1.

Calculate the emf of the cell in which the following reaction takes place: Ni_((S))+2Ag^(+)(0.002M) to Ni^(2+)(0.160M)+2Ag_((S))""[E_(cell)^(Theta)=1.05V]

Answer»

Solution :* In given CELL FOLLOWING reaction is possible
`Ni_((S))+2Ag^(+)(0.002M)toNi^(2+)(0.160M)+2Ag_((S))`
Where, n=2, `[Ni^(2+)]=0.160M,[Ag^(+)]=0.002M,E_(cell)^(Theta)=1.05V`.
* Note: Here, solid Ni and Ag is neglected
`THEREFORE E_(cell)=E_(cell)^(Theta)-(0.059)/(n)"log"([Ni^(2+)])/([Ag^(+)]^(2))`
`=1.05-(0.059)/(2)"log"((0.160)/((0.002)^(2)))`
`=1.05-0.0295log(40000)`
`=1.05-0.0295xx4.6021`
`=1.05-0.1358`
=0.9142
`~~0.91V`


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