An alloy of Pb-Ag weighing `1.08g` was dissolved in dilute `HNO_(3)` and the volume made to 100 mL.A ? Silver electrode was dipped in the solution and the emf of the cell dipped in the solution and the emf of the cell set-up as `Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq.)|Ag(s)` was `0.62 V` . If `E_("cell")^(@)` is `0.80 V`, what is the percentage of Ag in the alloy ? (At `25^(@)C, RT//F=0.06`)

Options

A. 25
B. 2.5
C. 10
D. 1

Answer

D

Explanation

The cell reaction is
`H_(2)+2Ag^(+)to2H^(+)+2Ag`
`E_(cell)=E_(cell)^(@)-(2.303RT)/(2F)"log"(1)/([Ag^(+)]^(2))`
`0.62=0.0.80+0.06log[Ag^(+)]`
or `log[Ag^(+)]=(-0.18)/(0.06)=-3`
`therefore[Ag^(+)]="antilog"(-3)=1.0xx10^(-3)M`
`=(1.0xx10^(-3))xx108" g "L^(-1)=0.108gL^(-1)`.
`therefore` Amount of Ag present in 100 ml solution
`=0.0108g`
`therefore%` of `Ag=(0.0108)/(1.08)xx100=1%`