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`)

Answer

A

Explanation

`H_(2)+2Ag^(+)to2H^(+)+2Ag`
`E_(cell)=E_(cell)^(o)-(2.303RT)/(nF)"log"(1)/([Ag^(+)]^(2))`
`0.62=0.80+0.06log[Ag^(+)]`
or log`[Ag^(+)]=(-0.18)/(0.06)=-3`
or `[Ag^(+)]=1.0xx10^(-3)M`
`=1.0xx10^(-3)xx108=0.108g" "L^-1`
`therefore`Amount of Ag in 100 mL solution=0.0108g
`therefore%Ag=(0.0108)/(1.08)xx100=1`