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

Options

A. 25
B. 2.5
C. 10
D. 1

Answer

D

Explanation

Overall cell reaction is
`H_(2)(g) + 2Ag^(+) hArr 2 Ag(s) + 2H^(+)(aq.)`
`E = E^(@) = (0.06 xx 2.303)/(2) "log " ([H^(+)]^(2))/([Ag^(+)]^(2) pH_(2))`
`0.62 = 0.80 + (2 xx 0.06 xx 2.303)/(2) " log" [Ag^(+)]`
`[Ag^(+)] = 0.05 M`
Number of moles of `Ag^(+)` in 100 mL
`= (MV)/(1000) = (0.50 xx 100)/(1000) = 0.005`
Mass of silver `= 0.05 xx 108 g`
Percentage of A in 1.80 g of alloy `= (0.005 xx 108 xx 100)/(1.80) = 50%`