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 emf of the cell set up pt (s) `H_(2)(g)|H^(+)(1M)||Ag^(+)(aq)|Ag(s)` was 0.62 V if `E_(cell)^(@)=0.80 V` what is the percentage of Ag in the alloy ? At `25^(@)C RT/F=0.6`

Options

A. 25
B. 2.5
C. 10
D. 50

Answer

B

Explanation

`Pt(s) H_(2)(g) |H^(+) (1 M) || Ag^(+) (aq)|Ag(s)`
EMF of cell `=0.62 V E_(cell)^(@)=0.80 V`
`H_(2) rarr 2H^(+) +2e^(-)`
`2Ag^(+) +2e^(-) rarr 2AG`
`H_(2) +2Ag^(+) rarr 2Ag+ 2H^(+)`
`E_(cell) =E^(@) -(2.033 RT)/(2F)log [(H^(+))]/(Ag^(+))^(2)`
`E_(cell) =E^(@) -(2.303 RT)/(2F) log (1)/(Ag^(+))^(2)`
`0.62 =0.80 +(2xx2.303xx0.06)/(2)log Ag^(+)`
`therefore` mole of `Ag^(+)` in 100 mL `=0.05 xx100/1000 xx108`
% of Ag in 1.08 g alloy `=(0.05 xx100xx108 )/(1000xx1.08)xx100=50%`