Question

JEE ADVANCED CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

consider a 70% efficient hydrogen - oxygen fuel cell working under standard conditions at 1 bar and 298K. its cell reaction is ...

H₂ (g) + 1/2 O₂ (g) ⇔H₂O (l)

The work derived from the cell on the consumption of 1.0 × 10¯³ mol of H₂ is used to compress 1 mol of monoatomic ideal gas in a thermally insulted container.

we have to find change in the temperature of the ideal gas..

solution : The standard reduction potential for the two half cells are given below.

O₂ + 4H⁺ + 4e¯⇒2H₂O, E⁰ = 1.23 V

2H⁺ + 2e¯ ⇒2H₂, E⁰ = 0.00V

here cell potential, ∆E⁰ = 1.23 - 0.00 = 1.23 V

now using formula, ∆G⁰ = -nF∆E⁰

= - (4 - 2) × 96500 × 1.23 J

= -2 × 96500 × 1.23

so work derived from this fuel = 70 % × -∆G × 10¯³ J

again, it has given that vessel is thermally insulted so, q = 0.

so, w = ∆U

for monoatomic,

∆U = nCv∆T = 1 × (3R/2) × ∆T

now work derived from the cell = w = ∆U

⇒70/100 × 2 × 96500 × 1.23 × 10¯³ = 1 × 3 × 8.314/2 × ∆T

⇒∆T = 13.32

Therefore the change in the temperature of the ideal gas is 13.32