CUET · CHEMISTRY · PYQ PAPER 2025
Calculate the Gibbs free energy for the given reaction, with the standard electrode potential equal to 1.1 V
\(Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Zn ^{2+}( aq )+ Cu ( s )\)
- A \(-106.15 kJ / mol\)
- B \(212.27 kJ / mol\)
- C \(-212.27 kJ / mol\)
- D \(-312.4 kJ / mol\)
Answer & Solution
Correct Answer
(C) \(-212.27 kJ / mol\)
Step-by-step Solution
Detailed explanation
\(n = 2\) \(\Delta G = -nFE^\circ = -(2)(96485 \,C/mol)(1.1 \,V) = -212267 \,J/mol = -212.27 \,kJ/mol\)
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