An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute `HNO_(3)` and the volume made to 100 mL. A silver electrode was depped in the solution and the emf of the cell set-up
`Pt(s), H_(2) (g) | H^(+) (1 M)|| Ag^(+) (aq.) | Ag(s)` was 0.62 V. If `E_("cell")^(@)` is 0.80V, what is teh precentage of Ag in the alloy ?
(At `25^(@)C, RT//F = 0.60`)

Options

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

Answer

D

Explanation

The cell reaction is
`H_(2)+2Ag^(+)rarr2H^(+)+2Ag`
`E_("cell")=E_("cell")^(@)-(2.303RT)/(2F)"log"(1)/([Ag^(+)]^(2))`
` 0.62=0.80+0.06"log"[Ag^(+)]`
` "log"[Ag^(+)]=(-0.18)/(0.06)=-3`
`[Ag^(+)]= "antilog"(-3) =1.0xx10^(-3)M`
`=1.0xx10^(-3)xx1 08L^(-1)`
`=0.108g L^(-1)`
Amount of Ag present in 100mL solution
=0.0108g
`therefore %` of Ag in the alloy `=(0.0108)/(1.08)xx100=1%`