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 set up `Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq)|Ag(s)` was 0.62V. If `E_("cell")^(@)=0.80V` what is the percentage of Ag in the alloy? [At `25^(@)C`, RT/F = 0.06]

Options

A. 25
B. `2.50`
C. 1
D. 50

Answer

D

Explanation

`Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq)|Ag(s)`
EMF of cell = `0.62V,E_("cell")^(@)=0.80V`
`H_(2)rarr2H^(+)+2e^(-)` (at anode)
`(2Ag^(+)+2e^(-)2Ag" "" "("at cathode"))/(H_(2)+2Ag^(+)rarr2Ag+2H^(+)"(overall reaction)")`
`E_("cell")=E^(@)-(2.303RT)/(2F)log""([H^(+)]^(2))/([Ag^(+)]^(2)[H_(2)])`
`E_("cell")=E^(@)-(2.303RT)/(2F)log""(1)/([Ag^(+)]^(2))`
`0.62=0.80+(2.303xx0.06)/(2)log""(1)/([Ag^(+)]^(2))`
`0.62=0.80+(2xx2.303xx0.06)/(2)log[Ag^(+)]`
`-0.80=0.1382log[Ag^(+)]`
`[Ag^(+)]=0.05m`
`therefore` Mole of `Ag^(+)` in `100mL=0.05xx(100)/(1000)`
Wt. of `Ag^(+)` in `100mL=0.05xx(100)/(1000)xx108`
% of Ag in 1.08 g alloy
`=(0.05xx100xx108)/(1000xx1.08)xx100=50%`