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. 50

Answer

C

Explanation

`0.62=0.8-(0.06)/(1)"log"([H^(+)])/([Ag^(+)])`
`-0.18=-0.06"log" (1)/([Ag^(+)])rArr[Ag^(+)]=10^(-3)M`
w.t of `Ag=10^(-3)xx108=0.108g`
w.t percent of `Ag=(0.108)/(1.08)xx100=10`