JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\(Cu ( s )+ Sn ^{2+}(0.001 M ) \rightarrow Cu ^{2+}(0.01 M )+ Sn ( s )\) The Gibbs free energy change for the above reaction at \(298\, K\) is \(x \times 10^{-1} \,k\,J\, mol ^{-1}\); The value of \(x\) is ..... [nearest integer]\(\left [\text { Given : } E _{ Cu ^{2} / / Cu }=0.34\, V ; E _{ Sn ^{2} / Sn }^{\ominus}=-0.14 \,V ; F=96500\, C\, mol ^{-1}\right]\)
- A \(123\)
- B \(983\)
- C \(552\)
- D \(631\)
Answer & Solution
Correct Answer
(C) \(552\)
Step-by-step Solution
Detailed explanation
\(Cu _{( s )} + Sn ^{2+}(0.001\, M ) \rightarrow Cu ^{2+}(0.01\, M )+ Sn _{( s )}\) \(E _{\text {cell }} = E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}\) \(=-0.14-(0.34)\) \(=-0.48 \,V\)…
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