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JEE Advanced · Chemistry · 17. Electrochemistry

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 1.0×10-3 mol of H2(g) is used to compress 1.00 mol 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. \(O _2(g)+4 H ^{+}( aq )+4 e ^{-} \rightarrow 2 H _2 O ( l ),\) \(E ^0=1.23 V\) 2H+(aq)+2e-H2(g),  E0=0.00 V Use F=96500 C mol-1,R=8.314 J mol-1K-1.

  1. A 13.32
  2. B 12.54
  3. C 10.88
  4. D 15.36
Verified Solution

Answer & Solution

Correct Answer

(A) 13.32

Step-by-step Solution

Detailed explanation

For given reaction H2g+12O2g2e-H2Ol E°=1.23 V \(\Delta G ^{\circ}=- nFE _{\text {cell }}^{\circ}=[-2 \times 96500 \times 1.23]\) \(1 \times 10^{-3} \times 0.7=-166.173 J\)
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