Verified Solution • 2025–26 Session moderate

QJEEAdv20Ch71b3: JEE Advanced 2020 Chemistry - Class 12 JEEAdv20Ch71b3 Solution: Consider a 70% efficient hydrogen-oxygen fuel cell working…

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

H2(g)+12O2(g)⟶H2O(l)

The work derived from the cell on the consumption of 1.0 ×\times 10 −- 3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?

The standard reduction potentials for the two half-cells are given below :

O2(g)+4H+(aq)+4e−⟶2H2O(l),

Eo=1.23V{E^o} = 1.23V

2H+(aq)+2e−⟶H2(g),

Eo=0.00 V{E^o} = 0.00\,V

Use, F=96500 C mol−1F = 96500\,C\,mo{l^{ - 1}} , R=8.314 J mol−1 K−1R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}} .

Step-by-Step Verified Solution

Vessel is insulated, thus q = 0

For the given reaction : H2(g)+1/2O2(g)⟶H2O(l);Eo=1.23−0.00=1.23V

Δ\Delta Go = −- nFEo

= −- 2 ×\times 96500 ×\times 1.23 J/mol

Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 ×\times 10 −- 3 mol of H2(g)

= 2 ×\times 96500 ×\times 1.23 ×\times 0.7 ×\times 1 ×\times 10 −- 3

= 166.17 J

This work done = change in internal energy (for monoatomic gas,

CV′m=3R/2{C_{V'm}} = 3R/2 ),

166.17=nCV,mΔT166.17 = n{C_V}{,_m}\Delta T

⇒ΔT=166.17×21×3×8.314\Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}

⇒13.32 K\Rightarrow 13.32\,K

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