An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO_(3) and the value made to 100 mL. A silver electrode was dipped in the solution and EMF of the cell set up Pt_((s)),H_(2(g))|H^(+)(1M)||Ag_((aq))^(+)|Ag_((s)). was 0.62V. The percentage of Ag in the alloy is [E_(cell)^(@)=0.80V,2.303RT//F=0.06" at "25^(@)C]

An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO_(3) and

1.	An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO_(3) and the value made to 100 mL. A silver electrode was dipped in the solution and EMF of the cell set up Pt_((s)),H_(2(g))\|H^(+)(1M)\|\|Ag_((aq))^(+)\|Ag_((s)). was 0.62V. The percentage of Ag in the alloy is [E_(cell)^(@)=0.80V,2.303RT//F=0.06" at "25^(@)C]
Answer» SOLUTION :`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`