An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO_(3) and th volume made to 100 ml. A silver electrode was dipped in the in the solution and the EMF of the cell set up Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq)|Ag(s) was 0.62 V. if E_(cell)^(@)=0.80V, what is the percentage of Ag in the alloy? [At 25^(@)C,2.303RT//F=0.6]

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 th volume made to 100 ml. A silver electrode was dipped in the in the solution and the EMF of the cell set up Pt(s),H_(2)(g)\|H^(+)(1M)\|\|Ag^(+)(aq)\|Ag(s) was 0.62 V. if E_(cell)^(@)=0.80V, what is the percentage of Ag in the alloy? [At 25^(@)C,2.303RT//F=0.6]
Answer» 25 2.5 10 1 SOLUTION :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%`