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Q.

The Gibb’s energy change ( in J ) for the given reaction, at  [Cu2+]=[Sn2+]=1M   and  298 K is  Cu(s)+Sn(aq)2+Cu(aq)2++Sn(s)ESn2+/Sn0=0.16v,ECu2+/Cu0=0.34 V,F=96500Cmol1

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a

96500 J

b

48,250 J

c

193000 J

d

24,125 J

answer is A.

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Detailed Solution

E0cell=Ecathode0EAnode0

              =0.160.34=0.50vΔG=nFEcell0        =2×96500×0.50=96500J

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