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JEE Advanced Chemistry Electrochemistry 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Chemistry Question (2025) — Solution

Question

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is X F 10^3 volts, where F is the Faraday constant. The value of X is . Use : Standard Gibbs energies of formation at 298 K are : _f G_ CO _2 ^ =-394 ~kJ ~mol ^ -1 ; _f G_ water ^ =-237 ~kJ ~mol ^ -1 ; _f G_ butane ^ =-18 ~kJ ~mol ^ -1

Options

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

Answer

A. A

Step-by-step solution

aligned & C _4 H _ 10 ( ~g )+ 13 2 O _2( ~g ) 4 CO _2( ~g )+5 H _2 O (l) \\ & _ r G ^ =4 _ f G _ CO _2 ^ +5 _ f G _ H _2 O ^ - _ f G _ C _4 H _ 10 ^ \\ & =4 (-394)+5(-237)+18 \\ & =-2743 ~kJ / mol \\ & _ r G ^ =- nFE ^ \\ & -2743 1000=-26 FE ^ \\ & E ^ = 105.5 ~F 10^3=105.50 aligned

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Related: Chemistry — Electrochemistry · All PYQ Banks