Q.4 For the electrochemical reaction
Cu2+(aq) + Zn(s) ⇌ Cu(s) + Zn2+(aq), the equilibrium constant at 25 °C is 1.7 × 1037.
The change in standard Gibbs free energy (ΔG°) for this reaction at that temperature will be ______ kJ mol−1
(up to one decimal place). (Given: R = 8.314 J K−1 mol−1)
Step-by-Step Calculation
The standard Gibbs free energy change relates to the equilibrium constant by ΔG° = -RT ln K. At 25°C (T = 298 K) and R = 8.314 J K−1 mol−1, ln(1.7 × 1037) ≈ 85.73. Thus, ΔG° = -(8.314)(298)(85.73) ≈ -212393 J mol−1 or -212.4 kJ mol−1 (to one decimal place).
- Identify n=2 electrons transferred, but use direct K formula since given.
- Compute ln K precisely via code for accuracy.
- Convert J to kJ and round as specified.
Why This Value?
A large K (1.7 × 1037) indicates strong favoritism toward products, yielding large negative ΔG°. This matches E°cell ≈ 1.10 V for Zn-Cu cell, where ΔG° = -nFE°cell ≈ -212 kJ mol−1. Slight variation from exact 1037 due to 1.7 factor.
Common Mistakes Explained
Users often confuse with E° method or wrong T/R units. No options provided, but typical distractors include:
- Using log10: -2.303RT log K ≈ -212.4 kJ mol−1 (same result).
- Forgetting negative sign: +212.4 kJ mol−1 (non-spontaneous error).
- T=25 K instead of 298 K: massively wrong magnitude.
Correct approach always starts with given K and ΔG° = -RT ln K.
