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JEE Advanced Chemistry Electrochemistry 2020 JEE Advanced 2020 (Paper 1)

JEE Advanced Chemistry Question (2020) — Solution

Question

Consider a 70 % efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298   K . Its cell reaction is H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O ( l ) The work derived from the cell on the consumption of 1 . 0 × 10 - 3   mol of H 2 ( 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 - → 2 H 2 O ( l ) ,    E 0 = 1 . 23   V , 2 H + ( aq ) + 2 e - → H 2 ( g ) ,    E 0 = 0 . 00   V Use F = 96500   C   mol - 1 , R = 8 . 314   J   mol - 1 K - 1 .  

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For given reaction  H 2 g + 1 2 O 2 g ⟶ 2 e - H 2 O l    E ° = 1 . 23   V ΔG ° = - nFE cell  ° = [ - 2 × 96500 × 1 . 23 ]   1 × 10 - 3 × 0 . 7 = - 166 . 173   J W = 166 . 173   J W = nRΔT γ - 1 166 . 173 = 1 × 8 . 314 × ΔT 5 3 - 1 = 8 . 314 × 3 2 ΔT ΔT = 166 . 173 × 2 8 . 314 × 3 = 13 . 32

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