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An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO3 and the volume made to 100 mL. A silver electrode was dipped in the solution and EMF of the cell set up was Pt(s),H2(g)H+(1M)∥Ag(aq.)+Ag(s)0.62 V. The percentage of Ag in the alloy is: Ecell ∘=0.80V,2.303RT/F=0.06 at 25∘C

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Detailed Solution

H2+2Ag+⟶2H++2AgEcell =Ecell ∘−2.303RTnFlog⁡1Ag+20.62=0.80+0.06log⁡Ag+or log⁡Ag+=−0.180.06=−3Ag+=1.0×10−3M=1.0×10−3×108=0.108gL−1∴ Amount of Ag in 100 mL solution = 0.0108 g∴ % Ag=0.01081.08×100=1

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