Calculate the emf of the cell in which the following reaction takes place

Asked by Ujjwal Kumar | 22nd Sep, 2010, 05:28: PM

Expert Answer:

Please try to type the question  in right way and the equation should be like this:

Ni+2Ag+-------------> Ni2+ +2Ag

Oxidation is taking place at Ni electrode and reduction is at Ag.

Hence,

Ni|Ni2+||Ag| Ag

Ni + 2Ag+......>Ni2+ + 2Ag 

here, n=2

 and, Eo=1.05 V

and by using the formula:

E = Eo - (.0592/2) log10 ((Ni2+)/(Ag+)2 )

 

E=1.05-.0592/2 log(.160/(.002)2)

 

Solve for ' E'

 

 

Answered by  | 22nd Sep, 2010, 08:44: PM

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