An alloy of `Pb-Ag` weighing 54 mg was dissolved in desired amount of `HNO_(3)` & volume was made upto `500ml`. An Ag electrode was dipped in solution and then connected to standard hydrogen electrode anode. Then calculate `%` of Ag in alloy.
Given: `E_(cell) = 0.5V, E_(Ag^(+)//Ag)^(@) = 0.8V (2.303RT)/(F) = 0.06`

Answer

1

Explanation

`Ag^(+) +1e^(-) rarr Ag`
`E_(cell) = 0.5 = 0.8 +(0.06)/(1) log(Ag^(+))`
`log [Ag^(+)] = (-0.30)/(0.006) =- 5`
`[Ag^(+)] = 10^(-5) mol//L`
moles of `Ag^(+)` in `500 mL = (10^(-5))/(2)`
Mass of `Ag = (10^(-5))/(2) xx 108`
`% Ag = ((10^(-5))/(2)xx108)/(54 xx 10^(-3)) xx 100 = 1`