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Q.

Consider the following cell reaction :
Cd(s)+Hg 2SO4(s)+95H 2O(l) CdSO4.95H 2O (s)+2Hg(l)
The value of  Ecell0 is 4.315V at 250C  .If ΔH0=825.2KJ  mol1,  the standard entropy change ΔS0  in  JK1 is ……………..( Nearest integer) [Given: Faraday constant =96487  C  mol1 ]
 

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answer is 25.

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Detailed Solution

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ΔG0=nFE0=ΔH0TΔS0        =ΔH0+nFE0T    =(825.2  X  103)+(2  X96487X4.315)298=825.2  X  103+832.682  X103298       =7.483  X  103298=25.1JK1mol1.

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