1.

An alloy of Pb Ag weighing 1.08 g was dissolved in dilute HNO_(3)and the volume made to 100 mLA 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)^(@)=0.80V what is the percentage of Ag in thealloy ? At 25^(@)C RT/F=0.6

Answer»

25
2.5
10
50

Solution :`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`moleof `Ag^(+)`in 100 mL `=0.05 xx100/1000 xx108`
%of Agin 1.08 g alloy`=(0.05 xx100xx108 )/(1000xx1.08)xx100=50%`


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