^(-1)`) at 298 K are `-237.3, -394.2" and " -8.2 " for " H_(2)O(l) , CO_(2)(g) ` and pentane (g) respectively . The value of `E_("cell")^(0) ` for the pentane - oxygen fuel cell is :

Options

A. `1.968 V`
B. `2.0968 V`
C. `1.0968 V`
D. `0.968 V`

Answer

c

Explanation

The balanced equation for pentane oxygen cell reaction will be :
`C_(5)H_(12) + 8O_(2) to 5CO_(2) + 6H_(2)O, n = 32 ` `(. :. 16(O) + 32e^(-) to 16O^(-2))`
`DeltaG_(f)^(0) = [5 xx DeltaG_(f)^(0)(CO_(2))] + [6 xx DeltaG_(f)^(0)(H_(2)O)`
` - [DeltaG_(f)^(0)(C_(5)H_(12)) + 8 DeltaG_(f)^(0) (O_(2))]`
` = [5 xx (-394.5) + 6xx (-237.2)] - [(-8.2) + 0]`
` = - 1972 - 1423.2 + 8.2`
` = - 3387 KJ mol^(-1)`
`DeltaG^(0) = - nF E_("cell")^(0) `
` -3387000 = - 32 xx 96500 xx E_("cell")^(0) `
`E_("cell")^(0) = 1.0968 V`