The relationship between the 1/2 cell potential and concentration for dilute solutions is given by:
#E=E^(@)+(RT)/(nF)xx2.303log([["oxidised"]]/[["reduced"]])#
Where #n# is the number of moles of electrons transferred which, in this case = 1.
For the #"Ag"^+"/Ag"# 1/2 cell this becomes:
#E=E^@+(RT)/(F)xx2.303log[Ag^+]#
Putting in the values #rArr#
#0.62=(+0.8)+0.06log[Ag^+]#
#:.log[Ag^+]=-0.18/0.06=-3.0#
From which:
#[Ag^+]=0.001"mol/l"#
To find the number of moles #n# we know that:
#c=n/v#
#:.n=cxxv=0.001xx100/1000=0.0001#
To convert this to grams #m_(Ag)# we multiply by the mass of 1 mole of silver:
#A_r[Ag]=108#
#:. m_(Ag)=0.0001xx108=0.0108"g"#
So the percentage by mass of silver in the alloy is given by:
#%"Ag"=(0.0108)/(0.108)xx100=10#
So the answer is #(3)#.
