Question

Standard free energies of formation (in J/mol) at 298 K are –x, –y and –z for H2O(), CO2(g) and CH4(g) respectively. The value of E°cell (in volts) for methane – oxygen fuel cell is

Answer 1

Answer:

Pentane −C5H12

C5H12+8O2→5CO2+6H2O

ΔG∘(298K)=5(−394.4)+6(−237.2)−(−8.2)

=−3387×103 J/mole

ΔG∘=nFEcell∘

n=5×4+6×2=32

ΔEcell∘=(32×96500)(3387×103)

=1.0968V

Answer 2

Answer:

The value of  $E^{o}_{cell}$  (in volts) is (-2x-y+z)1.29×10⁻⁶V for methane-oxygen fuel cell.

Explanation:

The combustion reaction for methane:

         $CH_{4}+2O_{2}$   →   $2H_{2}O+CO_{2}$

n is no. of electrons $=1*4+2*2=8$

The change in standard free energy:

ΔG° =  $G^{o}_{product}-G^{o}_{reactant}$

ΔG° $=[2(-x)+(-y)]-[-z]$

ΔG° $=(-2x-y+z )Jmol^{-1}$

We know that ΔG° $=-nFE^{o}_{cell}$

⇒  $E^{o}_{cell}=$ ΔG°/nF      where $F=96500Cmol^{-1}$

⇒  $E^{o}_{cell}= \frac{-2x-y+z}{(8)(96500)}$

⇒  $E^{o}_{cell}=(-2x-y+z)1.29*10^{-6}V$

Therefore $E^{o}_{cell}$ will be equal to (-2x-y+z)1.29×10⁻⁶V for methane-oxygen fuel cell.