JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The Gibbs energy change (in \(J\)) for the given reaction at \(\left[ Cu ^{2+}\right]=\left[ Sn^{2+}\right]=1\, M\) and \(298 \,K\) is: \(Cu ( s )+ Sn ^{2+}( aq ) \rightarrow Cu ^{2+}( aq )+ Sn ( s )\) \(\left( E _{ Sn ^{2+} \mid Sn }^{0}=-0.16\, V , E _{ Cu ^{2+} \mid Cu }^{0}=0.34\, V \right.\) Take \(F=96500\, C\, mol ^{-1}\) )
- A \(96500\)
- B \(96455\)
- C \(96530\)
- D \(96570\)
Answer & Solution
Correct Answer
(A) \(96500\)
Step-by-step Solution
Detailed explanation
\(\Delta G =\Delta G ^{\circ}+ RT \ln \left[\frac{ Sn ^{+2}}{ Cu ^{+2}}\right]\) \(=-2 \times 96500[(-0.16)-0.34]+ RT \ln \left(\frac{1}{1}\right)\) \(=96500 J\)
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