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

`0.0339 %`

Explanation

`{:("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 %`