Calculate the e.m.f. of the cell in which the following reaction takes place.
`Ni_((s)) + 2Ag_((0.002M))^(+) to Ni_((0.160M))^(2+) + 2Ag_((s)) , "Given " E_("cell")^@ = 1.05 V `

Explanation

Data: `[Ni^(2+)]=0.100M[Ag^(2+)]=0.002M`
`Ni(s)+2Ag^(+) to Ni^(2+)+2Ag(s)`
n=2
`E_("cell")^(0)=1.05V`
Formula `E_("cell")=E_("cell")^(0)-(0.0591)/(2)log""([Ni^(2+)])/([Ag^(+)]^(2))`
`E_("cell")=1.05-(0.0591)/(2)log""([0.100])/([0.002]^(2))`
`E_("cell")=1.05-0.1299=0.92V`