Question

Standard free energies of formation (in kJ/mol) at 298 K are
– 237.2, – 394.4 and – 8.2 for H₂O(l), CO₂(g) and pentane (g),
respectively. The value E°cell for the pentane-oxygen fuel
cell is :
(a) 1.968 V (b) 2.0968 V
(c) 1.0968 V (d) 0.0968 V

Answer 1

answer : option (c) 1.0968V

given, (1/2)H2 + (1/2)O2 ⇒H2O ; ∆H1 = -237.2 kJ/mol

C + (1/2)O2 ⇒CO2 ; ∆H2 = -394.4 kJ/mol

5C + 6H2 ⇒C5H12 ; ∆H3 = -8.2 kJ/mol

now combustion of pentane is ...

C5H12 + 16O2 ⇒5CO2 + 6H2O + ∆E

here ∆E = 5∆H2 + 6∆H1 - ∆H3

= 5 × -394.4 + 6 × -237.2 - (-8.2)

= -3387 kJ/mol

number of electrons involved in pentane - oxygen fuel = 5 × 4 + 12 × 1 = 32

applying formula, ∆E = nFE°_(cell)

⇒-3387 × 10³ = 32 × 96500 C × E°_(cell)

⇒E°_(cell) = -3387 × 10³/(32 × 96500)

= -1.0968 V

so, magnitude of ∆E°_(cell) is 1.0968 V

hence option (c) is correct choice.