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. `2.50`
C. 10
D. `1.0`

Answer

D

Explanation

Consider the concentration of `Ag^(+)` ions is X
Cell reaction is `(1)/(2) H_(2) (g) + Ag^(+)rightarrow H^(+) (aq)_ Ag(s)`
` E_(cell) = E_(cell)^(@)- (0.059)/(1) log [H^(+)]/[Ag^(+)]sqrtPH_(2)` or `0.62 = 0.8- (0.059)/(1)log (1)/[Ag^(+)]`
`(0.059)/(1) log ([Ag^(+) ] = -0.18Rightarrow log ([Ag^(+)] -3 Rightarrow~~ [Ag^(+)]) = 10^(-3)`
So mass of Ag in solution = `(100)/(1000) xx10^(-3)xx108 = 0.0108g`
Percentage of Ag in Alloy = `(0.010.8xx100)/(1.08) = 1%`