An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is volts, where is the Faraday constant. The value of is ____.
Use : Standard Gibbs energies of formation at 298 K are : ;
;
Use : Standard Gibbs energies of formation at 298 K are : ;
;
Answer: 105.5
Solution
Step 1: Write the combustion reaction:
Step 2: Calculate ΔᵣG°:
Step 3: Calculate number of electrons transferred:
Butane oxidation: C₄H₁₀ → 4CO₂ involves transfer of 26 electrons
(Each C goes from -2.5 to +4, total change = 26e⁻)
Step 4: Use the relationship ΔᵣG° = -nFE°:
Therefore, X = 105.50
Step 2: Calculate ΔᵣG°:
Step 3: Calculate number of electrons transferred:
Butane oxidation: C₄H₁₀ → 4CO₂ involves transfer of 26 electrons
(Each C goes from -2.5 to +4, total change = 26e⁻)
Step 4: Use the relationship ΔᵣG° = -nFE°:
Therefore, X = 105.50
JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer