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Calculate the e.m.f. of the cell in whic...

Calculate the e.m.f. of the cell in which the following reaction takes place :
`Ni(s) +2Ag^(+)(0.002 M)to Ni^(2+)(0.160 M)+2Ag(s)`
Given `E_(cell)^(@)`=1.05 v

Text Solution

Verified by Experts

`E_(cell)=E_(cell)^(@)-(0.0591)/(n)log[(Ni^(+2))]/([Ag^(+)]^(2)`
`=1.05 -(0.0561)/(2)log(0.160)/(0.002^(2)`
`=1.05-(0.0591)/(2)xx4.6`
=1.05-0.14=0.91 V
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