An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute `HNO_(3)` and the volume made to 100 mL. A silver electrode was dipped in the solution and the e.m.f. 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=0.06`)

Options

A. 25
B. `2.50`
C. 10
D. 1

Answer

D

Explanation

Cell reaction is
`H_(2)+2Ag^(+) rarr 2Ag + 2H^(+)`
`E= E^(Ɵ) -(2.303RT)/(2F) log. ([H^(+)]^(2))/([Ag^(+)]^(2)P_(H_(2)))`
`0.62=0.80 - (0.06)/(2) log . (1)/([Ag^(+)]^(2))`
`0.62= 0.80 +(2xx0.06)/(2) log [Ag^(+)]`
`0.06 log [Ag^(+)]= 0.62 - 0.80= -0.18`
`log[Ag^(+)]= (-0.18)/(0.06)= -3`
`[Ag^(+)]=10^(-3)M`
`:.` Weight of Ag in 1000 mL solution
`=10^(-3)xx108=0.108`g
Weight of Ag in 100 mL solution
` = (0.108)/(1000)xx100=0.0108g`
`:.` % of Ag`=(0.108)/(1.08)xx100=1%`