1.

Calculate the emf of the cell in which the following reaction takes place: Ni(s)+2Ag^(+)(0.002M)toNi^(2+)(0.160M)+2Ag(s) Given that E_(cell)^(@)=1.05V

Answer»

SOLUTION :Applying NERNST equation to the GIVEN cell REACTION,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Ag^(+)]^(2))=1.05V-(0.0591)/(2)"log"(0.160)/((0.002)^(2))=1.05-(0.0591)/(2)log(4XX10^(4))`
`=1.05-(0.0591)/(2)(4.6021)=1.05-0.14V=0.91V`


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