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​Q15. Calculate the emf of the following cell at 25° C.

         Ag (s) Ag+ (10–3 M) || Cu2+ (10–1M) | Cu (s) 

       [Given ,  E ° cell = + 0.46 V and log 10n
 

Dear Student,

At cathode : Cu2+ + 2e-  Cu(s)At anode    : 2Ag+ +2e- 2Ag(s)So, overall reaction is 2Ag(s) + Cu2+  2Ag+ +Cu(s)Ecell = E°cell -0.0591nlog [Ag+]2[Cu2+]         =0.46-0.05912log (10-3)2(10-1)          =0.46-0.02955log (10-5)          =0.46+0.14775          =0.60775 V  

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