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

(a) `E_("cell") = E_("cell")^(0) - (2.303 RT)/(nF)log_(10).([Ni^(2+)])/([Ag^(+)]^(2))`
`E_("cell") = 1.05 - (2.303 xx 0.314 xx 298)/(2 xx 96500)log.(0.16)/((0.002)^(2))`
`E_("cell") = 0.91 V`.
(b) (i) The law states that limiting molar conductivity of a electrolyte can be represented as the sum of the individual contribution of anion and cation of the electrolyte.
(ii) Limiting molar conductivity is molar conductivity at infinite dilution.
Or
When the concentration approaches Zero.