The emf of the cell reaction, Zn(s)+Cu2+(aq.)→Zn2+(aq.)+Cu(s) is 1.1V. If enthalpy of the reaction is −216.7kJmol−1, calculate the entropy change for the reaction(in J/Kmole)
A
−14.76
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B
−29.5
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C
14.76
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D
29.5
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Solution
The correct option is A−14.76 −ΔGo=n×F×Eo=2×96500×1.1=212.3kJ ΔGo=−212.3kJmol−1 ΔGo=ΔHo−TΔSo ΔSo=ΔHo−ΔGoT =−216.7−(−212.3)298 =−0.01476kJK−1mol−1 =−1476JK−1mol−1