CBSE Class 12-science Answered
Calculate the emf of the cell in which the following reaction takes place
Asked by Ujjwal Kumar | 22 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 | 22 Sep, 2010, 08:44: PM
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