1.05 g of a lead ore containing impurity of Ag was dissolved in quantity of HNO_(3) and the volume was made 350 mL . A Ag electrode was dipped in the solution and E_(cell) of Pt(H_(2))|H^(+) (1 M)||Ag^(+)|Ag was 0.503 V at 298 K. calculate % of lead in the ore. E_(Ag^(+)//Ag)^(@)=0.80 V

1.05 g of a lead ore containing impurity of Ag was dissolved in quanti

1.	1.05 g of a lead ore containing impurity of Ag was dissolved in quantity of HNO_(3) and the volume was made 350 mL . A Ag electrode was dipped in the solution and E_(cell) of Pt(H_(2))\|H^(+) (1 M)\|\|Ag^(+)\|Ag was 0.503 V at 298 K. calculate % of lead in the ore. E_(Ag^(+)//Ag)^(@)=0.80 V
Answer» Solution :`{:("Anode "1/2H_(2)(G) rarr H^(+) (1 M)+e^(-)),("CATHODE "Ag^(+)(x)+e^(-) rarrAg(s)),(ulbar(1/2H_(2)(g)+Ag^(+) (x) hArr Ag(s)+H^(+) (1 M))""(n=1)):}` `E^(@)=0.80-0=0.80` volt `Q=([H^(+)])/([Ag^(+)])=1/x` `E=E^(@)-0.0591/n log_(10) Q` `0.503=0.80-0.0591/1 log_(10) (1/x)` `x=9.43xx10^(-6) M` Number of moles of `Ag^(+)` in 350 mL `=(MV)/1000=(9.43xx10^(-6)xx350)/(1000)` `=3.3xx10^(-6)` Mass of `Ag=3.3xx10^(-6)xx108=3.56xx10^(-4)g` % Ag in the ore`=(3.56xx10^(-4))/1.05xx100=0.0339 %`